这是什么意思[：-ge：意外的运营商，在bash？ [英] What does this mean [: -ge: unexpected operator, in bash?

问题描述

我很新的shell脚本和学习它。我得到了code监测的Apache。该文件的名称是 test.sh 。我改变了code位。

I'm very new to shell scripting and learning it. I got the code for monitoring apache. The name of the file is test.sh. I changed the code a bit.

我一直在寻找，当我做的：

What I was looking for is, when I do :

./test.sh -H localhost -wr 2 -cr 5 -arg cpu_load

这应该测试其 cpu_load 的Apache。即我试图控制监视我的 -arg 参数的Apache。

It should test apache for its cpu_load. i.e I tried to control monitoring apache with my -arg parameter.

但是，这似乎并不奏效。

But that doesn't seem to be working.

当我运行这样的：

./test.sh -H localhost -wr 2 -cr 5 -arg cpu_load

我得到的错误：

./test.sh: 282: [: -ge: unexpected operator
./test.sh: 286: [: -ge: unexpected operator

下面是code的某些部分：

Here is some part of the code :

#!/bin/sh


while test -n "$1"; do
    case "$1" in
        --help|-h)
            print_help
            exit $ST_UK
            ;;
        --version|-v)
            print_version $PROGNAME $VERSION
            exit $ST_UK
            ;;
        --hostname|-H)
            hostname=$2
            shift
            ;;
        --port|-P)
            port=$2
            shift
            ;;
        --timeout|-t)
            timeout=$2
            shift
            ;;
        --remote-server|-R)
            remote_srv=1
            ;;
        --binary_path|-b)
            path_binary=$2
            shift
            ;;
        --pid_path|-p)
            path_pid=$2
            shift
            ;;
        --pid_name|-n)
            name_pid=$2
            shift
            ;;
        --status-page|-s)
            status_page=$2
            shift
            ;;
        --secure|-S)
            secure=1
            ;;
        --warning-req|-wr)
            warn_req=$2
            shift
            ;;
        --critical-req|-cr)
            crit_req=$2
            shift
            ;;
    --userargument|-arg)
       user_arg=$3
       shift
           ;;
        *)
            echo "Unknown argument: $1"
            print_help
            exit $ST_UK
            ;;
    esac
    shift
done

#other codes

    if [ ${wclvls_req} = 1 ]
    then
        if [ ${user_arg} -ge ${warn_req} -a ${user_arg} -lt ${crit_req} ]
        then
            echo "WARNING - ${output} | ${perfdata}"
            exit $ST_WR
        elif [ ${user_arg} -ge ${crit_req} ]
        then
            echo "CRITICAL - ${output} | ${perfdata}"
        exit $ST_CR
        else
            echo "OK - ${output} | ${perfdata}"
            exit $ST_OK
        fi
    else
        echo "OK - ${output} | ${perfdata}"
        exit $ST_OK
    fi

fi

如果我做了错误的？

Where I'm making the mistake?

先谢谢了。

推荐答案

你的一个变量（ user_arg ， warn_req 等）在如果条件可能是空的。

One of your variables ( user_arg, warn_req etc ) in the if condition might be empty.

更好的编写方式是引用了变量（如果你想比为整数可能你的情况不及格）：

Better way to write that is with quoting the variables as (which may fail in your case if you want to compare as integers):

if [ "${user_arg}" -ge "${warn_req}" -a "${user_arg}" -lt "${crit_req}" ]

或者另一种方式是指定的默认​​值，所以如果变量为空或未定义，如果不如下失败。

Or another way is to specify the default values so that if variable is null or undefined if won't fail as below.

if [ ${user_arg:-0} -ge ${warn_req:-0} -a ${user_arg:-0} -lt ${crit_req:-0} ]

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