CODEWITHSUNDEEP
javascriptjquery
Miha Šušteršič
Miha Šušteršič

Reputation: 10052

writing a for loop with .class:eq()

so, I have an array of 15 links, and I want to pass them as img src into my html elements. My elements are all of class. card, and I figured I'd write a loop instead of hard coding it (I have to do this 45 times with different links).

so it looks like this

var picks1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
for(x=0;x<picks1.length;x++) {
        $(".card:eq(x)").src = "picks1[x]";
    }

is there a way to do it like this? I figure this isn't working because of the quotation marks at .card:eq(x). (there are 45 picks arrays and their .length changes from 1-15). Is there another way to do this?

This is a part of a bigger webpage, can post source if needed, basically what it does it read a log file that another program makes, then makes links to images on a third party website from it and puts them in arrays.

Upvotes: 0

Views: 79

Answers (5)

Vaibhav
Vaibhav

Reputation: 1477

Lets say you have a pickArray which has 45 arrays with length 1-15.You can write like this

var i=0, length = pickArray.length, j=0, arrLen=0, arr;
  for(;i<length;i++){
    arr = pickArray[i];
    arrLen = arr.length;
    for(j=0;j<arrLen;j++) {
        $(".card:eq("+j+")").src = arr[j];
    }
 }

Upvotes: 1

Arun P Johny
Arun P Johny

Reputation: 388396

Although all the above answers are right, I think caching the value of $('.card') will be better

var picks1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
$(".card").slice(0, picks1.length).attr('src', function (i) {
    return picks1[i]
})

Note: The .slice(0, picks1.length) is not required if the length of $(".card") and picks1 will be same

Upvotes: 2

Utkarsh Dixit
Utkarsh Dixit

Reputation: 4275

Your are setting value of picks1[x] as string , try to remove quotes. Use the code below

var picks1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
for(x=0;x<picks1.length;x++) {
        $(".card").eq(x).src = picks1[x];
    }

Or use it with attr() function

$(".card").eq(x).attr('src',picks1[x]);

Hope this helps you

Upvotes: 0

Vladu Ionut
Vladu Ionut

Reputation: 8193

var picks1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
var cards = $(".card");
for (x = 0; x < picks1.length; x++) {
    cards.eq(x).attr('src', picks1[x]);
}

Upvotes: 0

empiric
empiric

Reputation: 7878

You have the use the +-operator to insert a variable into your selector-string:

var picks1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
for(x=0;x<picks1.length;x++) {
    $(".card:eq(" + x + ")").src = picks1[x];
    // OR
    $(".card").eq(x).attr('src', picks1[x]);
}

Upvotes: 0

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