问题
This question already has an answer here:
- Split string into array without deleting delimiter? 5 answers
How can I split a string without removing the delimiters?
Let\'s say I have a string:
var string = \"abcdeabcde\";
When I do
var newstring = string.split(\"d\"), I get something like this:
[\"abc\",\"eabc\",\"e\"]
But I want to get this:
[\"abc\",\"d\",\"eabc\",\"d\",\"e\"]
When I tried to do my \"split2\" function, I got all entangled in splice() and indexes and \"this\" vs \"that\" and ... aargh! Help! :D
回答1:
Try this:
- Replace all of the "d" instances into ",d"
- Split by ","
var string = "abcdeabcde";
var newstringreplaced = string.replace(/d/gi, ",d");
var newstring = newstringreplaced.split(",");
return newstring;
Hope this helps.
回答2:
Try:
"abcdeabcde".split(/(d)/);
回答3:
var parts= string.split('d');
for (var i= parts.length; i-->1;)
parts.splice(i, 0, 'd');
(The reversed loop is necessary to avoid adding ds to parts of the list that have already had ds inserted.)
回答4:
I like Kai's answer, but it's incomplete. Instead use:
"abcdeabcde".split(/(?=d)/g) //-> ["abc", "deabc", "de"]
This is using a Lookahead Zero-Length Assertion in regex, which makes a match not part of the capture group. No other tricks or workarounds needed.
回答5:
Can be done in one line:
var string = "abcdeabcde";
string.split(/(d)/);
["abc", "d", "eabc", "d", "e"]
回答6:
split - split is used to create separate lines not for searching.
[^d] - find a group of substrings not containing "d"
var str = "abcdeabcde";
str.match(/[^d]+|d/g) // ["abc", "d", "eabc", "d", "e"]
or
str.match(/[^d]+/g) // ["abc", "eabc", "e"]
or
str.match(/[^d]*/g) // ["abc", "", "eabc", "", "e", ""]
Read "RegExp Object" if you do not want problems with the "javascript".
回答7:
This is my version for regexp delimiters. It has same interface with String.prototype.split; it will treat global and non global regexp with no difference. Returned value is an array that odd member of it are matched delimiters.
function split(text, regex) {
var token, index, result = [];
while (text !== '') {
regex.lastIndex = 0;
token = regex.exec(text);
if (token === null) {
break;
}
index = token.index;
if (token[0].length === 0) {
index = 1;
}
result.push(text.substr(0, index));
result.push(token[0]);
index = index + token[0].length;
text = text.slice(index);
}
result.push(text);
return result;
}
// Tests
assertEquals(split("abcdeabcde", /d/), ["abc", "d", "eabc", "d", "e"]);
assertEquals(split("abcdeabcde", /d/g), ["abc", "d", "eabc", "d", "e"]);
assertEquals(split("1.2,3...4,5", /[,\.]/), ["1", ".", "2", ",", "3", ".", "", ".", "", ".", "4", ",", "5"]);
assertEquals(split("1.2,3...4,5", /[,\.]+/), ["1", ".", "2", ",", "3", "...", "4", ",", "5"]);
assertEquals(split("1.2,3...4,5", /[,\.]*/), ["1", "", "", ".", "2", "", "", ",", "3", "", "", "...", "4", "", "", ",", "5", "", ""]);
assertEquals(split("1.2,3...4,5", /[,\.]/g), ["1", ".", "2", ",", "3", ".", "", ".", "", ".", "4", ",", "5"]);
assertEquals(split("1.2,3...4,5", /[,\.]+/g), ["1", ".", "2", ",", "3", "...", "4", ",", "5"]);
assertEquals(split("1.2,3...4,5", /[,\.]*/g), ["1", "", "", ".", "2", "", "", ",", "3", "", "", "...", "4", "", "", ",", "5", "", ""]);
assertEquals(split("1.2,3...4,5.", /[,\.]/), ["1", ".", "2", ",", "3", ".", "", ".", "", ".", "4", ",", "5", ".", ""]);
assertEquals(split("1.2,3...4,5.", /[,\.]+/), ["1", ".", "2", ",", "3", "...", "4", ",", "5", ".", ""]);
assertEquals(split("1.2,3...4,5.", /[,\.]*/), ["1", "", "", ".", "2", "", "", ",", "3", "", "", "...", "4", "", "", ",", "5", "", "", ".", ""]);
assertEquals(split("1.2,3...4,5.", /[,\.]/g), ["1", ".", "2", ",", "3", ".", "", ".", "", ".", "4", ",", "5", ".", ""]);
assertEquals(split("1.2,3...4,5.", /[,\.]+/g), ["1", ".", "2", ",", "3", "...", "4", ",", "5", ".", ""]);
assertEquals(split("1.2,3...4,5.", /[,\.]*/g), ["1", "", "", ".", "2", "", "", ",", "3", "", "", "...", "4", "", "", ",", "5", "", "", ".", ""]);
// quick and dirty assert check
function assertEquals(actual, expected) {
console.log(JSON.stringify(actual) === JSON.stringify(expected));
}
回答8:
Try this:
var string = "abcdeabcde";
var delim = "d";
var newstring = string.split(delim);
var newArr = [];
var len=newstring.length;
for(i=0; i<len;i++)
{
newArr.push(newstring[i]);
if(i != len-1)newArr.push(delim);
}
回答9:
Try this
"abcdeabcde".split("d").reduce((result, value, index) => {
return (index !== 0) ? result.concat(["d", value]) : result.concat(value)
}, [])
回答10:
function split2(original){
var delimiter = "d", result = [], tmp;
tmp = original.split(delimiter);
tmp.forEach(function(x){result.push(x); result.push(delimiter); });
return result;
}
回答11:
Try this
var string = "abcdeabcde";
var res = string.replace( /d/g, 'd\0' ).split(/(?=d)|\0/);
console.log( res );
//["abc", "d", "eabc", "d", "e"]
回答12:
Just override the String.prototype.split function with this one :
String.prototype._split = String.prototype.split;
String.prototype.split = function(delimiter, keepDelimiters) {
if (!keepDelimiters) {
return this._split(delimiter);
} else {
var res = [];
var start = 0, index;
while ((index = this.indexOf(delimiter, start)) != -1) {
res.push(this.substr(start, index - start));
res.push(delimiter);
start = index + 1;
}
res.push(this.substr(start));
return res;
}
};
var str = "abcdeabcde";
alert(str.split("d", true));
alert(str.split("d"));
// output :
//
// [ 'abc', 'd', 'eabc', 'd', 'e' ]
// [ 'abc', 'eabc', 'e' ]
This method will not fail, do not use regular expressions and is consistent with the original String.split function. It is also consistent with the str.split(/(d)/); solution, but will not fail in IE.
The only restriction is that, if the second parameter is used (the split replacement drop-in function), then the delimiter cannot be a regular expression. But if you think about it a little, it's not a real restriction since you don't need the second parameter if using regular expression...
来源：https://stackoverflow.com/questions/4514144/js-string-split-without-removing-the-delimiters