I have following JS
var slider = new Slider("#testslider", {
tooltip: 'always'
});
To style my bootstrap slider by ID. I want to style every slider in that area so I tried to put them all in one DIV like:
<div id="testslider">
//all sliders
</div>
That will only style the first slider.
I tried [id^="testslider"]
and named every slider in this scheme: testslider1, testslider2 etc.
But that will only style the first slider.
Now with document.querySelectorAll([id^="testslider"]
); I could select all IDs but I can't figure out how to put that into
var slider = new Slider("#testslider", {
tooltip: 'always'
});
Now with document.querySelectorAll([id^="testslider"]); I could select all IDs but I can't figure out how to put that into
var slider = new Slider("#testslider", {
tooltip: 'always'
});
I suppose you absolutely need the id
to instanciate the tooltip plugin...
Use a loop to go through each elements, get it's specific id
to instancitate the slider:
const allSliders = Array.from(document.querySelectorAll('[id^="testslider"]'));
console.log("allSliders length:", allSliders.length)
allSliders.forEach((element) => {
console.log("id for tooltip:", "#"+element.id) // To check if the right ids are used
new Slider("#"+element.id, { // The specific id is used here
tooltip: 'always'
});
})