Return a JSON object using PHP json_encode() & MySQL to pass to jQuery function

Question

I'm trying to create a json object from MySQL results, but not getting the result I need.

Here is the PHP

$json = array();
$result = mysqli_query ($connection, $query);
    echo '['; 

        while($row = mysqli_fetch_array ($result))     
        {
            echo '{';
            echo '"latitude":"'.$row['lat'].'",';
            echo '"longitude":"'.$row['lng'].'",';
            echo '"icon":'.'"./images/'.$row['busColor'].'.png"';
            echo '}';    
        }
        echo ']';

        $jsonstring = json_encode($json);
        echo $jsonstring;

        die();

It outputs this

[{"latitude":"39.976257","longitude":"-83.003464","icon":"./images/pink.png"}][]

But I want this

[{"latitude":"39.976257","longitude":"-83.003464","icon":"./images/pink.png"}]

once I get the result I need to pass the object to a jQuery plugin function if that makes any difference

$.getJSON('myJsonURL, function(myMarkers){
  $("#map").goMap({
    markers: myMarkers
  });
});

Thanks

isnt the output that you want an object? not an array?.. – Branddd Nov 20 '20 at 19:00 — Branddd, Nov 20 '20 at 19:00

score 26 · Accepted Answer · answered Dec 22 '10 at 08:35

26

I guess the correct way to do this would be:

$json = array();
$result = mysqli_query ($connection, $query);
while($row = mysqli_fetch_array ($result))     
{
    $bus = array(
        'latitude' => $row['lat'],
        'longitude' => $row['lng'],
        'icon' => './images/' . $row['busColor'] . '.png'
    );
    array_push($json, $bus);
}

$jsonstring = json_encode($json);
echo $jsonstring;

die();

answered Dec 22 '10 at 08:35

RabidFire

6,280
1
28
24

This works perfectly and demonstrates how to do what David explained. The output looks like this `[{"latitude":"39.992989","longitude":"-249.090057","icon":".\/images\/pink.png"}`. So json_encode automatically escapes the slashes? Is there a way to suppress this? I don't need to, but just curious. – JCam Dec 22 '10 at 10:09
Yes, the slashes are added by default. You could try `urlencode('./images/')` and the `urldecode(json_encode($json))` - but that's kinda messy! – RabidFire Dec 22 '10 at 10:13

score 8 · Answer 2 · answered Dec 22 '10 at 08:26

8

you output your json by hand and then you call json_encode on an empty array() - $json

json_encode() outputs [] on you pass an empty array so your last [] comes from here
$jsonstring = json_encode($json); echo $jsonstring;
Edit: More about json_encode json_encode php manual

answered Dec 22 '10 at 08:26

cristian

8,676
3
38
44

@OctoPaul I went this route as it was the path of least resistance for me. I changed the last `echo` to `echo '},';`, because JSLint didn't like the syntax. I get one extra comma before the closing `]`, but jQuery parses it though. – JCam Dec 22 '10 at 08:41
Look at the other two answers, it is better to put all your data in an array and call json_encode on that array - this way your json will be valid – cristian Dec 22 '10 at 09:00
Thanks Paul, I updated the correct answer. Your method got me where I wanted to go, so it is still a valid answer. – JCam Dec 22 '10 at 10:14

score 5 · Answer 3 · answered Dec 22 '10 at 08:25

5

You start by defining an array.

You then generate some JSON manually.

You then convert the array to JSON and output it.

Replace all the echo statements in and at the edge of your while loop with code to generate an associative array containing the data, and then insert it into $json.

answered Dec 22 '10 at 08:25

Quentin

914,110
126
1,211
1,335

I should probably go this route, but I am a relative noob with PHP so I went the easier route for now. Especially since it's 3:43AM in my part of town. Thanks for the info. +1 – JCam Dec 22 '10 at 08:44
That *is* the easier route. Bashing strings together is going to generate invalid JSON, sooner or later. – Quentin Dec 22 '10 at 08:47
I figured as much, but I was looking for a quick and dirty fix after hacking at it for a while. I've resorted to your method using the example @RabidFire provided. – JCam Dec 22 '10 at 10:12

Return a JSON object using PHP json_encode() & MySQL to pass to jQuery function

3 Answers3

Linked