PHP : got a json parse error

Question

I want to import data from the database into my javascript code, using ajax through jQuery, but I got a json parse error and I really don't know what it does come from. Could you help me ?

The goal is to build buildings on a map, and i take the geometry elements such as coordinates and shape parameters from the database.

in the JS file, put:

 $.ajax({ 
 type: "GET",
 url: "ajax_processor.php",
 dataType : "html",
 error:function(msg, string){ 
 alert( "Error !: " + string );
 }
 success:function(returnData){ 
 var data = $.parseJSON(returnData); 
 for(var ID_geometryElement in data) {
 addComp($data[bldg], 
 $data[iZ], // zone where is the building
 $data[iType], //type of the geometric element
 $data[x0],
 $data[y0],//coordinates top-left
 $data[p], // geometric parameters
 );
 }
 }
 });

});

in the PHP file:

try {
 $bdd = new PDO('mysql:host=localhost;dbname=building_geometry','admin','password');
}
 $reponse = $bdd->query('SELECT * FROM ID_GEOMETRYELEMENT');
 $donnees = $reponse->fetch(); 
 header('Content-Type: application/json'); 
 echo json_encode($response); 
 ?> 

• Pedro Estrada: What are you getting back from the ajax?
• Elon Than: First of all you have to fix one thing in JS code. If you're writing for(var ID_geometryElement in data) { then you getting each item in ID_geometryElement variable, not in $data (and from where you get $ sign in JS var?).
• LaurianeF: What do you mean by that ? I want the ajax to execute the function addComp() and to fill the data with those it queried from the database, but I got an error before, in the json parsing.
• Spudley: Is your php program outputting anything else other than the output of a single call to json_encode()? What does the browser show in the http response data? Also, in your JS code, you appear to have a variable named data, but it looks like you're referring to it by the name $data. That looks like a bug.
• Orangepill: you have a try without a catch....

Answer 1

You're trying to json_encode the database statement handle your query returned. That is NOT something you can encode. The code should be

echo json_encode($donnees);
 ^^^^^^^^--- the actual data

If you'd done even the most basic debugging, e.g. console.log(returnData) in JS, you'd have seen you're not getting anything valid returned from the script.

• LaurianeF: Since that's my very first time in coding in javascript, i didn't know there was this debug tool. I'll try that, thanks !
• LaurianeF: Thanks, that works now using Firebug, I got the json parsed correctly.

Answer 2

dataType : "html",

must be

dataType : "json",

if you expect JSON in the response

EDIT: See http://api.jquery.com/jQuery.post/

dataType

Type: String

The type of data expected from the server. Default: Intelligent Guess (xml, json, script, text, html).

• Marc B: that does nothing. datatype json tells jquery you're EXPECTING json back from the server, causing it to decode the string automatically. but OP's doing .parseJSON anyways, so type:html is 'ok'.
• LaurianeF: Oh yeah, I already changed that, but I have still the error in json parse

Answer 3

$response is just a cursor, you need to return the fetched data.

try {
 $bdd = new PDO('mysql:host=localhost;dbname=building_geometry','admin','password');
}
$reponse = $bdd->query('SELECT * FROM ID_GEOMETRYELEMENT');
$donnees = $reponse->fetch(); 
header('Content-Type: application/json'); 
echo json_encode($donnees); 

?>