Looking for help with JavaScript:
"a a a a".replace(/(^|\s)a(\s|$)/g, '$1')
I thought the result would be '', but it's showing 'a a' instead. Can someone explain what I'm missing?
To clarify, I want to remove all instances of 'a' that have whitespace around them (as a complete token)
The reason behind this is that the regular expression /(^|\s)a(\s|$)/g matches the character before and after each "a" in the string.
For example, in the string "a a a a", the regex will match:
Update: It's a bit tricky but it works without using regular expressions:
var a = "a a a a";
// Handling the case at the beginning 'a '
var startI = a.indexOf("a ");
if (startI === 0){
var off = a.charAt(startI + 2) !== "a" ? 2 : 1; // checking if "a" comes next to keep the space before
a = a.slice(startI + off);
}
// Handling the case in the middle ' a '
var iOf = -1;
while ((iOf = a.indexOf(" a ")) > -1){
var off = a.charAt(iOf + 3) !== "a" ? 3 : 2; // same check here
a = a.slice(0, iOf) + a.slice(iOf+off, a.length);
}
// Handling the case at the end ' a'
var endI = a.indexOf(" a");
if (endI === a.length - 2){
a = a.slice(0, endI);
}
a; // should be an empty string ""
The first occurrence of "a " is matched initially. After that, it attempts to match against "a a a", skipping the first "a" and matching the following "a ". The next attempt is made to match against "a", but it fails to match.
The resulting output will be "a a".
To achieve your desired output, you can use the following code:
"a a a a".replace(/(?:\s+a(?=\s))+\s+|^a\s+(?=[^a]|$|a\S)|^a|\s*a$/g, '')
Explaining why the given regex doesn't work properly, it is important to note that the surrounding spaces are being consumed as part of the match. Here's a more simplified explanation:
The regex pattern tries to find a space or start of the string followed by an 'a', and then a space or end of the string.
Breaking down the string with character indexes:
a a a a
0123456
Starting at index 0, we find a match because there is an 'a' followed by a space. Moving to index 2, we don't have a match since 'a' is not preceded by a space or start of string. The same goes for index 6, where 'a' is not preceded by what the regex expects.
The suggested regex /(^|\s+)a(?=\s|$)/g works differently due to the ?= quantifier. It checks if the following character meets the condition without actually consuming it as part of the match.
(^|\s)b(?=\s|$)
Give this a shot. Replace with $1. Check out the live example.
Try using this alternative method:
"b b b b".replace(/(^|\s*)b(\s|$)/g, '$1')
This code will replace every instance of "b" with an empty space.
Best regards
One option is to divide the string, filter it, and then piece it back together:
"a ba sl lf a df a a df r a".split(/\s+/).filter(function (x) { return x != "a" }).join(" ")
>>> "ba sl lf df df r"
"a a a a".split(/\s+/).filter(function (x) { return x != "a" }).join(" ")
>>> ""
Alternatively, in ECMAScript 6:
"a ba sl lf a df a a df r a".split(/\s+/).filter(x => x != "a").join(" ")
>>> "ba sl lf df df r"
"a a a a".split(/\s+/).filter(x => x != "a").join(" ")
>>> ""
Assuming there are no leading or trailing spaces. You can modify the filter to x && x != 'a' if you wish to eliminate this assumption.
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