Javascript 如何使正则表达式变成非贪婪的？
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How to make Regular expression into non-greedy?
提问by Rueta
I'm using jQuery. I have a string with a block of special characters (begin and end). I want get the text from that special characters block. I used a regular expression object for in-string finding. But how can I tell jQuery to find multiple results when have two special character or more?
我正在使用 jQuery。我有一个带有特殊字符块（开始和结束）的字符串。我想从那个特殊字符块中获取文本。我使用正则表达式对象进行字符串内查找。但是，当有两个或更多特殊字符时，如何告诉 jQuery 查找多个结果？
My HTML:
我的 HTML：
<div id="container">
<div id="textcontainer">
Cu?c chi?n pháp ly gi?a [|c? th?|nghi?m|] th? tr??ng [|test2|?ay là test l?n 2|] ch?ng khoán [|M?|day la nuoc my|] và ngan hàng ??u t? quy?n l?c nh?t Ph? Wall m?i ch? b?t ??u.
</div>
</div>
and my JavaScript code:
和我的 JavaScript 代码：
$(document).ready(function() {
var takedata = $("#textcontainer").text();
var test = 'abcd adddb';
var filterdata = takedata.match(/(\[.+\])/);
alert(filterdata);
//end write js
});
My result is: [|c? th?|nghi?m|] th? tr??ng [|test2|?ay là test l?n 2|] ch?ng khoán [|M?|day la nuoc my|]. But this isn't the result I want :(. How to get [text] for times 1 and [demo] for times 2 ?
我的结果是：[|c? th?|nghi?m|]th? tr??ng [|test2|?ay là test l?n 2|] ch?ng khoán [|M?|day la nuoc my|]. 但这不是我想要的结果:(。如何获得第 1 次的 [文本] 和第 2 次的 [演示]？
I've just done my work after searching info on internet ^^. I make code like this:
我刚刚在互联网上搜索信息后完成了我的工作^^。我制作这样的代码：
var filterdata = takedata.match(/(\[.*?\])/g);
- my result is : [|c? th?|nghi?m|],[|test2|?ay là test l?n 2|]this is right!. but I don't really understand this. Can you answer my why?
- 我的结果是：[|c? th?|nghi?m|],[|test2|?ay là test l?n 2|]没错！但我真的不明白这一点。你能回答我的为什么吗？
回答by Asaph
The non-greedy regex modifiers are like their greedy counter-parts but with a ?immediately following them:
非贪婪的正则表达式修饰符就像它们贪婪的对应部分，但?紧跟在它们之后：
* - zero or more
*? - zero or more (non-greedy)
+ - one or more
+? - one or more (non-greedy)
? - zero or one
?? - zero or one (non-greedy)
回答by polygenelubricants
You are right that greediness is an issue:
你是对的，贪婪是一个问题：
--A--Z--A--Z--
^^^^^^^^^^
A.*Z
If you want to match both A--Z, you'd have to use A.*?Z(the ?makes the *"reluctant", or lazy).
如果您想同时匹配两者A--Z，则必须使用A.*?Z（?使*“不情愿”或懒惰）。
There are sometimes better ways to do this, though, e.g.
不过，有时有更好的方法来做到这一点，例如
A[^Z]*+Z
This uses negated character class and possessive quantifier, to reduce backtracking, and is likely to be more efficient.
这使用否定字符类和所有格量词来减少回溯，并且可能更有效。
In your case, the regex would be:
在您的情况下，正则表达式将是：
/(\[[^\]]++\])/
UnfortunatelyJavascript regex doesn't support possessive quantifier, so you'd just have to do with:
不幸的是Javascript 正则表达式不支持所有格量词，所以你只需要做：
/(\[[^\]]+\])/
See also
也可以看看
- regular-expressions.info/Repetition
- See: An Alternative to Laziness
- Flavors comparison
Quick summary
快速总结
* Zero or more, greedy
*? Zero or more, reluctant
*+ Zero or more, possessive
+ One or more, greedy
+? One or more, reluctant
++ One or more, possessive
? Zero or one, greedy
?? Zero or one, reluctant
?+ Zero or one, possessive
Note that the reluctant and possessive quantifiers are also applicable to the finite repetition {n,m}constructs.
请注意，不情愿和所有格量词也适用于有限重复{n,m}结构。
Examples in Java:
Java 中的示例：
System.out.println("aAoZbAoZc".replaceAll("A.*Z", "!")); // prints "a!c"
System.out.println("aAoZbAoZc".replaceAll("A.*?Z", "!")); // prints "a!b!c"
System.out.println("xxxxxx".replaceAll("x{3,5}", "Y")); // prints "Yx"
System.out.println("xxxxxx".replaceAll("x{3,5}?", "Y")); // prints "YY"
回答by iangraham
I believe it would be like this
我相信会是这样
takedata.match(/(\[.+\])/g);
the gat the end means global, so it doesn't stop at the first match.
在g末意味着全球性的，所以它不会在第一场比赛停止。
