如何到grep在shell程序中的变量？ [英] how to grep a variable in the shell program?

问题描述

#!/bin/bash

for ((var=0; var<20; var++))
do
echo " Number is: $(grep 'Multiple_Frame = echo **$var**'  20mrf.txt | wc -l)" >>statisic.txt 

done

这shell程序不能产生正确的结果，这可能错误变量的原因在第二个grep命令返回。

This shell program cannot produce correct result which maybe the reason of wrong variable returning in the second grep command.

我怎么可以grep第二回声句子中的变量？根据无功变化给grep不同的东西？

How can I grep a variable within the second echo sentence? to grep different things according to the var changing?

非常感谢！

推荐答案

正如其他人所指出的，问题是变量的单引号prevent扩张。然而，使用 $（）允许您使用双引号：

As others have stated, the problem is that the single quotes prevent expansion of the variable. However, using $() allows you to use double quotes:

echo " Number is: $(grep "Multiple_Frame = echo **$var**"  20mrf.txt | wc -l)" >>statisic.txt

虽然我怀疑这样的事情是你的意思：

although I suspect something like this is what you meant:

echo " Number is: $(grep "Multiple_Frame = $var"  20mrf.txt | wc -l)" >>statisic.txt

您也应该知道，的grep 有一个选项可以输出在计数这样你就可以省略厕所：

You should also be aware that grep has an option to output the count so you can omit wc:

echo " Number is: $(grep -c "Multiple_Frame = $var"  20mrf.txt)" >>statisic.txt

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