I'm trying to figure out how to use a variable containing an ANSI-C quoting string as an argument for a subsequent bash command.
The string in the variable itself is a list of files (can virtually be a list of anything).
For example, I have a file containing a list of other files, for example test.lst containing :
>$ cat test.lst
a.txt
b.txt
c.txt
I need to pass the file content as a single string so I'm doing :
test_str=$(cat test.lst)
then converts to ANSI-C quoting string:
test_str=${test_str#Q}
So at the end I have :
>$ test_str=$(cat test.lst)
>$ test_str=${test_str#Q}
>$ echo $test_str
$'a.txt\nb.txt\nc.txt'
which is what I'm looking for.
Then problem arises when I try to reuse this variable as a string list in another bash command.
For example direct use into a for loop :
>$ for str in $test_str; do echo $str; done
$'a.txt\nb.txt\nc.txt'
What I expect at this step is that it prints the same thing as the content of the original test.lst
I also tried expanding it back but it leaves leading $' and trailing '
>$ str=${test_str#E}
>$ echo $str
$'a.txt b.txt c.txt'
I also tried printf and some other stuffs to no avail. What is the correct way to use such ANSI-C quoting variable into a bash command ?
How about:
eval echo "${test_str}"
I believe that an ANSI-C quoted string is meant to be evaluated by bash command line parser.
In the first place, why do you need quoting? Just keep the data untouched stored as elements of an array:
mapfile -t filelist < test.lst
# iterate through the list
for file in "${filelist[#]}"; do printf '%s\n' "$file"; done
Related
I need to output a variable value to a file in a unix script. My problem it that the variable contains multiple lines. I need those to be output as '\n' literals in the file (a java options file), but I'm using echo and they always get processed into real new lines.
echo "-dmyproperty=$MULTILINE_VAR" >> jvm.options
I've tried echo options like -e o -E but they don't seem to do anything. Can anyone help?
You can use bash parameter substitution with an ANSI-C quoted newline
$ var="line1
line2
line3"
$ echo "${var//$'\n'/\\n}"
line1\nline2\nline3
I have a for loop that takes a CIDR that contains an illegal character for file names ("/").
part of the command is to save the output with the name of the CIDR.
for i in $(more subnets.lst);do shodan download $i-shodan net:$i;done
the download argument is followed by the result of the more command, which are the CIDR (192.168.21.0/24).
is there a way in bash to rename a variable while the loop is running ?
I remember doing this years back in batch files by subtracting from the str length, but that won't help me as I just need to replace the "/" with a "-"(or any other compliant char.
You can do this with using bash Parameter Expansion:
$ echo $var
192.168.21.0/24
$ echo "${var//\//-}"
192.168.21.0-24
So in your command, just use "${i//\//-}" whenever needed without changing the original value of $i. If you want to set variable i to the new value:
i="${i//\//-}"
On a side note, use while loop to read lines from a file, not cat, more or brothers, like:
while IFS= read -r line; do ....; done
I worked it out for myself:
for i in $(more subnets-test);do shodan download $(echo $i | tr "/" "-") net:$i;done
I have a file stv.txt containing some names
For example stv.txt is as follows:
hello
world
I want to generate another file by using these names and adding some extra text to them.I have written a script as follows
for i in `cat stvv.txt`;
do printf 'if(!strcmp("$i",optarg))' > my_file;
done
output
if(!strcmp("$i",optarg))
desired output
if(!strcmp("hello",optarg))
if(!strcmp("world",optarg))
how can I get the correct result?
This is a working solution.
1 All symbols inside single quotes is considered a string. 2 When using printf, do not surround the variable with quotes. (in this example)
The code below should fix it,
for i in `cat stvv.txt`;
printf 'if(!strcmp('$i',optarg))' > my_file;
done
basically, break the printf statement into three parts.
1: the string 'if(!strcmp('
2: $i (no quotes)
3: the string ',optarg))'
hope that helps!
To insert a string into a printf format, use %s in the format string:
$ for line in $(cat stvv.txt); do printf 'if(!strcmp("%s",optarg))\n' "$line"; done
if(!strcmp("hello",optarg))
if(!strcmp("world",optarg))
The code $(cat stvv.txt) will perform word splitting and pathname expansion on the contents of stvv.txt. You probably don't want that. It is generally safer to use a while read ... done <stvv.txt loop such as this one:
$ while read -r line; do printf 'if(!strcmp("%s",optarg))\n' "$line"; done <stvv.txt
if(!strcmp("hello",optarg))
if(!strcmp("world",optarg))
Aside on cat
If you are using bash, then $(cat stvv.txt) could be replaced with the more efficient $(<stvv.txt). This question, however, is tagged shell not bash. The cat form is POSIX and therefore portable to all POSIX shells while the bash form is not.
The script that I have to write must find the directories from the $PATH variable and print only the ones that end with an i.
How am I thinking about doing it
Get each directory from the variable with a for loop.
Find the length of each directory and get the last character from each using a substring
Use an If condition to print the directories that end with an i
Problems
The directories are not separated with a new line and I can't read them using a for loop.
Any ideas on how to get over this problem,or can you think of something more appropriate.
You can use this BASH one-liner for that job:
(IFS=':'; for i in $PATH; do [[ -d "$i" && $i =~ i$ ]] && echo "$i"; done)
IFS=':' sets input field separator to :
$PATH is iterated in a for loop
Each path element is tested if it is a directory and if it is ending with i using BASH regex
If test passes then it is pritned
Use bash's parameter expansion to replace all delimiters.
${parameter//pat/string}
For example,
mypaths="${PATH//:/ }"
will split the path by directory, so then you can run:
for directory in $mypaths
do
...
done
You can change the Inter Field Separator (IFS) to colon then path is dissected auto_magically. ;-)
IFS=:
for i in $PATH
do
echo $i | egrep -e 'i$'
done
grep 'i$' <<<"${PATH//:/$'\n'}"
The $PATH entries are split into individual lines by replacing : instances with newlines ($'\n') in a parameter expansion; $'\n' is an ANSI C-quoted string.
The resulting strings is passed to the stdin of grep as a here-string
(<<<...).
grep is then used to match only those lines that end in ($) the letter i.
To match case-insensitively, use grep -i 'i$'.
A demonstration:
$ (PATH='/ends/in_i:/usr/bin:/also/ends_in_i'; grep 'i$' <<<"${PATH//:/$'\n'}")
/ends/in_i
/also/ends_in_i
I would like to copy the contents of a variable (here called var) into a file.
The name of the file is stored in another variable destfile.
I'm having problems doing this. Here's what I've tried:
cp $var $destfile
I've also tried the same thing with the dd command... Obviously the shell thought that $var was referring to a directory and so told me that the directory could not be found.
How do I get around this?
Use the echo command:
var="text to append";
destdir=/some/directory/path/filename
if [ -f "$destdir" ]
then
echo "$var" > "$destdir"
fi
The if tests that $destdir represents a file.
The > appends the text after truncating the file. If you only want to append the text in $var to the file existing contents, then use >> instead:
echo "$var" >> "$destdir"
The cp command is used for copying files (to files), not for writing text to a file.
echo has the problem that if var contains something like -e, it will be interpreted as a flag. Another option is printf, but printf "$var" > "$destdir" will expand any escaped characters in the variable, so if the variable contains backslashes the file contents won't match. However, because printf only interprets backslashes as escapes in the format string, you can use the %s format specifier to store the exact variable contents to the destination file:
printf "%s" "$var" > "$destdir"
None of the answers above work if your variable:
starts with -e
starts with -n
starts with -E
contains a \ followed by an n
should not have an extra newline appended after it
and so they cannot be relied upon for arbitrary string contents.
In bash, you can use "here strings" as:
cat <<< "$var" > "$destdir"
As noted in the comment by Ash below, #Trebawa's answer (formulated in the same room as mine!) using printf is a better approach than cat.
All of the above work, but also have to work around a problem (escapes and special characters) that doesn't need to occur in the first place: Special characters when the variable is expanded by the shell. Just don't do that (variable expansion) in the first place. Use the variable directly, without expansion.
Also, if your variable contains a secret and you want to copy that secret into a file, you might want to not have expansion in the command line as tracing/command echo of the shell commands might reveal the secret. Means, all answers which use $var in the command line may have a potential security risk by exposing the variable contents to tracing and logging of the shell.
For variables that are already exported, use this:
printenv var >file
That means, in case of the OP question:
printenv var >"$destfile"
Note: variable names are case sensitive.
Warning: It is not a good idea to export a variable just for the sake of printing it with printenv. If you have a non-exported script variable that contains a secret, exporting it will expose it to all future child processes (unless unexported, for example using export -n).
If I understood you right, you want to copy $var in a file (if it's a string).
echo $var > $destdir
When you say "copy the contents of a variable", does that variable contain a file name, or does it contain a name of a file?
I'm assuming by your question that $var contains the contents you want to copy into the file:
$ echo "$var" > "$destdir"
This will echo the value of $var into a file called $destdir. Note the quotes. Very important to have "$var" enclosed in quotes. Also for "$destdir" if there's a space in the name. To append it:
$ echo "$var" >> "$destdir"
you may need to edit a conf file in a build process:
echo "db-url-host=$POSTGRESQL_HOST" >> my-service.conf
You can test this solution with running before export POSTGRESQL_HOST="localhost"