Json display all data using for loop

Question

My json is

[{"dir":"down","sym":"oo","b":"1.35553","a":"1.35556"},
{"dir":"down","sym":"pp","b":"1.64442","a":"1.64451"},
{"dir":"down","sym":"qq","b":"104.603","a":"104.609"},
{"dir":"up","sym":"rr","b":"0.88186","a":"0.88191"},
{"dir":"up","sym":"ss","b":"1.09667","a":"1.09672"}]

using this jquery

<script type="text/javascript">
$(document).ready(function() {
setInterval(function() {
$.ajax({
    cache: false,
        url: 'quotes1.php',
        dataType: 'json',
    success: function(json){
        for(var i=0;i<json.length;i++){
            $('#divToRefresh').html(json[i].dir + '/' + json[i].sym+ '/' + json[i].b+ '/' + json[i].a);
        }
    }
});
}, 1000);  
});
</script>

But it displays 1st data only

up/oo/1.35553/1.35556

How can I display all the data?

Pranav C Balan · Accepted Answer · 2014-01-21 07:59:49Z

3

Use jQuery $.each() and append() method for that

success: function(json){
    $.each(json,function(i,val){
        $('#divToRefresh').append(val.dir + '/' + val.sym+ '/' + val.b+ '/' + val.a);
    });
}

html() will replace existing data hence it only display the last value , but append() will help to concatenate data.

$.each is better than the for loop.

edited Jan 21, 2014 at 7:59

answered Jan 21, 2014 at 7:54

Pranav C Balan

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Comments

Satpal · Accepted Answer · 2014-01-21 07:53:23Z

1

Use .append() instead of .html(). You can also use $.each()

success: function(json){
    $.each(json,function(i,value){
        $('#divToRefresh').append(value.dir + '/' + value.sym+ '/' + value.b+ '/' + value.a);
    });
}

answered Jan 21, 2014 at 7:53

Satpal

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Comments

Karthick Kumar · Accepted Answer · 2014-01-21 07:55:06Z

0

 $('#divToRefresh').append(json[i].dir + '/' + json[i].sym+ '/' + json[i].b+ '/' + json[i].a);

.html() will replace content .append() will concate the content

answered Jan 21, 2014 at 7:55

Karthick Kumar

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Comments

Code Lღver · Accepted Answer · 2014-01-21 07:55:16Z

0

Instead of the below code:

for(var i=0;i<json.length;i++){
            $('#divToRefresh').html(json[i].dir + '/' + json[i].sym+ '/' + json[i].b+ '/' + json[i].a);
        }

Use the below code:

$.each(JSON.parse(json), function(key,val){
  $('#divToRefresh').append(val.dir + '/' + val.sym+ '/' + val.b+ '/' + val.a);
});

answered Jan 21, 2014 at 7:55

Code Lღver

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1 Comment

Ashley Medway Over a year ago

This will only ever display the last value? html will replace the content for every item in the loop

Maurice Perry · Accepted Answer · 2014-01-21 07:56:56Z

0

That's strange: I would have thought that it would rather display the last...

Try this:

    for(var i=0;i<json.length;i++){
        $('#divToRefresh').append("<p>" + json[i].dir + '/' + json[i].sym+ '/' + json[i].b+ '/' + json[i].a + "</p>");
    }

answered Jan 21, 2014 at 7:56

Maurice Perry

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Comments

Steve Bals · Accepted Answer · 2014-01-21 08:36:45Z

0

<script type="text/javascript">
$(document).ready(function() {

setInterval(function() {
$.ajax({
    cache: false,
        url: 'quotes1.php',
        dataType: 'json',
        success: function(json){
        var dd="";
        for(var i=0;i<json.length;i++){
            dd += json[i].dir + '-' + json[i].sym+ '-' + json[i].bid+ '-' + json[i].ask+'</br>';
        }
        $('#divToRefresh').html(dd);
    }
});
}, 1000);  
});
</script>

answered Jan 21, 2014 at 8:36

Steve Bals

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