正则如何匹配某个以字符开始剩下的所有内容? - V2EX
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Air_Mu

正则如何匹配某个以字符开始剩下的所有内容?

  •  
  •   Air_Mu Jul 12, 2015 5016 views
    This topic created in 3942 days ago, the information mentioned may be changed or developed.
    想弄个油猴脚本解决一些问题,需要用到正则,GOOGLE第一个正则30分钟入门的网站研究了一下。

    我想匹配内容中第一个“_”或者“-”开始之后的所有内容
    试过这样 RegExp("(_.*|\-.*)") 勉强成功,但如果后面的内容出现换行符就不行了,换行后的内容就无法匹配。 求解
    10 replies    2015-07-12 19:08:47 +08:00
    imn1
        1
    imn1  
       Jul 12, 2015
    (?:[_-])[\s\S]*
    yangg
        2
    yangg  
       Jul 12, 2015
    (str.match(/[_\-]([\s\S]*)/) || [0, null)[1]
    Biwood
        3
    Biwood  
       Jul 12, 2015
    (?:[^_-]).*
    q84629462
        4
    q84629462  
       Jul 12, 2015 via Android
    需要匹配多行内容不是添加一个flag的事?
    Biwood
        5
    Biwood  
       Jul 12, 2015   1
    要匹配多行的话就把 . 换成[\s\S],应该是这样 (?:[^_-])[\s\S]*
    Air_Mu
        6
    Air_Mu  
    OP
       Jul 12, 2015
    @Biwood 不行啊 只匹配到"-"本身
    ditsing
        7
    ditsing  
       Jul 12, 2015
    @Biwood 这个表情不错拿走了....
    Biwood
        8
    Biwood  
       Jul 12, 2015   1
    @Air_Mu 亲测可行,如果用 new RegExp 的方式创建正则对象的话,记得先给这段字符转义, escapeRegExp 你明白的吧

    Air_Mu
        9
    Air_Mu  
    OP
       Jul 12, 2015
    @Biwood 非常感谢 不用new RegExp就成功了,困扰了我好久。
    zjuster
        10
    zjuster  
       Jul 12, 2015
    .是不包括换行符的,所以要换个思路。
    \s是所有空格,\S是所有非空格,那么使用[\s\S]*应该可以了。
    学习了。
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