Passing Data MySQL -> PHP -> JSON

Question

I'm trying to make a php web service that can get data from a MySQL DB on my webspace at school, and parse the data to JSON. Then I call this php function from jQuery, but it seems php returns an empty array.

Here is the php code:

    <?php



// Create connection
$con=mysqli_connect("mysqlstudent","sylvainvansteela","zei8quea0eeP","sylvainvansteela");

if (mysqli_connect_errno($con))
 {

 echo "Failed to connect to MySQL: " . mysqli_connect_error();

 } else {

$mysqlstring = "SELECT * FROM customers";

$result = mysqli_query($con,$mysqlstring);
$rows = array();

while($r = mysql_fetch_array($result)) {
  $rows["id"] = $r[0];
  $rows["email"] = $r[1];
}


header("Content-type: application/json");
echo json_encode($rows);

}


?>

And here is the jQuery code:

function getCustomers(){

    var url = 'http://student.howest.be/sylvain.vansteelandt/fedex/server/getcustomers.php';

    $.getJSON(url,function(data){

    if(data){
        alert(data);
        console.log(data.length);
    } else {
        alert('error');
    }



});

};

getCustomers();

You are using mysqli_query, but then mysql_fetch_.... fix that! — Eleazan
– Eleazan, Commented Jul 18, 2013 at 10:21

Yotam Omer · Accepted Answer · 2013-07-18 10:21:41Z

1

I would recommend changing the following part of you PHP:

$rows = array();

$i = 0; // add this
while($r = mysqli_fetch_array($result)) { // mysqli!
  $rows[$i]["id"] = $r[0];
  $rows[$i]["email"] = $r[1];
  $i++; // don't forget to increment
}

answered Jul 18, 2013 at 10:21

Yotam Omer

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3 Comments

svdotbe Over a year ago

thanks for the adjustments. But now I'm getting a 500 error from the server instead of an empty array.

svdotbe Over a year ago

I'm also trying to figure out why the jquery code only works when uploaded to my webserver, it doesn't work when i debug it in Adobe Edge Animate

Yotam Omer Over a year ago

@user1422136 this is most likely to security. If you have a local server installed such as WAMP or XAMPP try opening the files from localhost to overcome this.

Code Lღver · Accepted Answer · 2013-07-18 10:22:36Z

0

You need to make change here:

$inc = 0;
while($r = mysqli_fetch_array($result)) { //here you are using the mysql it should be mysqli
  $rows[$inc]["id"] = $r[0];
  $rows[$inc]["email"] = $r[1];
  $inc++;
}

If it is possible then use the mysqli_fetch_assoc.

answered Jul 18, 2013 at 10:22

Code Lღver

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Comments

Allen Chak · Accepted Answer · 2013-07-18 10:47:33Z

0

$rows = array();
while($r = mysqli_fetch_array($result)) {
    $row = array();
    $row['id'] = $r[0];
    $row['email'] = $r[1];
    array_push($rows, $row);
}
echo json_encode($rows);

// OR simplify
$rows = array();
while($r = mysqli_fetch_array($result)) {
    array_push($rows, array('id'=>$r[0], 'email'=>$r[1]));
}
echo json_encode($rows);

answered Jul 18, 2013 at 10:47

Allen Chak

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