Shell 脚本变量中未捕获计数

Question

我试图通过运行命令捕获活动进程的数量，并试图在 shell 脚本的变量中捕获结果，但不幸的是，没有捕获到任何东西。代码如下：

#!/bin/ksh

## Checking whether or not the Previous Build is Completed

count_build_status=`ps -ef | grep BDD_PreCheck.sh | grep -c FT_BGmgmt` | tee -a ${logFile}

    echo "The Value of Count Build Status is $count_build_status"

        if [[ "${count_build_status}" != "0" ]]

        then

            echo INFO -  The previous build has not ended yet. Please Wait for some time or contact the Administrator | tee -a $logFile

        exit 1;

        fi

exit 0;

这里，ps -ef | grep BDD_PreCheck.sh | grep -c FT_BGmgmt单独执行结果为0，但'count_build_status'中存储的值为null。

有人可以帮忙吗？

Answer 1

示例：将 cmd 行的输出捕获到变量并附加带有 tee 的日志文件：

var=`ps -ef | grep 0 | grep -c 1 | tee -a log`
echo $var

Answer 2

这里的问题是反引号，如果没有在反引号内完成发球，它不会为变量赋值。它应该包含在 logFile 的末尾，如下面的代码所示。

#!/bin/bash

logFile=log.txt

## Checking whether or not the Previous Build is Completed

count_build_status=`ps -ef | grep BDD_PreCheck.sh | grep -c FT_BGmgmt | tee -a ${logFile}`

    echo "The Value of Count Build Status is $count_build_status"

        if [[ "${count_build_status}" != "0" ]]

        then

            echo INFO -  The previous build has not ended yet. Please Wait for some time or contact the Administrator | tee -a $logFile

        exit 1;

        fi

exit 0;

Shell 脚本变量中未捕获计数

Count is not Captured in Shell Script Variable

unix

variables

bash

shell

sh