bash 如何在变量中 grep 复杂的字符串？
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How can I grep complex strings in variables?
提问by Craig
I am trying to grep for a small string in a much larger string. Both strings are being stored as variables and here is a code example:
我试图在一个更大的字符串中搜索一个小字符串。两个字符串都被存储为变量，这里是一个代码示例：
#!/bin/bash
long_str=$(man man)
shrt_str="guide"
if grep -q $shrt_str $long_str ; then
echo "Found it!"
fi
I don't think variable expansion is working the way I expect it to. I have tried [ ]and [[ ]], also quoting the variables and piping the output to /dev/nullbut no matter what I do it won't work.
我认为变量扩展并没有像我期望的那样工作。我已经尝试过，[ ]并且[[ ]]还引用了变量并将输出管道传输到/dev/null但无论我做什么都行不通。
Does anyone have any ideas?
有没有人有任何想法？
回答by ghostdog74
if echo "$long_str" | grep -q "$shrt_str";then
echo "found"
fi
or
或者
echo "$long_str" | grep -q "$shrt_str" && echo "found" || echo "not found"
But since you are using bash shell, then use shell internals. No need to call external commands
但是既然您使用的是 bash shell，那么请使用 shell 内部结构。无需调用外部命令
shrt_str="guide"
case "$long_str" in
*"$shrt_str"* ) echo "Found";;
* ) echo "Not found";;
esac
回答by Ignacio Vazquez-Abrams
grep is for files or stdin. If you want to use a variable as stdin then you need to use bash's herestring notation:
grep 用于文件或标准输入。如果你想使用一个变量作为标准输入，那么你需要使用 bash 的 herestring 表示法：
if grep -q "$shrt_str" <<< "$long_str" ; then
回答by Alok Singhal
You want
你要
if echo $long_str | grep -q $shrt_str; then
回答by Paused until further notice.
Another Bash-specific technique:
另一种 Bash 特定的技术：
if [[ $long =~ $short ]] # regex match
then
echo "yes"
fi
But if you don't need the long string in a variable:
但是，如果您不需要变量中的长字符串：
if man man | grep $short; then ...
but I'm assuming that was just for the purpose of having an example.
但我假设这只是为了举个例子。
