3 ways to judge variables as numbers in a shell script _linux shell

Method 1: Add a variable to an integer value by using expr, or an integer if it is performed correctly, or an error is executed, $? will be a value that is non 0

Copy Code code as follows:

Expr $args + 0 &>/dev/null

Method 2: Print variables replaced by SED in the way that the number in the variable is NULL, if the variable is empty after the substitution, then integer

Copy Code code as follows:

echo $args | Sed ' s/[0-9]//g '

If you judge a negative number, then filter the minus with SED.

Copy Code code as follows:

echo $args | Sed ' s/[0-9]//g ' | Sed ' s/-//g '

The following script uses two functions to achieve numerical judgment. The code is simple, without comment.

Copy Code code as follows:

#!/bin/bash
Usage () {
Cat <<eof
Useage:sh $0ARGS1 ARGS2
Exit 1
Eof
}
Checkint () {
Expr $1+ 0&>/dev/null
[$-ne 0] && {echo "Args must be integer!"; Exit 1; }
}
CheckInt1 () {
Tmp= ' echo $1|sed ' s/[0-9]//g '
[-N ' ${tmp} ']&& {echo ' Args must be integer! '; Exit 1; }
}
[$#-ne 2]&&usage
Args1=$1
Args2=$2
Checkint $args 1
CheckInt1 $args 2
if[$args 1-gt $args 2];then
echo "Yes, $args 1 greate than $args 2"
Else
echo "No, $args 1 less than $args 2"
Fi