如果字符串没有后跟字母数字[重复],如何为字符串grep

时间:2022-02-05 18:47:42

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这个问题在这里已有答案:

I need the below requirement

我需要以下要求

string123

stringsecond

string-third

-string-four

string_five

/string/

sixstring

this should not print

output

string-third

-string-four

string_five

/string/

So as per the output,the command should not grep string followed with alphanumeric only. I am using -w option with grep ,but it is eliminating underscore(_) strings also.

因此,根据输出,命令不应该是grep字符串,后面只有字母数字。我在grep中使用-w选项,但它也消除了下划线(_)字符串。

1 个解决方案

#1


0  

Use a negated character class at the end of your regex.

在正则表达式的末尾使用否定的字符类。

If you're matching "string", then append a class like [^0-9a-zA-Z_] to the end.

如果您匹配“字符串”,则将类似[^ 0-9a-zA-Z_]的类附加到末尾。

#1


0  

Use a negated character class at the end of your regex.

在正则表达式的末尾使用否定的字符类。

If you're matching "string", then append a class like [^0-9a-zA-Z_] to the end.

如果您匹配“字符串”,则将类似[^ 0-9a-zA-Z_]的类附加到末尾。