Here is the case: I want to find the elements which match the regex...
targetText = "SomeT1extSomeT2extSomeT3extSomeT4extSomeT5extSomeT6ext"
and I use the regex in javascript like this
reg = new RegExp(/e(.*?)e/g);
var result = reg.exec(targetText);
and I only get the first one, but not the follow.... I can get the T1 only, but not T2, T3 ... ...
function doSomethingWith(result) {
console.log(result)
}
const targetText = "SomeT1extSomeT2extSomeT3extSomeT4extSomeT5extSomeT6ext"
const reg = /e(.*?)e/g;
let result;
while ((result = reg.exec(targetText)) !== null) {
doSomethingWith(result);
}
/e(.*?)e/.exec(...) will give an infinite loop.
while(result = reg.exec(targetText)) { leveraging the "false" resolve.
Commented
Mar 13, 2018 at 16:10
reg.lastIndex = result.index + 1;) if you are not using the g (global) flag in your regex. The global flag will automatically move to the next match, however, the disadvantage is that if you want to re-use a global regex, you must reset reg.lastIndex = 0; before you match a new string.
Three approaches depending on what you want to do with it:
Loop through each match: .match
targetText.match(/e(.*?)e/g).forEach((element) => {
// Do something with each element
});
Loop through and replace each match on the fly: .replace
const newTargetText = targetText.replace(/e(.*?)e/g, (match, $1) => {
// Return the replacement leveraging the parameters.
});
Loop through and do something on the fly: .exec
const regex = /e(.*?)e/g; // Must be declared outside the while expression,
// and must include the global "g" flag.
let result;
while(result = regex.exec(targetText)) {
// Do something with result[0].
}
== in place of = in the while expression.
= the assignment can be surrounded by parentheses like so: while ((result = regex.exec(targetText)))
match is the whole RegExp match, and $1 is the first capture group (), in that example, only one.
Commented
Jan 31, 2020 at 22:09
Try using match() on the string instead of exec(), though you could loop with exec as well. Match should give you the all the matches at one go. I think you can omit the global specifier as well.
reg = new RegExp(/e(.*?)e/);
var matches = targetText.match(reg);
String.prototype.match — developer.mozilla.org/en/docs/Web/JavaScript/Reference/…
Commented
Dec 7, 2015 at 9:56
String.match, while RegExp.exec does.
I kept getting infinite loops while following the advice above, for example:
var reg = /e(.*?)e/g;
var result;
while((result = reg.exec(targetText)) !== null) {
doSomethingWith(result);
}
The object that was assigned to result each time was:
["", "", index: 50, input: "target text", groups: undefined]
So in my case I edited the above code to:
const reg = /e(.*?)e/g;
let result = reg.exec(targetText);
while(result[0] !== "") {
doSomethingWith(result);
result = reg.exec(targetText);
}
targetText = "SomeT1extSomeT2extSomeT3extSomeT4extSomeT5extSomeT6ext"
reg = new RegExp(/e(.*?)e/g);
var result;
while (result = reg.exec(targetText))
{
...
}
You could also use the String.replace method to loop through all elements.
result = [];
// Just get all numbers
"SomeT1extSomeT2extSomeT3ext".replace(/(\d+?)/g, function(wholeMatch, num) {
// act here or after the loop...
console.log(result.push(num));
return wholeMatch;
});
console.log(result); // ['1', '2', '3']
Greetings
I did this in a console session (Chrome).
> let reg = /[aeiou]/g;
undefined
> let text = "autoeciously";
undefined
> matches = text.matches(reg);
(7) ["a", "u", "o", "e", "i", "o", "u"]
> matches.forEach(x =>console.log(x));
a
u
o
e
i
o
u
Although @tvanfosson suggested in his answer, I'm adding the modified version for someone who only needs to get all matchings.
var targetText = "SomeT1extSomeT2extSomeT3extSomeT4extSomeT5extSomeT6ext";
var reg = /e(.*?)e/g;
var ss = targetText.match(reg); // ss = "eT1e,eT2e,eT3e,eT4e,eT5e,eT6e"
I was actually dealing with this issue. I prefer Lambda functions for about everything.
reg = /e(.*?)e/gm;
targetText.match(reg).forEach(element => console.log(element));
