Question

这是我到目前为止所做的：/(^|[\s])#\d+/g

我的测试字符串为："#123 it should match this: #1234, but not this: http://example.org/derp#6326 . How about on a new line?\n\n#1284"

当我尝试匹配时，我得到以下匹配：

"#123"
" #1234"
"\n#1284"（假装那里有一个实际的换行符）

我尝试通过将?:添加到分组来更改正则表达式，并使用括号括起我想要的内容：/(?:^|[\s])(#\d+)/g但是，这不起作用，并提供相同的匹配

如何才能匹配# +号码，之前没有任何内容？

Answer 1

实际上你确实抓住了你想要的东西，你只需要看看捕捉小组内部的内容，而不是整场比赛......

尝试

var myString = "#123 it should match this: #1234, but not this: http://example.org/derp#6326 . How about on a new line?\n\n#1284";
var myRegex = /(?:^|\s)(#\d+)/g;
var allMatches = [];
while((result = myRegex.exec(myString)) != null) {
    var match = result[1]; // get the first capturing group of the current match
    allMatches.push(match);
}

您可以清楚地看到正则表达式捕获的内容here

Answer 2

内存捕获可以解决问题：

var re = /(?:^|[\s])(#\d+)/g;

while (match = re.exec(str)) {
    console.log(match[1]);
}

Demo

Answer 3

subject= "#1234, but not this: http://example.org/derp#6326";
match = subject.match(/^\s*?#(\d+)/m);
if (match != null) {
    var result =  match[1]
}

Assert position at the beginning of a line (at beginning of the string or after a line break character) «^»
Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s*?»
   Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “#” literally «#»
Match the regular expression below and capture its match into backreference number 1 «(\d+)»
   Match a single digit 0..9 «\d+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»

正则表达匹配：被空间或起点所包围，但不匹配

3 个答案: