bash 将变量传递给 grep 命令
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Pass variable to grep command
提问by MSK
I want to pass a variable in my grepcommand in Linux bash script. Variable is a text file from Internet and i want to find some words in it.
我想grep在 Linux bash 脚本的命令中传递一个变量。变量是来自 Internet 的文本文件，我想在其中查找一些单词。
I have tried the following command in my bash:
我在我的 bash 中尝试了以下命令：
cat "$var" | grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)'echo "$var" | grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)'grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)' <<< "$var"
cat "$var" | grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)'echo "$var" | grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)'grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)' <<< "$var"
but i dont get a right Result. How can i do it?
但我没有得到正确的结果。我该怎么做？
Here is my bash:
这是我的 bash：
#!/bin/sh
urltext=$(curl -s https://example.com)
string=$(grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)' "$urltext" | tr '.' '\n' )
cat $string
回答by Idriss Neumann
What's supposed to be contained in the variable ?
变量中应该包含什么？
It's a file ?
是文件吗？
grep <options> <expression> "$var"
It's a string ?
是字符串吗？
echo "$var"|grep <options> <expression>
grep <options> <expression> <(echo "$var")
NB : try -eoption if there are several lines in $var
注意：-e如果 $var 中有几行，请尝试选项
