如何替换所有BUT字符串中第一次出现的模式

quick question: my pattern is an svg string and it looks like l 5 0 l 0 10 l -5 0 l 0 -10 To do some unittest comparison against a reference I need to ditch all but the first l I 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位测试比较我需要抛弃除了第一个l我知道我可以抛弃它们所有并预先设置'l'，或者我可以使用子串。但我想知道这有一个javascript正则表达式成语吗？

5 个解决方案

#1

You can try a negative lookahead, avoiding the start of the string:

您可以尝试使用负向前瞻，避免字符串的开头：

/(?!^)l/g

See if online: jsfiddle

看看是否在线：jsfiddle

#2

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来替换所有东西 - 但是第一个模式匹配。但是，您可以通过将函数作为第二个参数传递给replacemethod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 $0 $1 $2 $3 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.

为每个匹配执行函数（fn_replaceBy）。在第一次匹配时，函数立即返回匹配的字符串（没有任何反应），并设置一个标志。每个其他匹配将根据函数中描述的逻辑进行替换：通常，您使用$ 0 $ 1 $ 2等等来引用组。在fn_replaceBy中，函数参数等于：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 replace allows very powerful applcations, such as an intelligent HTML parser.

匹配的子字符串将被函数fn_replaceBy的返回值替换。使用函数作为替换的第二个参数允许非常强大的应用程序，例如智能HTML解析器。

另请参见：MDN：String.replace>将函数指定为参数

#3

 "l 5 0 l 0 10 l -5 0 l 0 -10".replace(/^\s+/, '').replace(/\s+l/g, '')

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，并删除任何空格后跟'l'。

#4

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，但您可以使用任意（如占位符）和链替换来替换第一个匹配项，以实现其余的逻辑：

'-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

Produces:

生产：

-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/出现是字符串中第一个字符的示例的人来说，这仅仅扩展了所接受的答案。

#5

-1

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，''）

#1