bash 如何将grep输出中的值存储在变量中
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How to store value from grep output in variable
提问by rut2
I am working on one bash script, in which I have to use the regular expression to match string and then store the output in a variable to reuse it.
我正在编写一个 bash 脚本，其中我必须使用正则表达式来匹配字符串，然后将输出存储在一个变量中以重用它。
here is my script,
这是我的脚本，
#!/bin/sh
NAME="MET-3-get-code-from-string"
por="$($NAME | grep -P -o -e '(?<=MET-).*?(\d+)')" #this should store 3 in variable por
echo $por
I tried this many ways, but I am getting error :
./check.sh: MET-3-get-issue-id-from-branch-name: not found
我尝试了很多方法，但出现错误：
./check.sh: MET-3-get-issue-id-from-branch-name: not found
if I run individual grep command then yes, it is working properly. But I am not able to store output.
如果我运行单独的 grep 命令，那么是的，它工作正常。但我无法存储输出。
I also tried :
我也试过：
por=$($NAME | grep -P -o -e '(?<=MET-).*?(\d+)')
por=$NAME | grep -P -o -e '(?<=MET-).*?(\d+)'
and many other similar references.
以及许多其他类似的参考资料。
but it's not working. can anyone please help me on this. I have not much experience in bash.
但它不起作用。任何人都可以帮我解决这个问题。我在 bash 方面没有太多经验。
thank you.
谢谢你。
采纳答案by Arjun Mathew Dan
Change
改变
por="$($NAME | grep -P -o -e '(?<=MET-).*?(\d+)')"
to
到
por="$(echo "$NAME" | grep -P -o -e '(?<=MET-).*?(\d+)')"
Also, you are missing a closing double quote (maybe just a typo, should be NAME="MET-3-get-code-from-string")
此外，您缺少一个结束双引号（可能只是一个错字，应该是NAME="MET-3-get-code-from-string"）
