CODEWITHSUNDEEP
javascriptjquery
user7739271
user7739271

Reputation:

Lazy syntax to loop through jquery objects

These are some basic ways I have used to declare an jquery object while looping:

let $divs = $('.mydiv');

for (let i = 0; i < $divs.length; i++) {
    let $div = $($divs[i]);
}

// or

for (let $div of Array.from($divs)) {
    $div = $($div);
}

// or

Array.from($divs).forEach(function ($div) {
    $div = $($div);
});

I want to save the line $div = $($div) (for writing code faster), like these:

for (let $($div) of Array.from($divs)) {
    // ...
}

// or

Array.from($divs).forEach(function ($div = $($div)) {
    // ...
});

Is there a way to achieve my goal?

UPDATE:

Because $('.mydiv') can return more than 1 element, so I need a loop to work with them. But using a loop, it doesn't return an jquery object for per element, I have to wrap it by $(...)

Upvotes: 0

Views: 89

Answers (1)

Mulan
Mulan

Reputation: 135396

Not really an answer to not really a question

After you have an array of $-wrapped elements, what do you plan on doing with it?

$('.mydiv')                        // = $-wrapped html collection
  .map((_, e) => $(e))             // = array of $-wrapped elements
  .each((_, e) => console.log(e))  // = log each $-wrapped element
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<div class="mydiv">1</div>
<div class="mydiv">2</div>
<div class="mydiv">3</div>

That's really no different than this script below, so what's the point?

$('.mydiv')                          // = $-wrapped html collection
  .each((_, e) => console.log($(e))) // = log each $-wrapped element
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<div class="mydiv">1</div>
<div class="mydiv">2</div>
<div class="mydiv">3</div>

Upvotes: 1

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