I have come across the following regular expression in a piece of client side javascript:
我在一段客户端javascript中遇到了以下正则表达式:
([^?=&]+)(=([^&]*))?
When I run it through a regular expression tester (firefox add-on), I get the values of the query string of a url highlighted in one colour, the question mark and ampersands not highlighted at all and the rest in another colour.
当我在一个正则表达式测试器(firefox扩展)中运行它时，我得到一个用一种颜色突出显示的url的查询字符串的值，问号和&完全没有突出显示，其余的则用另一种颜色。
I'm not sure whether it is matching one colour or the other or both, but then I use the replace option and nothing gets replaced.
我不确定它是匹配一种颜色还是另一种颜色还是两者都匹配，但是我使用了替换选项，没有任何东西被替换。
That's only the beginning of my question. The piece of code in full is this:
这只是我问题的开始。完整的代码是:
var linkObj = new Object();
jQuery(this).attr('href').replace(
new RegExp( "([^?=&]+)(=([^&]*))?", "g" ),
function( jQuery0, jQuery1, jQuery2, jQuery3 ){
linkObj[ jQuery1 ] = jQuery3;
}
);
What are those 4 parameters in the function (jQuery0, jQuery1, jQuery2, jQuery3)?
函数中的这4个参数是什么(jQuery0、jqueryy1、jQuery2、jQuery3)?
Any help would be appreciated.
如有任何帮助，我们将不胜感激。
Thank you.
谢谢你！
1 个解决方案
#1
3
Appears to break down the GET portion of a URL.
似乎要分解URL的GET部分。
([^?=&]+) # 1 or more characters, exclusive of ?, = or &
(= # = sign
([^&]*) # any character not a &, 0 or more times
)? # but optional
The function is a callback which replace can accept with regular expression calls. Each argument is a different value found within the regex match. The function can then return the result that the replace method should substitute with). So, given ?foo=bar:
函数是一个回调，它可以用正则表达式调用来代替。每个参数都是regex匹配中发现的不同值。然后，函数可以返回替换方法应该替换为)的结果。所以,鉴于? foo = bar:
- jQuery0 - full match
- jQuery0——完全匹配
- jQuery1 - First capture (e.g.
foo) - jQuery1 -第一次捕获(例如foo)
- jQuery2 - Second capture (
=bar) - jQuery2 -第二次捕获(=bar)
- jQuery3 - Third (nested) capture (
bar) - jQuery3 -第三(嵌套)捕获(bar)
#1
3
Appears to break down the GET portion of a URL.
似乎要分解URL的GET部分。
([^?=&]+) # 1 or more characters, exclusive of ?, = or &
(= # = sign
([^&]*) # any character not a &, 0 or more times
)? # but optional
The function is a callback which replace can accept with regular expression calls. Each argument is a different value found within the regex match. The function can then return the result that the replace method should substitute with). So, given ?foo=bar:
函数是一个回调，它可以用正则表达式调用来代替。每个参数都是regex匹配中发现的不同值。然后，函数可以返回替换方法应该替换为)的结果。所以,鉴于? foo = bar:
- jQuery0 - full match
- jQuery0——完全匹配
- jQuery1 - First capture (e.g.
foo) - jQuery1 -第一次捕获(例如foo)
- jQuery2 - Second capture (
=bar) - jQuery2 -第二次捕获(=bar)
- jQuery3 - Third (nested) capture (
bar) - jQuery3 -第三(嵌套)捕获(bar)