How can I return an array object to a jquery $.ajax call?

Question

I am trying to return an array object to my Jquery ajax call, but it seems to return as a String with Array length =1

PHP code:

$results = mysqli_query($link, $sql);

if ($results->num_rows > 0) {

    while ($row = $results->fetch_assoc()) {


        $array = array($row['eventDate'], $row['time'] ,['data' => $row['time']])   ;

        echo json_encode($array) ;
    }
} else {
    echo "0 results";
}

Response in Developer tools shows:

["12_03_2015","0:00:00",{"data":"0:00:00"}]["12_03_2015","0:00:00",{"data":"0:00:00"}]["12_03_2015","0:00:00",{"data":"0:00:00"}]

My Ajax Call:

function trend(data){

$.ajax({
    url: "trendData.php",
    cache: true,
    type: "POST",
    data: {arr:data},

    success: function (data) {
        originalData=[];

        originalData.push(data)
        trender(data)
    }
});

}

In the Preview window for Network, it shows an Array:

I cannot get an array of , in this example, 3 objects and get the array.length of 3

Any help will be much appreciated.

When I debug data.length, it returns 1 I for some reason cannot return the array of 3 — motti10
– motti10, Commented Dec 4, 2015 at 16:41
Because it's an associative array look at starting [[ you should do data[0].length — michelem
– michelem, Commented Dec 4, 2015 at 16:42
It also looks like the code is printing out 3 separate arrays, as opposed to a collection of all records. How is that not a JSON syntax error I wonder.. — Flosculus
– Flosculus, Commented Dec 4, 2015 at 16:43

maxhb · Accepted Answer · 2015-12-04 16:57:09Z

3

You should collect all your data and return it as one json result and add a correct http-header (suggested by @Flosculus) so that the json format will be recognized:

$results = mysqli_query($link, $sql);
$data = array();

if ($results->num_rows > 0) {
    while ($row = $results->fetch_assoc()) {
        $data[] = array($row['eventDate'], $row['time'] ,['data' => $row['time']])   ;        
    }

    header('Content-Type: application/json');
    echo json_encode($data) ;
} else {
    echo "0 results";
}

edited Dec 4, 2015 at 16:57

answered Dec 4, 2015 at 16:44

maxhb

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4 Comments

Flosculus Over a year ago

Thats good, I'd also change the else statement to something more compatible with Javascript, so if (data.length > 0) {

motti10 Over a year ago

I have added a screen shot to the original question Still getting length = 1

Flosculus Over a year ago

@user5451365 Try putting header('Content-Type: application/json') at the top of your php file.

motti10 Over a year ago

That did it. Thank you so much. I had no idea about the header.

Adzimzf · Accepted Answer · 2017-06-02 01:12:18Z

0

I think you have to use JSON.parse(), if you're using jquery so your js script will look like

$.ajax({
url: "trendData.php",
cache: true,
type: "POST",
data: {arr:data},

success: function (data) {
    var data = JSON.parse(data);

    //see length of array
    console.log(data.length);
}
});

answered Jun 2, 2017 at 1:12

Adzimzf

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