jquery div replace from array

Question

I have a two-dimensional array which I get back as JSON from a call.

and I also have a 3-by-3 div.

as follows:

<div id="type0">
    <div id="image00"></div>
    <div id="image01"></div>
    <div id="image02"></div>
</div><br />

<div id="type1">
    <div id="image10"></div>
    <div id="image11"></div>
    <div id="image12"></div>
</div><br />

<div id="type2">
    <div id="image20"></div>
    <div id="image21"></div>
    <div id="image22"></div>
</div><br />

and i have the loop below to iterate the data.

$.each(res, function (i, tickets) {
     $.each(tickets, function (j, ticket) {

         console.log('i:' + i);
         console.log('j:' + j); 



     });
 });

How can I replace the content of the div above every time this script is invoked?

Essentially, I want to fill up the 3-by-3 div.

So, with the inner loop, I guess I need something like:

$("#type1").$("#image1").html(image here);

What's the correct way to do this?

i and j gives u, 0,1,2 and 0,1,2.

or is there a better way to do this?

One issue that should be corrected is that the value of ID must be unique within the HTML document. You have id="image0" (and so on) set three times. — honyovk
– honyovk, Commented Jul 15, 2012 at 22:43

Ram · Accepted Answer · 2012-07-15 22:52:14Z

3

you can try eq() and start with attribute selector:

$.each(res, function (i, tickets) {
     $.each(tickets, function (j, ticket) {
         $('div[id^=img]').eq(j).text(ticket)
     });
 });

$.each(res, function (i, tickets) {
     var ind = i;
     $.each(tickets, function (j, ticket) {
         $('#image'+ind+j).text(ticket)
     });
 });

edited Jul 15, 2012 at 22:52

answered Jul 15, 2012 at 22:44

Ram

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3 Comments

DarthVader Over a year ago

how does i adds in here? how can i make image00 ? you are just doing j

Tats_innit Over a year ago

@DarthVader :)) just what Darth Vador will say :P [quote]: "I find your lack of faith disturbing." eh ! :)

Ram Over a year ago

@DarthVader I have updated ma answer, hope that it can help you.

Musa · Accepted Answer · 2012-07-15 22:49:14Z

1

Use the ids as selectors since they are unique

$.each(res, function (i, tickets) {
     $.each(tickets, function (j, ticket) {
         $(('#image'+i)+j).html(image here)
     });
 });

answered Jul 15, 2012 at 22:49

Musa

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