N methods for shell to determine a variable as a number

Method 1: Use expr to calculate the sum of the variable and an integer. If normal execution is performed, the variable is an integer. Otherwise, an error occurs. $? Replace a non-0 value expr $ args + 0 & amp; & gt;/dev/null. Method 2: print the variable by replacing it with sed. Replace the number in the variable with null, if the variable is null after replacement, the value is an integer echo $ args | sed & #39; s/[0-9] // g & #39; if a negative number is determined, use sed to filter the negative number echo $ args | s

Method 1: Use expr to calculate the sum of the variable and an integer. If the value is normal, it is an integer. Otherwise, an error occurs. $? The value is not 0.

Expr $ args + 0 &>/dev/null

Method 2: The print variable is replaced by sed, and the number in the variable is replaced by null. If the variable is empty after replacement, it is an integer.

Echo $ args | sed's/[0-9] // G'

If a negative number is determined, use sed to filter the negative number.

Echo $ args | sed's/[0-9] // G' | sed's/-// G'

The following script uses two functions to determine the value. The code is very simple, so no comments are added.

#!/bin/bashusage(){cat <
 
  /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]&&usageargs1=$1args2=$2checkInt $args1checkInt1 $args2if[ $args1 -gt $args2 ];thenecho "yes,$args1 greate than $args2"elseecho "no,$args1 less than $args2"fi
 

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