What are the ways to do numeric testing in bash?

Suppose I have variable "x" in bash. How can I test if it's some number?

I tried if x=5; then echo "it is 5"; fi but that doesn't work,

then I tried if x==5; then echo "it is 5"; fi but that also doesn't work,

then I tried if [x==5]; then echo "it is 5"; fi but that also doesn't work,

then I tried if [[x==5]]; then echo "it is 5"; fi but th开发者_JAVA百科at also doesn't work,

then I tried if [[x=5]]; then echo "it is 5"; fi but that also doesn't work,

then I tried if [x=5]; then echo "it is 5"; fi but that also doesn't work,

then I tried if [x -eq 5]; then echo "it is 5"; fi but that also doesn't work,

then I tried if [[x -eq 5]]; then echo "it is 5"; fi and it finally worked.

I am lost. Bash seems to be hell. Why so many ways didn't work? And only the last one did?

Can't bash adopt if x==5; then ...; fi syntax?

Thanks, Boda Cydo

Try:

if [ "$x" = 5 ]; then echo "it is 5"; fi

Note that [ is a command in itself; you may find that it's symlinked to test on your system. So, man test tells you how to use it. In particular, it is important to put a space after it.

Bash cannot adopt the "if x==5" syntax to have the semantics that you want because it is already valid syntax with a different meaning. 'if x==5; then ...; fi' currently sets x to the value '=5' and then executes the commands following then.

In terms of your first question, I advise avoiding the bracket notation completely. Just do

if test "$x" = 5; then
...
fi

Also note that 'if test x = 5' will always be false, since this is a comparison of the literal string 'x' and the literal string '5'. You need to use $x to get the value of the variable.

Variables in bash should have a $ before them - if [${x} -eq 5] should have worked too.

sh syntax is actually amazingly flexible, and it is quite possible to get the syntax that you want. For example, if you want to be able to write "if x == 5", you just need to realize that sh will evaluate that string by invoking a command named 'x' with arguments '==' and '5', and react to the value returned by that function. The following is probably not a good idea for any sort of production setting, but it may be instructive to understand how this works:

#!/bin/sh

declare() {
    eval $1'() {
    case $# in
        0) test $'$1';;
        1) false;;
        *) case $1 in
            =) shift; 
            '$1'=$(expr $*);;
            ==) test "$'$1'" = $2;;
            .=) '$1'="$'$1'$2";;
            +=) '$1'=$(expr $'$1' + $2);;
            -=) '$1'=$(expr $'$1' - $2);;
            *) false;;
        esac
    esac
    }'
}

declare x
x = 5
if x == 4; then echo not printed; fi  
if x == 5; then echo x = $x; fi  # "x = 5"
x += 2;
if x == 5; then echo not printed; fi  
if x == 7; then echo x = $x; fi  # "x = 7"
x = foo
x .= bar
echo x = $x                      # "x = foobar"
x = 4 + 5
echo x = $x                      # "x = 9"

The output of the above script is:

x = 5
x = 7
x = foobar
x = 9

You might find chapter 7.1. Test Constructs in the Advanced Bash-Scripting Guide helpful.

Can't bash adopt if x==5; then ...; fi syntax?

In Bash, you can use this syntax which allows you to omit the dollar sign and avoid using -eq, so it comes closest to what you'd like to use:

if (( x == 5 ))

You can even leave out the spaces if you prefer:

if ((x==5))

you can use case/esac

case "$x" in
 5) echo "x is 5";;
esac