latent thorn
Let’s say I have a URL: https://test.com/en/editor?env=row&lat=24.89479&lon=-53.69694&s=8522367606782&zoomLevel=16
How can I remove the part from &s= to &zoomLevel (I want to remove „&s=8522367606782“ and send the URL without that part, how would I do that?
lime pagodaBOT
String.prototype.replace()
The replace() method returns a new string with one, some, or all matches of a pattern replaced by a replacement. The pattern can be a string or a RegExp, and the replacement can be a string or a function called for each match. If pattern is a string, only the first occurrence will be replaced. The original string is left unchanged.
silk portal
<string>.replace(“<that stuff>”, “”)
dire tartan
if the number is always changing, use the regex /&s=\d{13}/gm
assuming the number is always 13 long
silk portal
ah ofc, yh
latent thorn
How would I do it when the number is not always 13 long?
silk portal
I was just about to ask what each bit did 😂, thanks chewie
latent thorn
remove a part from a string
rotund root
let url = "https://test.com/en/editor?env=row&lat=24.89479&lon=-53.69694&s=8522367606782&zoomLevel=16";
let modifiedUrl = url.replace(/&s=[^&]*/, "");
or
let url = "https://test.com/en/editor?env=row&lat=24.89479&lon=-53.69694&s=8522367606782&zoomLevel=16";
let startIndex = url.indexOf("&s=")
let endIndex = url.indexOf("&zoomLevel")
let modifiedUrl = url.substring(0, startIndex) + url.substring(endIndex)
Such a shit solution lol
rotund root
but it works lol
dire tartan
The first one remove too much
And it's just overcomplicating lmao
rotund root
The first one remove too much
yeah you right first one removes to mutch /&s=[^&]*/ should do
dire tartan
now its just mine with a different identifier
rotund root
now its just mine with a different identifier
i wasnt trying to show yours is wrong lol or copie it i use second method myself most of the time for smaller aplicatins as its easyert to debug find issues