Forcing bash to expand variables in a string loaded from a file - linux

I am trying to work out how to make bash (force?) expand variables in a string (which was loaded from a file).
I have a file called "something.txt" with the contents:
hello $FOO world
I then run
export FOO=42
echo $(cat something.txt)
this returns:
hello $FOO world
It didn't expand $FOO even though the variable was set. I can't eval or source the file - as it will try and execute it (it isn't executable as it is - I just want the string with the variables interpolated).
Any ideas?

I stumbled on what I think is THE answer to this question: the envsubst command:
echo "hello \$FOO world" > source.txt
export FOO=42
envsubst < source.txt
This outputs: hello 42 world
If you would like to continue work on the data in a file destination.txt, push this back to a file like this:
envsubst < source.txt > destination.txt
In case it's not already available in your distro, it's in the
GNU package gettext.
#Rockallite
I wrote a little wrapper script to take care of the '$' problem.
(BTW, there is a "feature" of envsubst, explained at
https://unix.stackexchange.com/a/294400/7088
for expanding only some of the variables in the input, but I
agree that escaping the exceptions is much more convenient.)
Here's my script:
#! /bin/bash
## -*-Shell-Script-*-
CmdName=${0##*/}
Usage="usage: $CmdName runs envsubst, but allows '\$' to keep variables from
being expanded.
With option -sl '\$' keeps the back-slash.
Default is to replace '\$' with '$'
"
if [[ $1 = -h ]] ;then echo -e >&2 "$Usage" ; exit 1 ;fi
if [[ $1 = -sl ]] ;then sl='\' ; shift ;fi
sed 's/\\\$/\${EnVsUbDolR}/g' | EnVsUbDolR=$sl\$ envsubst "$#"

Many of the answers using eval and echo kind of work, but break on various things, such as multiple lines, attempting to escaping shell meta-characters, escapes inside the template not intended to be expanded by bash, etc.
I had the same issue, and wrote this shell function, which as far as I can tell, handles everything correctly. This will still strip only trailing newlines from the template, because of bash's command substitution rules, but I've never found that to be an issue as long as everything else remains intact.
apply_shell_expansion() {
declare file="$1"
declare data=$(< "$file")
declare delimiter="__apply_shell_expansion_delimiter__"
declare command="cat <<$delimiter"$'\n'"$data"$'\n'"$delimiter"
eval "$command"
}
For example, you can use it like this with a parameters.cfg which is really a shell script that just sets variables, and a template.txt which is a template that uses those variables:
. parameters.cfg
printf "%s\n" "$(apply_shell_expansion template.txt)" > result.txt
In practice, I use this as a sort of lightweight template system.

you can try
echo $(eval echo $(cat something.txt))

You don't want to print each line, you want to evaluate it so that Bash can perform variable substitutions.
FOO=42
while read; do
eval echo "$REPLY"
done < something.txt
See help eval or the Bash manual for more information.

Another approach (which seems icky, but I am putting it here anyway):
Write the contents of something.txt to a temp file, with an echo statement wrapped around it:
something=$(cat something.txt)
echo "echo \"" > temp.out
echo "$something" >> temp.out
echo "\"" >> temp.out
then source it back in to a variable:
RESULT=$(source temp.out)
and the $RESULT will have it all expanded. But it seems so wrong !
Single line solution that doesn't need temporary file :
RESULT=$(source <(echo "echo \"$(cat something.txt)\""))
#or
RESULT=$(source <(echo "echo \"$(<something.txt)\""))

If you only want the variable references to be expanded (an objective that I had for myself) you could do the below.
contents="$(cat something.txt)"
echo $(eval echo \"$contents\")
(The escaped quotes around $contents is key here)

If something.txt has only one line, a bash method, (a shorter version of Michael Neale's "icky" answer),
using process & command substitution:
FOO=42 . <(echo -e echo $(<something.txt))
Output:
hello 42 world
Note that export isn't needed.
If something.txt has one or more lines, a GNU sed evaluate method:
FOO=42 sed 's/"/\\\"/g;s/.*/echo "&"/e' something.txt

Following solution:
allows replacing of variables which are defined
leaves unchanged variables placeholders which are not defined. This is especially useful during automated deployments.
supports replacement of variables in following formats:
${var_NAME}
$var_NAME
reports which variables are not defined in environment and returns error code for such cases
TARGET_FILE=someFile.txt;
ERR_CNT=0;
for VARNAME in $(grep -P -o -e '\$[\{]?(\w+)*[\}]?' ${TARGET_FILE} | sort -u); do
VAR_VALUE=${!VARNAME};
VARNAME2=$(echo $VARNAME| sed -e 's|^\${||g' -e 's|}$||g' -e 's|^\$||g' );
VAR_VALUE2=${!VARNAME2};
if [ "xxx" = "xxx$VAR_VALUE2" ]; then
echo "$VARNAME is undefined ";
ERR_CNT=$((ERR_CNT+1));
else
echo "replacing $VARNAME with $VAR_VALUE2" ;
sed -i "s|$VARNAME|$VAR_VALUE2|g" ${TARGET_FILE};
fi
done
if [ ${ERR_CNT} -gt 0 ]; then
echo "Found $ERR_CNT undefined environment variables";
exit 1
fi

foo=45
file=something.txt # in a file is written: Hello $foo world!
eval echo $(cat $file)

$ eval echo $(cat something.txt)
hello 42 world
$ bash --version
GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin17)
Copyright (C) 2007 Free Software Foundation, Inc.

envsubst is a great solution (see LenW's answer) if the content you're substituting is of "reasonable" length.
In my case, I needed to substitute in a file's content to replace the variable name. envsubst requires that the content be exported as environment variables and bash has a problem when exporting environment variables that are more than a megabyte or so.
awk solution
Using cuonglm's solution from a different question:
needle="doc1_base64" # The "variable name" in the file. (A $ is not needed.)
needle_file="doc1_base64.txt" # Will be substituted for the needle
haystack=$requestfile1 # File containing the needle
out=$requestfile2
awk "BEGIN{getline l < \"${needle_file}\"}/${needle}/{gsub(\"${needle}\",l)}1" $haystack > $out
This solution works for even large files.

expenv () {
LF=$'\n'
echo "cat <<END_OF_TEXT${LF}$(< "$1")${LF}END_OF_TEXT" | bash
return $?
}
expenv "file name"

The following works: bash -c "echo \"$(cat something.txt)"\"

Related

Evaluation of curly braces in Linux

I’ve noticed that we can use curly braces to make some of the commands much shorter as it is evaluated into list of arguments.
Input:
echo a{,b,c}
Output:
a ab ac
How do I force the same behaviour when the arguments are passed from the file?
Input:
cat file.txt | xargs echo
Output:
a{,b,c}
Expected output - same as in the previous example.
That {} expansion is a bash / zsh feature, as such then you need to explicitly run it thru any of these shells, in your case would be (using -I<STRING> to let xargs replace it in the string before running it):
cat file.txt |xargs -I# bash -c 'echo #'
xargs calls the echo as found in the $PATH, not the shell's builtin echo.
check the list of bash expansions: brace expansion happens first, so it won't get a chance to expand in that pipeline.
You'll have to do something like
while read -r line; do eval echo "$line"; done < file.txt
which exposes you to all kinds of nasty attacks if someone puts something malicious in that file.
Other than asking why would you want to do this... I offer the following:
add the string to a file:
echo 'a{,b,c}' > /tmp/foo
put the string in a variable:
export thing=`cat /tmp/foo`
eval the string:
eval $thing
If you had a bunch of these in a file then run the file through a loop and eval the loop value:
echo 'a{,b,c}' >> /tmp/foo
echo 'a{,b,c}' >> /tmp/foo
echo 'a{,b,c}' >> /tmp/foo
for i in `cat /tmp/foo`; do eval echo $i; done

linux terminal execute echo function

when I read the book linux shell scripting cookbook
they say when you wanna print !,you shouldn't put it in double quote,or you can add \ before ! to escape it.
e.g.
$echo "Hello,world!"
bash: !:event not found error
$echo "Hello,world\\!"
Hello,world!
but in my situation(ubuntu14.04), I get the answer like that:
$echo "Hello,world!"
Hello,world!
$echo "Hello,world\\!"
Hello,world\!
So, why in my machine can't get the same answer?
Why the escape symbol \ was printed as a normal symbol?
When you're typing interactively to the shell, ! has special meaning, it's the history expansion character. To prevent this special meaning, you need to put it in single quotes or escape it.
echo 'Hello, world!'
echo "Hello, world\!'
The reason it's not happening on Ubuntu may be because it's running a newer version of bash, which is apparently more selective about when history expansion occurs. It seems to require ! to be followed by alphanumerics, not punctuation.
You don't need to do this in scripts, because history is not normally enabled there. It's just for interactive shells.
Create a shell script called file.sh:
#!/bin/bash
# file.sh: a sample shell script to demonstrate the concept of Bash shell functions
# define usage function
usage(){
echo "Usage: $0 filename"
exit 1
}
# define is_file_exits function
# $f -> store argument passed to the script
is_file_exits(){
local f="$1"
[[ -f "$f" ]] && return 0 || return 1
}
# invoke usage
# call usage() function if filename not supplied
[[ $# -eq 0 ]] && usage
# Invoke is_file_exits
if ( is_file_exits "$1" )
then
echo "File found"
else
echo "File not found"
fi
Run it as follows:
chmod +x file.sh
./file.sh
./file.sh /etc/resolv.conf

How do I echo "-e"?

I want to echo a string that might contain the same parameters as echo. How can I do it without modifying the string?
For instance:
$ var="-e something"
$ echo $var
something
... didn't print -e
A surprisingly deep question. Since you tagged bash, I'll assume you mean bash's internal echo command, though the GNU coreutils' standalone echo command probably works similarly enough.
The gist of it is: if you really need to use echo (which would be surprising, but that's the way the question is written by now), it all depends on what exactly your string can contain.
The easy case: -e plus non-empty string
In that case, all you need to do is quote the variable before passing it to echo.
$ var="-e something"
$ echo "$var"
-e something
If the string isn't eaxctly an echo option or combination, which includes any non-option suffix, it won't be recognized as such by echo and will be printed out.
Harder: string can be -e only
If your case can reduce to just "-e", it gets trickier. One way to do it would be:
$ echo -e '\055e'
-e
(escaping the dash so it doesn't get interpreted as an option but as on octal sequence)
That's rewriting the string. It can be done automatically and non-destructively, so it feels acceptable:
$ var="-e something"
$ echo -e ${var/#-/\\055}
-e something
You noticed I'm actually using the -e option to interpret an octal sequence, so it won't work if you intended to echo -E. It will work for other options, though.
The right way
Seriously, you're not restricted to echo, are you?
printf '%s\n' "$var"
The proper bash way is to use printf:
printf "%s\n" "$var"
By the way, your echo didn't work because when you run:
var="-e something"
echo $var
(without quoting $var), echo will see two arguments: -e and something. Because when echo meets -e as its first argument, it considers it's an option (this is also true for -n and -E), and so processes it as such. If you had quoted var, as shown in other answers, it would have worked.
Quote it:
$ var="-e something"
$ echo "$var"
-e something
If what you want is to get echo -e's behaviour (enable interpretation of backslash escapes), then you have to leave the $var reference without quotes:
$ var="hi\nho"
$ echo $var
hi
ho
Or use eval:
$ var="hi\nho"
$ eval echo \${var}
hi\nho
$ var="-e hi\nho"
$ eval echo \${var}
hi
ho
Since we're using bash, another alternative to echo is to simply cat a "here string":
$ var="-e something"
$ cat <<< "$var"
-e something
$ var="-e"
$ cat <<< "$var"
-e
$
printf-based solutions will almost certainly be more portable though.
Try the following:
$ env POSIXLY_CORRECT=1 echo -e
-e
Due to shell aliases and built-in echo command, using an unadorned
echo interactively or in a script may get you different functionality
than that described here. Invoke it via env (i.e., env echo ...)
to avoid interference from the shell.
The environment variable POSIXLY_CORRECT was introduced to allow the user to force the standards-compliant behaviour. See: POSIX at Wikipedia.
Or use printf:
$ printf '%s\n' "$var"
Source: Why is bash swallowing -e in the front of an array at stackoverflow SE
Use printf instead:
var="-e bla"
printf "%s\n" "$var"
Using just echo "$var" will still fail if var contains just a -e or similar. If you need to be able to print that as well, use printf.

How can I store a command in a variable in a shell script?

I would like to store a command to use at a later time in a variable (not the output of the command, but the command itself).
I have a simple script as follows:
command="ls";
echo "Command: $command"; #Output is: Command: ls
b=`$command`;
echo $b; #Output is: public_html REV test... (command worked successfully)
However, when I try something a bit more complicated, it fails. For example, if I make
command="ls | grep -c '^'";
The output is:
Command: ls | grep -c '^'
ls: cannot access |: No such file or directory
ls: cannot access grep: No such file or directory
ls: cannot access '^': No such file or directory
How could I store such a command (with pipes/multiple commands) in a variable for later use?
Use eval:
x="ls | wc"
eval "$x"
y=$(eval "$x")
echo "$y"
Do not use eval! It has a major risk of introducing arbitrary code execution.
BashFAQ-50 - I'm trying to put a command in a variable, but the complex cases always fail.
Put it in an array and expand all the words with double-quotes "${arr[#]}" to not let the IFS split the words due to Word Splitting.
cmdArgs=()
cmdArgs=('date' '+%H:%M:%S')
and see the contents of the array inside. The declare -p allows you see the contents of the array inside with each command parameter in separate indices. If one such argument contains spaces, quoting inside while adding to the array will prevent it from getting split due to Word-Splitting.
declare -p cmdArgs
declare -a cmdArgs='([0]="date" [1]="+%H:%M:%S")'
and execute the commands as
"${cmdArgs[#]}"
23:15:18
(or) altogether use a bash function to run the command,
cmd() {
date '+%H:%M:%S'
}
and call the function as just
cmd
POSIX sh has no arrays, so the closest you can come is to build up a list of elements in the positional parameters. Here's a POSIX sh way to run a mail program
# POSIX sh
# Usage: sendto subject address [address ...]
sendto() {
subject=$1
shift
first=1
for addr; do
if [ "$first" = 1 ]; then set --; first=0; fi
set -- "$#" --recipient="$addr"
done
if [ "$first" = 1 ]; then
echo "usage: sendto subject address [address ...]"
return 1
fi
MailTool --subject="$subject" "$#"
}
Note that this approach can only handle simple commands with no redirections. It can't handle redirections, pipelines, for/while loops, if statements, etc
Another common use case is when running curl with multiple header fields and payload. You can always define args like below and invoke curl on the expanded array content
curlArgs=('-H' "keyheader: value" '-H' "2ndkeyheader: 2ndvalue")
curl "${curlArgs[#]}"
Another example,
payload='{}'
hostURL='http://google.com'
authToken='someToken'
authHeader='Authorization:Bearer "'"$authToken"'"'
now that variables are defined, use an array to store your command args
curlCMD=(-X POST "$hostURL" --data "$payload" -H "Content-Type:application/json" -H "$authHeader")
and now do a proper quoted expansion
curl "${curlCMD[#]}"
var=$(echo "asdf")
echo $var
# => asdf
Using this method, the command is immediately evaluated and its return value is stored.
stored_date=$(date)
echo $stored_date
# => Thu Jan 15 10:57:16 EST 2015
# (wait a few seconds)
echo $stored_date
# => Thu Jan 15 10:57:16 EST 2015
The same with backtick
stored_date=`date`
echo $stored_date
# => Thu Jan 15 11:02:19 EST 2015
# (wait a few seconds)
echo $stored_date
# => Thu Jan 15 11:02:19 EST 2015
Using eval in the $(...) will not make it evaluated later:
stored_date=$(eval "date")
echo $stored_date
# => Thu Jan 15 11:05:30 EST 2015
# (wait a few seconds)
echo $stored_date
# => Thu Jan 15 11:05:30 EST 2015
Using eval, it is evaluated when eval is used:
stored_date="date" # < storing the command itself
echo $(eval "$stored_date")
# => Thu Jan 15 11:07:05 EST 2015
# (wait a few seconds)
echo $(eval "$stored_date")
# => Thu Jan 15 11:07:16 EST 2015
# ^^ Time changed
In the above example, if you need to run a command with arguments, put them in the string you are storing:
stored_date="date -u"
# ...
For Bash scripts this is rarely relevant, but one last note. Be careful with eval. Eval only strings you control, never strings coming from an untrusted user or built from untrusted user input.
For bash, store your command like this:
command="ls | grep -c '^'"
Run your command like this:
echo $command | bash
Not sure why so many answers make it complicated!
use alias [command] 'string to execute'
example:
alias dir='ls -l'
./dir
[pretty list of files]
I tried various different methods:
printexec() {
printf -- "\033[1;37m$\033[0m"
printf -- " %q" "$#"
printf -- "\n"
eval -- "$#"
eval -- "$*"
"$#"
"$*"
}
Output:
$ printexec echo -e "foo\n" bar
$ echo -e foo\\n bar
foon bar
foon bar
foo
bar
bash: echo -e foo\n bar: command not found
As you can see, only the third one, "$#" gave the correct result.
I faced this problem with the following command:
awk '{printf "%s[%s]\n", $1, $3}' "input.txt"
I need to build this command dynamically:
The target file name input.txt is dynamic and may contain space.
The awk script inside {} braces printf "%s[%s]\n", $1, $3 is dynamic.
Challenge:
Avoid extensive quote escaping logic if there are many " inside the awk script.
Avoid parameter expansion for every $ field variable.
The solutions bellow with eval command and associative arrays do not work. Due to bash variable expansions and quoting.
Solution:
Build bash variable dynamically, avoid bash expansions, use printf template.
# dynamic variables, values change at runtime.
input="input file 1.txt"
awk_script='printf "%s[%s]\n" ,$1 ,$3'
# static command template, preventing double-quote escapes and avoid variable expansions.
awk_command=$(printf "awk '{%s}' \"%s\"\n" "$awk_script" "$input")
echo "awk_command=$awk_command"
awk_command=awk '{printf "%s[%s]\n" ,$1 ,$3}' "input file 1.txt"
Executing variable command:
bash -c "$awk_command"
Alternative that also works
bash << $awk_command
As you don't specify any scripting language, I would recommand tcl, the Tool Command Language for this kind of purpose.
Then in the first line, add the appropriate shebang:
#!/usr/local/bin/tclsh
with appropriate location you can retrieve with which tclsh.
In tcl scripts, you can call operating system commands with exec.
#!/bin/bash
#Note: this script works only when u use Bash. So, don't remove the first line.
TUNECOUNT=$(ifconfig |grep -c -o tune0) #Some command with "Grep".
echo $TUNECOUNT #This will return 0
#if you don't have tune0 interface.
#Or count of installed tune0 interfaces.
First of all, there are functions for this. But if you prefer variables then your task can be done like this:
$ cmd=ls
$ $cmd # works
file file2 test
$ cmd='ls | grep file'
$ $cmd # not works
ls: cannot access '|': No such file or directory
ls: cannot access 'grep': No such file or directory
file
$ bash -c $cmd # works
file file2 test
$ bash -c "$cmd" # also works
file
file2
$ bash <<< $cmd
file
file2
$ bash <<< "$cmd"
file
file2
Or via a temporary file
$ tmp=$(mktemp)
$ echo "$cmd" > "$tmp"
$ chmod +x "$tmp"
$ "$tmp"
file
file2
$ rm "$tmp"
Be careful registering an order with the: X=$(Command)
This one is still executed. Even before being called. To check and confirm this, you can do:
echo test;
X=$(for ((c=0; c<=5; c++)); do
sleep 2;
done);
echo note the 5 seconds elapsed
It is not necessary to store commands in variables even as you need to use it later. Just execute it as per normal. If you store in variables, you would need some kind of eval statement or invoke some unnecessary shell process to "execute your variable".

How to properly handle wildcard expansion in a bash shell script?

#!/bin/bash
hello()
{
SRC=$1
DEST=$2
for IP in `cat /opt/ankit/configs/machine.configs` ; do
echo $SRC | grep '*' > /dev/null
if test `echo $?` -eq 0 ; then
for STAR in $SRC ; do
echo -en "$IP"
echo -en "\n\t ARG1=$STAR ARG2=$2\n\n"
done
else
echo -en "$IP"
echo -en "\n\t ARG1=$SRC ARG2=$DEST\n\n"
fi
done
}
hello $1 $2
The above is the shell script which I provide source (SRC) & desitnation (DEST) path. It worked fine when I did not put in a SRC path with wild card ''. When I run this shell script and give ''.pdf or '*'as follows:
root#ankit1:~/as_prac# ./test.sh /home/dev/Examples/*.pdf /ankit_test/as
I get the following output:
192.168.1.6
ARG1=/home/dev/Examples/case_Contact.pdf ARG2=/home/dev/Examples/case_howard_county_library.pdf
The DEST is /ankit_test/as but DEST also get manupulated due to '*'. The expected answer is
ARG1=/home/dev/Examples/case_Contact.pdf ARG2=/ankit_test/as
So, if you understand what I am trying to do, please help me out to solve this BUG.
I'll be grateful to you.
Thanks in advance!!!
I need to know exactly how I use '*.pdf' in my program one by one without disturbing DEST.
Your script needs more work.
Even after escaping the wildcard, you won't get your expected answer. You will get:
ARG1=/home/dev/Examples/*.pdf ARG2=/ankit__test/as
Try the following instead:
for IP in `cat /opt/ankit/configs/machine.configs`
do
for i in $SRC
do
echo -en "$IP"
echo -en "\n\t ARG1=$i ARG2=$DEST\n\n"
done
done
Run it like this:
root#ankit1:~/as_prac# ./test.sh "/home/dev/Examples/*.pdf" /ankit__test/as
The shell will expand wildcards unless you escape them, so for example if you have
$ ls
one.pdf two.pdf three.pdf
and run your script as
./test.sh *.pdf /ankit__test/as
it will be the same as
./test.sh one.pdf two.pdf three.pdf /ankit__test/as
which is not what you expect. Doing
./test.sh \*.pdf /ankit__test/as
should work.
If you can, change the order of the parameters passed to your shell script as follows:
./test.sh /ankit_test/as /home/dev/Examples/*.pdf
That would make your life a lot easier since the variable part moves to the end of the line. Then, the following script will do what you want:
#!/bin/bash
hello()
{
SRC=$1
DEST=$2
for IP in `cat /opt/ankit/configs/machine.configs` ; do
echo -en "$IP"
echo -en "\n\t ARG1=$SRC ARG2=$DEST\n\n"
done
}
arg2=$1
shift
while [[ "$1" != "" ]] ; do
hello $1 $arg2
shift
done
You are also missing a final "done" to close your outer for loop.
OK, this appears to do what you want:
#!/bin/bash
hello() {
SRC=$1
DEST=$2
while read IP ; do
for FILE in $SRC; do
echo -e "$IP"
echo -e "\tARG1=$FILE ARG2=$DEST\n"
done
done < /tmp/machine.configs
}
hello "$1" $2
You still need to escape any wildcard characters when you invoke the script
The double quotes are necessary when you invoke the hello function, otherwise the mere fact of evaluating $1 causes the wildcard to be expanded, but we don't want that to happen until $SRC is assigned in the function
Here's what I came up with:
#!/bin/bash
hello()
{
# DEST will contain the last argument
eval DEST=\$$#
while [ $1 != $DEST ]; do
SRC=$1
for IP in `cat /opt/ankit/configs/machine.configs`; do
echo -en "$IP"
echo -en "\n\t ARG1=$SRC ARG2=$DEST\n\n"
done
shift || break
done
}
hello $*
Instead of passing only two parameters to the hello() function, we'll pass in all the arguments that the script got.
Inside the hello() function, we first assign the final argument to the DEST var. Then we loop through all of the arguments, assigning each one to SRC, and run whatever commands we want using the SRC and DEST arguments. Note that you may want to put quotation marks around $SRC and $DEST in case they contain spaces. We stop looping when SRC is the same as DEST because that means we've hit the final argument (the destination).
For multiple input files using a wildcard such as *.txt, I found this to work perfectly, no escaping required. It should work just like a native bash app like "ls" or "rm." This was not documented just about anywhere so since I spent a better part of 3 days trying to figure it out I decided I should post it for future readers.
Directory contains the following files (output of ls)
file1.txt file2.txt file3.txt
Run script like
$ ./script.sh *.txt
Or even like
$ ./script.sh file{1..3}.txt
The script
#!/bin/bash
# store default IFS, we need to temporarily change this
sfi=$IFS
#set IFS to $'\n\' - new line
IFS=$'\n'
if [[ -z $# ]]
then
echo "Error: Missing required argument"
echo
exit 1
fi
# Put the file glob into an array
file=("$#")
# Now loop through them
for (( i=0 ; i < ${#file[*]} ; i++ ));
do
if [ -w ${file[$i]} ]; then
echo ${file[$i]} " writable"
else
echo ${file[$i]} " NOT writable"
fi
done
# Reset IFS to its default value
IFS=$sfi
The output
file1.txt writable
file2.txt writable
file3.txt writable
The key was switching the IFS (Internal Field Separator) temporarily. You have to be sure to store this before switching and then switch it back when you are done with it as demonstrated above.
Now you have a list of expanded files (with spaces escaped) in the file[] array which you can then loop through. I like this solution the best, easiest to program for and easiest for the users.
There's no need to spawn a shell to look at the $? variable, you can evaluate it directly.
It should just be:
if [ $? -eq 0 ]; then
You're running
./test.sh /home/dev/Examples/*.pdf /ankit_test/as
and your interactive shell is expanding the wildcard before the script gets it. You just need to quote the first argument when you launch it, as in
./test.sh "/home/dev/Examples/*.pdf" /ankit_test/as
and then, in your script, quote "$SRC" anywhere where you literally want the things with wildcards (ie, when you do echo $SRC, instead use echo "$SRC") and leave it unquoted when you want the wildcards expanded. Basically, always put quotes around things which might contain shell metacharacters unless you want the metacharacters interpreted. :)

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