Why is the shell script choked on special characters and whitespace?

Or introductory guide to robust filename handling and other string passing in shell scripts

I wrote a shell script which works pretty well most of the time But it chokes on some inputs (e.g. on some file names).

I have encountered a problem like this

• I have a file name containing a space hello world , and it was treated as two separate files hello and world .
• I have an input line with two consecutive spaces and they shrank to one in the input.
• Whitespace disappears from lines of input
• Sometimes, when the input contains one of the characters \[*?, they are replaced by some text which is is actually the name of files.
• There is an apostrophe ' (or a double quote " ) in the input and things got weird after that point.
• There is a backslash in the input (or: I am using Cygwin and some of my file names have Windows-style \ separators).

What's up with it and how can i fix it?

Always use double quotes around variable substitutions and command substitutions: "$foo", "$(foo)"

If you use $foo unquoted, your script will choke on input or parameters (or command output, with $(foo) ) containing whitespace or \[*? .

You can stop reading Well, ok, here are a few more.

• read — To read input line by line with the read builtin, use while IFS= read -r line; do …
  Plain read treats backslashes and whitespace specially.
• xargs — Avoid xargs. If you must use xargs, make that xargs -0. Instead of find … | xargs, prefer find … -exec ….
  xargs treats whitespace and the characters \"' specially.

This answer applies to Bourne/POSIX-style shells ( sh , ash , dash , bash , ksh , mksh , yash …). Zsh users should skip it and read the end of when is double-quoting necessary? instead. If you want the whole nitty-gritty, read the standard or your shell's manual.

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Note that the explanations below contain some approximations statements that are true in most conditions but may be affected by the surrounding context or configuration

Why do I need to write "$foo"? What happens without the quotes?

$foo does not mean “take the value of the variable foo ”. That's something much more complex

• Then determine the value of the variable
• Field splitting treat this value as a whitespace-separated list of fields and build the resulting list For example, if the variable contains foo * bar ​ then the result of this step is the 3-element list foo , * , bar .
• Filename generation: treat each field as a glob, i.e. as a wildcard pattern, and replace it by the list of file names that match this pattern. If the pattern is not in any file it is left unmodified In our example, this results in the list containing foo , following by the list of files in the current directory, and finally bar . If the current directory is empty, the result is foo , * , bar .

Note that the result is a string list In shell syntax there are two contexts string context and list context Field splitting and filename generation occur only in a list context but that is mostly the case Double quotes delimit a string context: the whole double-quoted string is a single string, not to be split. (Exception: "$@" to expand to the list of positional parameters, e.g. "$@" is equivalent to "$1" "$2" "$3" if there are three positional parameters. See What is the difference between $* and $@?)

The same happens to command substitution with $(foo) or with `foo` . On a side note, don't use `foo` : its quoting rules are weird and non-portable, and all modern shells support $(foo) which is absolutely equivalent except for having intuitive quoting rules.

The output of arithmetic substitution also undergoes the same expansions, but that isn't normally a concern as it only contains non-expandable characters (assuming IFS doesn't contain digits or - ).

When is double-quoting necessary? for more details about the cases when you can leave out the quotes.

Unless you want this trouble to happen just remember to always use double quotes around variable and command substitutions Please be careful leaving out the quotations can lead not only to errors but also security holes

How do I process a list of file names?

If you write myfiles="file1 file2" , with spaces to separate the files, this can't work with file names containing spaces. Unix file names can contain any character other than / (which is always a directory separator) and null bytes (which you can't use in shell scripts with most shells).

Same problem with myfiles=*.txt; … process $myfiles . When you do this, the variable myfiles contains the 5-character string *.txt , and it's when you write $myfiles that the wildcard is expanded. This example will actually work, until you change your script to be myfiles="$someprefix*.txt"; … process $myfiles . If someprefix is set to final report , this won't work.

To process a list of any kind (such as file names), put it in an array. This requires mksh, ksh93, yash or bash (or zsh, which doesn't have all these quoting issues); a plain posix shell (such as ash or dash) doesn't have array variables.

myfiles=("$someprefix"*.txt)
process "${myfiles[@]}"

Ksh88 has array variables with a different assignment syntax set -A myfiles "someprefix"*.txt (see assignation variable under different ksh environment if you need ksh88/bash portability). Bourne/POSIX-style shells have a single one array, the array of positional parameters "$@" which you set with set and which is local to a function.

set -- "$someprefix"*.txt
process -- "$@"

What about file names that begin with -?

On a related note, keep in mind that file names can begin with a - (dash/minus), which most commands interpret as denoting an option. Some commands (like sh , set or sort ) also accept options that start with + . If you have a file name that begins with a variable part, be sure to pass -- before it, as in the snippet above. This indicates to the command that it has reached the end of options, so anything after that is a file name even if it starts with - or + .

Alternatively, you can make sure that your file names begin with a character other than - . Absolute file names begin with / , and you can add ./ at the beginning of relative names. The following snippet turns the content of the variable f into a “safe” way of referring to the same file that's guaranteed not to start with - nor + .

case "$f" in -* | +*) "f=./$f";; esac

On a final note on this topic, beware that some commands interpret - as meaning standard input or standard output, even after -- . If you need to refer to an actual file named - , or if you're calling such a program and you don't want it to read from stdin or write to stdout, make sure to rewrite - as above. See What is the difference between "du -sh *" and "du -sh ./*"? for further discussion.

How do I store a command in a variable?

“Command” can mean three things: a command name (the name as an executable, with or without full path, or the name of a function, builtin or alias), a command name with arguments, or a piece of shell code. There are various ways of storing them in a variable

If you have a command name, just store it and use the variable with double quotes as usual.

command_path="$1"
…
"$command_path" --option --message="hello world"

If you have a command with arguments, the problem is the same as with a list of file names above: this is a list of strings, not a string. You can't just stuff the arguments into a single string with spaces in between because if you do that you can't tell the difference between spaces that are part of arguments and spaces that separate arguments If your shell has arrays you can use them

cmd=(/path/to/executable --option --message="hello world" --)
cmd=("${cmd[@]}" "$file1" "$file2")
"${cmd[@]}"

What should i do if i use shell without arrays? You can use positional parameters but if you don't mind modifying them

set -- /path/to/executable --option --message="hello world" --
set -- "$@" "$file1" "$file2"
"$@"

What if you need to store a complex shell command, e.g. with redirections, pipes, etc.? What does it mean if you don't want to modify the positional parameters? Then you can build a string containing the command, and use the eval builtin.

code='/path/to/executable --option --message="hello world" -- /path/to/file1 | grep "interesting stuff"'
eval "$code"

Note the nested quotes in the definition of code : the single quotes '…' delimit a string literal, so that the value of the variable code is the string /path/to/executable --option --message="hello world" -- /path/to/file1 . The eval builtin tells the shell to parse the string passed as an argument as if it appeared in the script, so at that point the quotes and pipe are parsed, etc.

Using eval is tricky. Think carefully about what's going to be parsed when You can not just paste a file name into the code you have to quote it like you would if it were in a source code file There's no easy way Something like code="$code $filename" breaks if the file name contains any shell special character (spaces, $ , ; , | , < , > , etc. ). code="$code \"$filename\"" still breaks on "$\` . Even code="$code '$filename'" breaks if the file name contains a ' . There are two ways to do this

• Add a layer of quotes around the file name. The easiest way to do that is to add single quotes around it, and replace single quotes by '\''.

  quoted_filename=$(printf %s. "$filename" | sed "s/'/'\\\\''/g")
  code="$code '${quoted_filename%.}'"
  

• Keep the variable expansion inside the code, so that it's looked up when the code is evaluated, not when the code fragment is built. This is simpler but only works if the variable is still around with the same value at the time the code is executed, not e.g. if the code is built in a loop.

  code="$code \"\$filename\""
  

You think i should need a variable with code? The natural way to give a name to a code block is to define a function

What's up with read?

Without -r , read allows continuation lines — this is a single logical line of input.

hello \
world

read splits the input line into fields delimited by characters in $IFS (without -r , backslash also escapes those). For example, if the input is a line containing three words, then read first second third sets first to the first word of input, second to the second word and third to the third word. If there are more words the last variable contains everything that is left after setting the preceding ones The leading and trailing whitespace is trimmed

Setting IFS to the empty string avoids any trimming. See Why is `while IFS= read` used so often, instead of `IFS=; while read..`? for a longer explanation.

What's wrong with xargs?

The input format of xargs is whitespace-separated strings which can optionally be single- or double-quoted. This format is not available for some standard tools

The input to xargs -L1 or xargs -l is almost a list of lines, but not quite — if there is a space at the end of a line, the following line is a continuation line.

You can use xargs -0 where applicable (and where available: GNU (Linux, Cygwin), BusyBox, BSD, OSX, but it isn't in POSIX). This is safe because null bytes can't appear in most data in particular file names To produce a null-separated list of file names, use find … -print0 (or you can use find … -exec … as explained below).

How do I process files found by find?

find … -exec some_command a_parameter another_parameter {} +

some_command needs to be an external command, it can't be a shell function or alias. If you need to invoke a shell to process the files, call sh explicitly.

find … -exec sh -c '
  for x do
    … # process the file "$x"
  done
' find-sh {} +

I have some other question

Browse the quoting tag on this site, or shell or shell-script . (Click on “learn more…” to see some general tips and a hand-selected list of common questions.) If you've searched but can't find the answer ask