使用bash正则表达式匹配行 [英] match a line using bash regex

问题描述

我想匹配包含一个单词但没有分号的行

I want to match a line that contains a word, but does not have semi-colon in it

这应该匹配:

class test

这不应该匹配

class test;

这也不应该匹配

class test; // test class

这是我期望的工作，但事实并非如此:

this is what I was expecting to work, but it doesn't:

pattern="class [^;]*"

if [[ $line =~ $pattern ]]

谢谢

推荐答案

您的正则表达式不是 anchored ，这意味着[^;]*仍将与所有字符匹配，直到可能的;(并因此整体匹配).如果将正则表达式锚定到行尾([^;]*$)，它将产生您想要的结果:

Your regular expression is not anchored which means that [^;]* will still match against all characters up to a possible ; (and thus match as a whole). If you anchor the regex against the end of the line ([^;]*$) it will produce the results you are after:

$ cat t.sh
#!/bin/bash

pattern='class [^;]*$'
while read -r line; do
    printf "testing '${line}': "
    [[ $line =~ $pattern ]] && echo matches || echo "doesn't match"
done <<EOT
class test
class test;
class test; // test class
EOT

---

$ ./t.sh
testing 'class test': matches
testing 'class test;': doesn't match
testing 'class test; // test class': doesn't match

TL; DR :换句话说，

班级考试； foo bar quux

class test; foo bar quux

即使字符串包含分号，

也会匹配您的正则表达式，这就是为什么它总是匹配的原因.锚确保只有在字符串的末尾没有分号 时，正则表达式才匹配.

matches your regex even though the string contains a semicolon which is why it always matches. The anchor makes sure that the regular expression only matches if there is no semicolon until the very end of the string.

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