CODEWITHSUNDEEP
phpjavascriptajaxjson
Ben
Ben

Reputation: 85

AJAX PHP JSON encode not returning all results

I'll be brief... I'm about to jump

not all results are working

in the document header I am including jquery-1.7.2.min.js -

my query:

"SELECT tableA.columns_xyz..., tableB.columns_xyz..., tableC.columns_xyz 
FROM tableB
JOIN tableC ON tableB.x = tableC.x
JOIN tableA ON tableB.x = tableA.x
WHERE tableA.x = '$query_value'
GROUP BY tableA.y
ORDER BY tableA.y DESC 
LIMIT 1";

PROBLEM 1:

change LIMIT 1 to LIMIT 5 or anything >1 and the query executes fine in phpmyadmin but not via ajax/json/php

the php:

$sql = above...
$result = $mysqli->query($sql);
while($row = mysqli_fetch_array($result)) {
echo json_encode(array($row['x'],$row['y'],....$row['z']));
$mysqli->close();

the javascript:

$.ajax ({
    url: 'script.php',
    type: "post",
    dataType: "json",
    data: {value : $("#searchbox").val()},
    success: function(data) {
        var a = parseFloat(data[0]);
        var b = parseFloat(data[1]);
        var c = data[2];

            document.getElementById("blah1").innerHTML = a;
            document.getElementById("blah2").innerHTML = b;
            document.getElementById("blah3").innerHTML = c;
       }
});

the html:

<td><span id="blah1"></span></td>
<td><span id="blah2"></span></td>
<td><span id="blah3"></span></td>

PROBLEM 2:

some elements are printing to page, some aren't. what am I doing wrong ?

PROBLEM 3:

I know I should be able to express document.getElementById("blah1").innerHTML; as $('#blah1'); but $('#blah1'); is just not working... I don't know why...

Upvotes: 1

Views: 1054

Answers (2)

Wern Ancheta
Wern Ancheta

Reputation: 23297

Store each row in an array then use json_encode once:

$store = array();
while($row = mysqli_fetch_array($result)) {
    $store[] = array('x'=>$row['x'], 'y'=>$row['y'], 'z'=>$row['z']);
}
echo json_encode($store);

Then you can use JSON.parse on the client-side:

var store = JSON.parse(data)

And then just loop through that:

for(var x in store){
    $('#blah1').html(store[x]['x']);
}

You're already using jquery so why don't you make full use of it by converting this:

document.getElementById("blah1");

To its jquery version:

$('#blah1');

Just make sure that those elements are unique since you're using ID's

Upvotes: 1

Jon Taylor
Jon Taylor

Reputation: 7905

Should you not be creating one object then json encoding the whole lot, rather than json encoding each row seperatley?

Im not great with sql but having a LIMIT greater than 1 means ur expecting more than one row back (potentially), correct?

If so then you are essentially returning multiple json responses in one response which I dont think you should be doing.

Upvotes: 1

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