The Metacharacters

Unix Shell provides various metacharacters which have special meaning while using them in any Shell Script and causes termination of a word unless quoted.

For example ? matches with a single charater while listing files in a directory and an * would match more than one characters. Here is a list of most of the shell special characters (also called metacharacters) −

A character may be quoted (i.e., made to stand for itself) by preceding it with a \.

Example

Following is the example which show how to print a * or a ? −

#!/bin/sh

echo Hello; Word

This would produce following result −

Hello
./test.sh: line 2: Word: command not found

shell returned 127

Now let us try using a quoted character −

#!/bin/sh

echo Hello\; Word

This would produce following result −

Hello; Word

The $ sign is one of the metacharacters, so it must be quoted to avoid special handling by the shell −

#!/bin/sh

echo "I have \$1200"

This would produce following result −

I have $1200

There are following four forms of quotings −

The Single Quotes

Consider an echo command that contains many special shell characters −

echo <-$1500.**>; (update?) [y|n]

Putting a backslash in front of each special character is tedious and makes the line difficult to read −

echo \<-\$1500.\*\*\>\; \(update\?\) \[y\|n\]

There is an easy way to quote a large group of characters. Put a single quote ( ') at the beginning and at the end of the string −

echo '<-$1500.**>; (update?) [y|n]'

Any characters within single quotes are quoted just as if a backslash is in front of each character. So now this echo command displays properly.

If a single quote appears within a string to be output, you should not put the whole string within single quotes instead you whould preceed that using a backslash (\) as follows −

echo 'It\'s Shell Programming'

The Double Quotes

Try to execute the following shell script. This shell script makes use of single quote −

VAR=ZARA
echo '$VAR owes <-$1500.**>; [ as of (`date +%m/%d`) ]'

This would produce following result −

$VAR owes <-$1500.**>; [ as of (`date +%m/%d`) ]

So this is not what you wanted to display. It is obvious that single quotes prevent variable substitution. If you want to substitute variable values and to make invert commas work as expected then you would need to put your commands in double quotes as follows −

VAR=ZARA
echo "$VAR owes <-\$1500.**>; [ as of (`date +%m/%d`) ]"

Now this would produce following result −

ZARA owes <-$1500.**>; [ as of (07/02) ]

Double quotes take away the special meaning of all characters except the following −

• $ for parameter substitution.

• Backquotes for command substitution.

• \$ to enable literal dollar signs.

• \` to enable literal backquotes.

• \" to enable embedded double quotes.

• \\ to enable embedded backslashes.

• All other \ characters are literal (not special).

Any characters within single quotes are quoted just as if a backslash is in front of each character. So now this echo command displays properly.

If a single quote appears within a string to be output, you should not put the whole string within single quotes instead you whould preceed that using a backslash (\) as follows −

echo 'It\'s Shell Programming'

The Back Quotes

Putting any Shell command in between back quotes would execute the command

Syntax:

Here is the simple syntax to put any Shell command in between back quotes −

Example

var=`command`

Example

Following would execute date command and produced result would be stored in DATA variable.

DATE=`date`

echo "Current Date: $DATE"

This would produce following result −

Current Date: Thu Jul  2 05:28:45 MST 2009