Today I tried a script in linux to get all files in one dir. It was pretty straightforward, but I found something interesting.
#!/bin/bash
InputDir=/home/XXX/
for file in $InputDir'*'
do
echo $file
done
The output is:
/home/XXX/fileA /home/XXX/fileB
But when I just input the dir directly, like:
#!/bin/bash
InputDir=/home/XXX/
for file in /home/XXX/*
do
echo $file
done
The output is:
/home/XXX/fileA
/home/XXX/fileB
It seems, in the first script, there was only one loop and all the file names were stored in the variable $file in the FIRST loop, separated by space. But in the second script, one file name was stored in $file just in one loop, and there were more than one loop. What is exactly the difference between these two scripts?
Thanks very much, maybe my question is a little bit naive..
The behavior is correct and "as expected".
for file in $InputDir'*' means assign "/home/XXX/*" to $file (note the quotes). Since you quoted the asterisk, it will not be executed at this time. When the shell sees echo $file, it first expands the variables and then it does glob expansion. So after the first step, it sees
echo /home/XXX/*
and after glob expansion, it sees:
echo /home/XXX/fileA /home/XXX/fileB
Only now, it will execute the command.
In the second case, the pattern /home/XXX/* is expanded before the for is executed and thus, each file in the directory is assigned to file and then the body of the loop is executed.
This will work:
for file in "$InputDir"*
but it's brittle; it will fail, for example, when you forget to add a / to the end of the variable $InputDir.
for file in "$InputDir"/*
is a little bit better (Unix will ignore double slashes in a path) but it can cause trouble when $InputDir is not set or empty: You'll suddenly list files in the / (root) folder. This can happen, for example, because of a typo:
inputDir=...
for file in "$InputDir"/*
Case matters on Unix :-)
To help you understand code like this, use set -x ("enable tracing") in a line before the code you want to debug.
The difference is the quoting of '*'. In the first case the loop only executes once, with $file equal to /home/XXX/* which then expands to all the files in the directory when passed to echo. In the second case it executes once per file, with $file equal to each file name in turn.
Bottom line - change:
for file in $InputDir'*'
to:
for file in $InputDir*
or, better, and to make it more readable - change:
InputDir=/home/XXX/
for file in $InputDir'*'
to:
InputDir=/home/XXX
for file in $InputDir/*
Related
I am trying to create a shell script for logs and trying to append data into a text file. I have write this sample "test.sh" code for testing:
#!/bin/sh -e
touch /home/sample.txt
SPTH = '/home/sample'.txt
echo "MY LOG FILE" >> "$SPTH"
echo "DUMP started at $(date +'%d-%m-%Y %H:%M:%S')" >> /home/sample.txt
echo "DUMP finished at $(date +'%d-%m-%Y %H:%M:%S')" >> /home/sample.txt
but in above code all lines are working correct except one line of code i.e.
echo "MY LOG FILE" >> "$SPTH"
It is giving error:
test.sh: line 6: : No such file or directory
I want to replace this full path of file "/home/sample.txt" to variable "$SPATH".
I am executing my shell script using
sh test.sh
What I am doing wrong.
Variable assignments in bash shell does not allow you to have spaces within. It will be actually interpreted as command with = and the subsequent keywords as arguments to the first word, which is wrong.
Change your code to
SPTH="/home/sample.txt"
That is the reason why SPTH was not assigned to the actual path you intended it to have. And you have no reason to have single-quote here and excluding the extension part. Using it fully within double-quotes is absolutely fine.
The syntax for the command line is that the first token is a command, tokens are separated by whitespace. So:
SPTH = '/home/sample'.txt
Has the command as SPTH, the second token is =, and so on. You might think this is daft, but most shells behave like this for historical reasons.
So you need to remove the whitespace:
SPTH='/home/sample'.txt
This question already has answers here:
Read user input inside a loop
(6 answers)
Closed 5 years ago.
First post here! I really need help on this one, I looked the issue on google, but can't manage to find an useful answer for me. So here's the problem.
I'm having fun coding some like of a framework in bash. Everyone can create their own module and add it to the framework. BUT. To know what arguments the script require, I created an "args.conf" file that must be in every module, that kinda looks like this:
LHOST;true;The IP the remote payload will connect to.
LPORT;true;The port the remote payload will connect to.
The first column is the argument name, the second defines if it's required or not, the third is the description. Anyway, long story short, the framework is supposed to read the args.conf file line by line to ask the user a value for every argument. Here's the piece of code:
info "Reading module $name argument list..."
while read line; do
echo $line > line.tmp
arg=`cut -d ";" -f 1 line.tmp`
requ=`cut -d ";" -f 2 line.tmp`
if [ $requ = "true" ]; then
echo "[This argument is required]"
else
echo "[This argument isn't required, leave a blank space if you don't wan't to use it]"
fi
read -p " $arg=" answer
echo $answer >> arglist.tmp
done < modules/$name/args.conf
tr '\n' ' ' < arglist.tmp > argline.tmp
argline=`cat argline.tmp`
info "Launching module $name..."
cd modules/$name
$interpreter $file $argline
cd ../..
rm arglist.tmp
rm argline.tmp
rm line.tmp
succes "Module $name execution completed."
As you can see, it's supposed to ask the user a value for every argument... But:
1) The read command seems to not be executing. It just skips it, and the argument has no value
2) Despite the fact that the args.conf file contains 3 lines, the loops seems to be executing just a single time. All I see on the screen is "[This argument is required]" just one time, and the module justs launch (and crashes because it has not the required arguments...).
Really don't know what to do, here... I hope someone here have an answer ^^'.
Thanks in advance!
(and sorry for eventual mistakes, I'm french)
Alpha.
As #that other guy pointed out in a comment, the problem is that all of the read commands in the loop are reading from the args.conf file, not the user. The way I'd handle this is by redirecting the conf file over a different file descriptor than stdin (fd #0); I like to use fd #3 for this:
while read -u3 line; do
...
done 3< modules/$name/args.conf
(Note: if your shell's read command doesn't understand the -u option, use read line <&3 instead.)
There are a number of other things in this script I'd recommend against:
Variable references without double-quotes around them, e.g. echo $line instead of echo "$line", and < modules/$name/args.conf instead of < "modules/$name/args.conf". Unquoted variable references get split into words (if they contain whitespace) and any wildcards that happen to match filenames will get replaced by a list of matching files. This can cause really weird and intermittent bugs. Unfortunately, your use of $argline depends on word splitting to separate multiple arguments; if you're using bash (not a generic POSIX shell) you can use arrays instead; I'll get to that.
You're using relative file paths everywhere, and cding in the script. This tends to be fragile and confusing, since file paths are different at different places in the script, and any relative paths passed in by the user will become invalid the first time the script cds somewhere else. Worse, you aren't checking for errors when you cd, so if any cd fails for any reason, then entire rest of the script will run in the wrong place and fail bizarrely. You'd be far better off figuring out where your system's root directory is (as an absolute path), then referencing everything from it (e.g. < "$module_root/modules/$name/args.conf").
Actually, you're not checking for errors anywhere. It's generally a good idea, when writing any sort of program, to try to think of what can go wrong and how your program should respond (and also to expect that things you didn't think of will also go wrong). Some people like to use set -e to make their scripts exit if any simple command fails, but this doesn't always do what you'd expect. I prefer to explicitly test the exit status of the commands in my script, with something like:
command1 || {
echo 'command1 failed!' >&2
exit 1
}
if command2; then
echo 'command2 succeeded!' >&2
else
echo 'command2 failed!' >&2
exit 1
fi
You're creating temp files in the current directory, which risks random conflicts (with other runs of the script at the same time, any files that happen to have names you're using, etc). It's better to create a temp directory at the beginning, then store everything in it (again, by absolute path):
module_tmp="$(mktemp -dt module-system)" || {
echo "Error creating temp directory" >&2
exit 1
}
...
echo "$answer" >> "$module_tmp/arglist.tmp"
(BTW, note that I'm using $() instead of backticks. They're easier to read, and don't have some subtle syntactic oddities that backticks have. I recommend switching.)
Speaking of which, you're overusing temp files; a lot of what you're doing with can be done just fine with shell variables and built-in shell features. For example, rather than reading line from the config file, then storing them in a temp file and using cut to split them into fields, you can simply echo to cut:
arg="$(echo "$line" | cut -d ";" -f 1)"
...or better yet, use read's built-in ability to split fields based on whatever IFS is set to:
while IFS=";" read -u3 arg requ description; do
(Note that since the assignment to IFS is a prefix to the read command, it only affects that one command; changing IFS globally can have weird effects, and should be avoided whenever possible.)
Similarly, storing the argument list in a file, converting newlines to spaces into another file, then reading that file... you can skip any or all of these steps. If you're using bash, store the arg list in an array:
arglist=()
while ...
arglist+=("$answer") # or ("#arg=$answer")? Not sure of your syntax.
done ...
"$module_root/modules/$name/$interpreter" "$file" "${arglist[#]}"
(That messy syntax, with the double-quotes, curly braces, square brackets, and at-sign, is the generally correct way to expand an array in bash).
If you can't count on bash extensions like arrays, you can at least do it the old messy way with a plain variable:
arglist=""
while ...
arglist="$arglist $answer" # or "$arglist $arg=$answer"? Not sure of your syntax.
done ...
"$module_root/modules/$name/$interpreter" "$file" $arglist
... but this runs the risk of arguments being word-split and/or expanded to lists of files.
I'am new in Linux and I want to write a bash script that can read in a file name of a directory that starts with LED + some numbers.(Ex.: LED5.5.002)
In that directory there is only one file that will starts with LED. The problem is that this file will every time be updated, so the next time it will be for example LED6.5.012 and counting.
I searched and tried a little bit and came to this solution:
export fspec=/home/led/LED*
LedV=`basename $fspec`
echo $LedV
If I give in those commands one by one in my terminal it works fine, LedV= LED5.5.002 but if i run it in a bash scripts it gives the result: LedV = LED*
I search after another solution:
a=/home/led/LED*
LedV=$(basename $a)
echo $LedV
but here again the same, if i give it in one by one it's ok but in a script: LedV = LED*.
It's probably something small but because of my lack of knowledge over Linux I cannot find it. So can someone tell what is wrong?
Thanks! Jan
Shell expansions don't happen on scalar assignments, so in
varname=foo*
the expansion of "$varname" will literally be "foo*". It's more confusing when you consider that echo $varname (or in your case basename $varname; either way without the double quotes) will cause the expansion itself to be treated as a glob, so you may well think the variable contains all those filenames.
Array expansions are another story. You might just want
fspec=( /path/LED* )
echo "${fspec[0]##*/}" # A parameter expansion to strip off the dirname
That will work fine for bash. Since POSIX sh doesn't have arrays like this, I like to give an alternative approach:
for fspec in /path/LED*; do
break
done
echo "${fspec##*/}"
pwd
/usr/local/src
ls -1 /usr/local/src/mysql*
/usr/local/src/mysql-cluster-gpl-7.3.4-linux-glibc2.5-x86_64.tar.gz
/usr/local/src/mysql-dump_test_all_dbs.sql
if you only have 1 file, you will only get 1 result
MyFile=`ls -1 /home/led/LED*`
My Aim -->
Files Listing from a command has to be read line by line and be used as part of another command.
Description -->
A command in linux returns
archive/Crow.java
archive/Kaka.java
mypmdhook.sh
which is stored in changed_files variable. I use the following while loop to read the files line by line and use it as part of a pmd command
while read each_file
do
echo "Inside Loop -- $each_file"
done<$changed_files
I am new to writing shell script but my assumption was that the lines would've been separated in the loop and printed in each iteration but instead I get the following error --
mypmdhook.sh: 7: mypmdhook.sh: cannot open archive/Crow.java
archive/Kaka.java
mypmdhook.sh: No such file
Can you tell me how I can just get the value as a string and not as a file what is opened. By the way, the file does exist which made me feel even more confused.(and later use it inside a command). I'd be happy with any kind of answer that helps me understand and resolve this issue.
Since you have data stored in a variable, use a "here string" instead of file redirection:
changed_files="archive/Crow.java
archive/Kaka.java
mypmdhook.sh"
while read each_file
do
echo "Inside Loop -- $each_file"
done <<< "$changed_files"
Inside Loop -- archive/Crow.java
Inside Loop -- archive/Kaka.java
Inside Loop -- mypmdhook.sh
Extremely important to quote "$changed_files" in order to preserve the newlines, so the while-read loop works as you expect. A rule of thumb: always quote variables, unless you knows exactly why you want to leave the quotes off.
What happens here is that the value of your variable $changed_files is substituted into your command, and you get something like
while read each_file
do
echo "Inside Loop -- $each_file"
done < archive/Crow.java
archive/Kaka.java
mypmdhook.sh
then the shell tries to open the file for redirecting the input and obviously fails.
The point is that redirections (e.g. <, >, >>) in most cases accept filenames, but what you really need is to give the contents of the variable to the stdin. The most obvious way to do that is
echo $changed_files | while read each_file; do echo "Inside Loop -- $each_file"; done
You can also use the for loop instead of while read:
for each_file in $changed_files; do echo "inside Loop -- $each_file"; done
I prefer using while read ... if there is a chance that some filename may contain spaces, but in most cases for ... in will work for you.
Rather than storing command's output in a variable use while loop like this:
mycommand | while read -r each_file; do echo "Inside Loop -- $each_file"; done
If you're using BASH you can use process substitution:
while read -r each_file; do echo "Inside Loop -- $each_file"; done < <(mycommand)
btw your attempt of done<$changed_files will assume that changed_files represents a file.
I'm trying to redirect(?) my standard error/output to a text file.
I did my research, but for some reason the online answers are not working for me.
What am I doing wrong?
cd /home/user1/lists/
for dir in $(ls)
do
(
echo | $dir > /root/user1/$dir" "log.txt
) > /root/Desktop/Logs/Update.log
done
I also tried
2> /root/Desktop/Logs/Update.log
1> /root/Desktop/Logs/Update.log
&> /root/Desktop/Logs/Update.log
None of these work for me :(
Help please!
Try this for the basics:
echo hello >> log.txt 2>&1
Could be read as: echo the word hello, redirecting and appending STDOUT to the file log.txt. STDERR (file descriptor 2) is redirected to wherever STDOUT is being pointed. Note that STDOUT is the default and thus there is no "1" in front of the ">>". Works on the current line only.
To redirect and append all output and error of all commands in a script, put this line near the top. It will be in effect for the length of the script instead of doing it on each line:
exec >>log.txt 2>&1
If you are trying to obtain a list of the files in /home/user1/lists, you do not need a loop at all:
ls /home/usr1/lists/ >Update.log
If you are attempting to run every file in the directory as an executable with a newline as its input, and collect the output from all these programs in Update.log, try this:
for file in /home/user1/lists/*; do
echo | "$file"
done >Update.log
(Notice how we avoid the useless use of ls and how there is no redirection inside the loop.)
If you want to create an empty file called *.log.txt for each file in the directory, you would do
for file in /home/user1/lists/*; do
touch "$(basename "$file")"log.txt
done
(Using basename to obtain the file name without the directory part avoids the cd but you could do it the other way around. Generally, we tend to avoid changing the directory in scripts, so that the tool can be run from anywhere and generate output in the current directory.)
If you want to create a file containing a single newline, regardless of whether it already exists or not,
for file in /home/user1/lists/*; do
echo >"$(basename "$file")"log.txt
done
In your original program, you redirect the echo inside the loop, which means that the redirection after done will not receive any output at all, so the created file will be empty.
These are somewhat wild guesses at what you might actually be trying to accomplish, but should hopefully help nudge you slightly in the right direction. (This should properly be a comment, I suppose, but it's way too long and complex.)