bash 正则表达式在 if 语句中匹配带有空格的字符串（使用引号？）
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Regex match a string with spaces (use quotes?) in an if statement
提问by Mint
How would I do a regex match as shown below but with quotes around the ("^This") as in the real world "This" will be a string that can have spaces in it.
我将如何进行如下所示的正则表达式匹配，但在 ("^This") 周围加上引号，就像在现实世界中一样，"This" 将是一个可以包含空格的字符串。
#!/bin/bash
text="This is just a test string"
if [[ "$text" =~ ^This ]]; then
echo "matched"
else
echo "not matched"
fi
I want to do something like
我想做类似的事情
if [[ "$text" =~ "^This is" ]]; then
but this doesn't match.
但这不匹配。
回答by too much php
You can use \before spaces.
您可以\在空格之前使用。
#!/bin/bash
text="This is just a test string"
if [[ "$text" =~ ^This\ is\ just ]]; then
echo "matched"
else
echo "not matched"
fi
回答by UlfR
I did not manage to inline the expression like this:
我没有设法像这样内联表达式：
if [[ "$text" =~ "^ *This " ]]; then
but if you put the expression in a variable you could use normal regex syntax like this:
但是如果你把表达式放在一个变量中，你可以使用像这样的普通正则表达式语法：
pat="^ *This "
if [[ $text =~ $pat ]]; then
Note that the quoting on $textand $patis unnessesary.
请注意，引用$text和$pat是不必要的。
Edit:A convinient oneliner during the development:
编辑：开发过程中方便的oneliner：
pat="^ *This is "; [[ " This is just a test string" =~ $pat ]]; echo $?
回答by ghostdog74
can you make your problem description clearer?
你能把你的问题描述更清楚吗？
text="This is just a test string"
case "$text" in
"This is"*) echo "match";;
esac
the above assume you want to match "This is" at exactly start of line.
以上假设您想在行首匹配“This is”。
回答by Kshitij Saxena -KJ-
Have you tried:
你有没有尝试过：
^[\s]*This
