javascript 如何替换所有但字符串中第一次出现的模式
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How to replace all BUT the first occurrence of a pattern in string
提问by dr jerry
quick question: my pattern is an svg string and it looks like l 5 0 l 0 10 l -5 0 l 0 -10To do some unittest comparison against a reference I need to ditch all but the first lI know i can ditch them all and put an 'l' upfront, or I can use substrings. But I'm wondering is there a javascript regexp idiom for this?
快速问题：我的模式是一个 svg 字符串，它看起来像l 5 0 l 0 10 l -5 0 l 0 -10要对参考进行一些单元测试比较，我需要l放弃所有，但首先我知道我可以放弃它们并在前面放置一个 'l'，或者我可以使用子字符串。但我想知道是否有一个 javascript regexp 成语？
回答by Mark Byers
回答by Rob W
There's no JS RegExp to replace everything-but-the-first-pattern-match. You can, however, implement this behaviour by passing a function as a second argument to the replacemethod.
没有 JS RegExp 可以替换除第一个模式匹配之外的所有内容。但是，您可以通过将函数作为第二个参数传递给replace方法来实现此行为。
var regexp = /(foo bar )(red)/g; //Example
var string = "somethingfoo bar red foo bar red red pink foo bar red red";
var first = true;
//The arguments of the function are similar to "l 5 0 l 0 10 l -5 0 l 0 -10".replace(/^\s+/, '').replace(/\s+l/g, '')
etc
var fn_replaceBy = function(match, group1, group2){ //group in accordance with RE
if (first) {
first = false;
return match;
}
// Else, deal with RegExp, for example:
return group1 + group2.toUpperCase();
}
string = string.replace(regexp, fn_replaceBy);
//equals string = "something foo bar red foo bar RED red pink foo bar RED red"
The function (fn_replaceBy) is executed for each match. At the first match, the function immediately returns with the matched string (nothing happens), and a flag is set.
Every other match will be replaced according to the logic as described in the function: Normally, you use $0 $1 $2, et cetera, to refer back to groups. In fn_replaceBy, the function arguments equal these: First argument = $0, second argument = $1, et cetera.
fn_replaceBy为每个匹配执行函数 ( )。在第一次匹配时，该函数立即返回匹配的字符串（什么也没发生），并设置一个标志。
将根据函数中描述的逻辑替换所有其他匹配项：通常，您使用$0 $1 $2等来引用组。在 中fn_replaceBy，函数参数等于这些：第一个参数 = $0，第二个参数 =$1等等。
The matched substring will be replaced by the return value of function fn_replaceBy. Using a function as a second parameter for replaceallows very powerful applcations, such as an intelligent HTML parser.
匹配的子字符串将被 function 的返回值替换fn_replaceBy。使用函数作为第二个参数replace允许非常强大的应用程序，例如智能 HTML 解析器。
See also: MDN: String.replace > Specifying a function as a parameter
回答by Mike Samuel
'-98324792u4234jkdfhk.sj.dh-f01' // construct valid float
.replace(/[^\d\.-]/g, '') // first, remove all characters that aren't common
.replace(/(?!^)-/g, '') // replace negative characters that aren't in beginning
.replace('.', '%FD%') // replace first occurrence of decimal point (placeholder)
.replace(/\./g, '') // now replace all but first occurrence (refer to above)
.replace(/%FD%(0+)?$/, '') // remove placeholder if not necessary at end of string
.replace('%FD%', '.') // otherwise, replace placeholder with period
makes sure the first 'l'is not preceded by space and removes any space followed by an 'l'.
确保第一个'l'前面没有空格，并删除后面跟有'l'.
回答by Erutan409
It's not the prettiest solution, but you could replace the first occurrence with something arbitrary (like a placeholder) and chain replacements to fulfill the rest of the logic:
这不是最漂亮的解决方案，但您可以用任意的东西（如占位符）和链替换来替换第一次出现，以实现其余的逻辑：
##代码##Produces:
产生：
-983247924234.01
-983247924234.01
This merely expands on the accepted answer for anyone looking for an example that can't depend on the first match/occurrence being the first character in the string.
对于任何寻找不能依赖于第一个匹配/出现是字符串中的第一个字符的示例的人来说，这只是扩展了公认的答案。
回答by Supernormal
I found this solution at https://www.regextester.com/99881, using a lookbehind pattern:
我在https://www.regextester.com/99881 上找到了这个解决方案，使用了后视模式：
/(?<=(.*l.*))l/g
/(?<=(.*l.*))l/g
Or more generally
或更一般地
/(?<=(.*MYSTRING.*))MYSTRING/g
/(?<=(.*MYSTRING.*))MYSTRING/g
where MYSTRINGis something that you want to remove.
MYSTRING您要删除的内容在哪里。
(This may also be a useful string for removing all but the first occurrence of "Re:" in an email subject string, by the way.)
（顺便说一下，这也可能是一个有用的字符串，用于删除电子邮件主题字符串中除第一次出现的“Re：”之外的所有内容。）
回答by Mic
Something like this?
像这样的东西？
"l 5 0 l 0 10 l -5 0 l 0 -10".replace(/[^^]l/g, '')
"l 5 0 l 0 10 l -5 0 l 0 -10".replace(/[^^]l/g, '')
