fetching data dynamically from sql

Question

My jQuery part isn't working, can sombody please explain why I do not get the desired output ? index.php

while ($row = mysql_fetch_array($res))
    array_push($result, array('text' =>$row[1]));
echo json_encode(array("result"=> $result));

getting.js

$.getJson("view_note.php", function(data){
            $("ul").empty();

            $.each(data.result,function(){

            $("ul").append("<li>Text: "+this['text']+"</li><br />";
            });
        });
    });

Output

You will have to add some debug output. On php side: what does $result finally contain? On JS side: what does data contain? On network level: what is transferred? Those are all things only you can find out, we do not have access to your system... — arkascha
– arkascha, Commented Dec 7, 2015 at 9:02
Lesson 1: Learn to debug my friend. You can use console.log(data); to view what kind of data are you receiving. Another thing I see in your code is you have to remove the <br /> — aceraven777
– aceraven777, Commented Dec 7, 2015 at 9:09
there are a syntex error in ur each loop. u can check it in below my answer. — harry
– harry, Commented Dec 7, 2015 at 9:23

harry · Accepted Answer · 2015-12-07 09:19:24Z

1

Try this:

$.getJson("view_note.php", function(data){
    $("ul").empty();

    $.each(data.result,function(){

        $("ul").append("<li>Text: "+this['text']+"</li><br />");
    });
});

answered Dec 7, 2015 at 9:19

harry

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Comments

Parth Trivedi · Accepted Answer · 2015-12-07 09:01:04Z

0

Use each loop as

$.each(data.result,function(i,item){
            $("ul").append("<li>Text: "+$(item).text+"</li>";
            });

also remove <br/>

answered Dec 7, 2015 at 9:01

Parth Trivedi

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Comments

Jomie · Accepted Answer · 2015-12-07 09:11:31Z

0

I think it's because you wrap the array 2 times.

echo json_encode(array("result"=> $result));

to

echo json_encode($result);

answered Dec 7, 2015 at 9:11

Jomie

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1 Comment

Jomie Over a year ago

The variable $result is already an array from what I see coz you use array_push to insert the data and then you use echo json_encode(array("result"=> $result)); to put it inside another array. That's why you got the answer still in {}

anon · Accepted Answer · 2015-12-07 09:29:31Z

0

The answer is the syntax (AFAIK), this is your code formatted and with syntax fixed:

$.getJson("view_note.php", function(data){
    $("ul").empty();

    $.each(data.result,function(){
        $("ul").append("<li>Text: "+this['text']+"</li><br />");
    });
});

I'd prefer this to be a comment but I cannot because of the size and formatting constraints

answered Dec 7, 2015 at 9:29

anon

Comments

Jai · Accepted Answer · 2015-12-07 09:38:22Z

0

You can change to this:

var data = [{
  "result": [{
    "text": "My first note."
  }, {
    "text": "My Second note."
  }]
}]
$.each(data[0]['result'], function(i, item) {
  $("ul").append("<li>Text: " + item.text + "</li>");
  //-----------add index ----------^-----put closing-^-----here.
  // also no need to add <br> tag at the end.
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<ul></ul>

edited Dec 7, 2015 at 9:38

answered Dec 7, 2015 at 9:32

Jai

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