Unix – How to read a file into a variable in shell

shshellunix

I want to read a file and save it in variable, but I need to keep the variable and not just print out the file.
How can I do this? I have written this script but it isn't quite what I needed:

#!/bin/sh
while read LINE  
do  
  echo $LINE  
done <$1  
echo 11111-----------  
echo $LINE

In my script, I can give the file name as a parameter, so, if the file contains "aaaa", for example, it would print out this:

aaaa
11111-----

But this just prints out the file onto the screen, and I want to save it into a variable!
Is there an easy way to do this?

Best Answer

In cross-platform, lowest-common-denominator sh you use:

#!/bin/sh
value=`cat config.txt`
echo "$value"

In bash or zsh, to read a whole file into a variable without invoking cat:

#!/bin/bash
value=$(<config.txt)
echo "$value"

Invoking cat in bash or zsh to slurp a file would be considered a Useless Use of Cat.

Note that it is not necessary to quote the command substitution to preserve newlines.

See: Bash Hacker's Wiki - Command substitution - Specialities.

Related Solutions

Bash – How to check if a directory exists in a Bash shell script

To check if a directory exists in a shell script, you can use the following:

if [ -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY exists.
fi

Or to check if a directory doesn't exist:

if [ ! -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY doesn't exist.
fi

However, as Jon Ericson points out, subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check. E.g. running this:

ln -s "$ACTUAL_DIR" "$SYMLINK"
if [ -d "$SYMLINK" ]; then 
  rmdir "$SYMLINK" 
fi

Will produce the error message:

rmdir: failed to remove `symlink': Not a directory

So symbolic links may have to be treated differently, if subsequent commands expect directories:

if [ -d "$LINK_OR_DIR" ]; then 
  if [ -L "$LINK_OR_DIR" ]; then
    # It is a symlink!
    # Symbolic link specific commands go here.
    rm "$LINK_OR_DIR"
  else
    # It's a directory!
    # Directory command goes here.
    rmdir "$LINK_OR_DIR"
  fi
fi

Take particular note of the double-quotes used to wrap the variables. The reason for this is explained by 8jean in another answer.

If the variables contain spaces or other unusual characters it will probably cause the script to fail.

Python – How to execute a program or call a system command

Use the subprocess module in the standard library:

import subprocess
subprocess.run(["ls", "-l"])

The advantage of subprocess.run over os.system is that it is more flexible (you can get the stdout, stderr, the "real" status code, better error handling, etc...).

Even the documentation for os.system recommends using subprocess instead:

The subprocess module provides more powerful facilities for spawning new processes and retrieving their results; using that module is preferable to using this function. See the Replacing Older Functions with the subprocess Module section in the subprocess documentation for some helpful recipes.

On Python 3.4 and earlier, use subprocess.call instead of .run:

subprocess.call(["ls", "-l"])

Best Answer

Related Solutions

Bash – How to check if a directory exists in a Bash shell script

Python – How to execute a program or call a system command

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