create assciative array while loop without quotes around values

Question

I have this PHP code that creates an array to display as JSON:

$return_arr = array();
$sql="SELECT * from prices ";
$rs=mysql_query($sql,$conn) or die(mysql_error());
while($result=mysql_fetch_array($rs)) {
    $return_arr[] = array('label' => $result["product"], 'id' => $result["sequence"]);
}

currently, it displays like:

var data = [{"label":"VoIP Telephone Numbers","id":"3"},{"label":"VoIP Port Submit","id":"4"}];

but i need to get the quotes (") removed from the label and id

http://stackoverflow.com/questions/13523729/how-to-json-encode-php-array-but-the-keys-without-quotes — halabuda, Jun 02 '14 at 16:46
Do you mean removed from the keys or the values? You cannot have the value of label without quotes (its a string) and the keys should be in quotes also — Steve, Jun 02 '14 at 16:46

score 0 · Answer 1 · answered Jun 02 '14 at 16:39

This is the regex I use for this task:

$not_valid_json = preg_replace(array('%([{,])(\s*)"([a-z0-9]+)"\s*:%i','%\\\\/%','%"%'), array('$1$3:','/','\''), json_encode($value));

Be aware though that this is not really valid JSON anymore according to the spec, but it will still evaluate in JavaScript.

Also note that this doesn't always work great if the value being encoded contains quotes.

score 0 · Answer 2 · answered Jun 02 '14 at 16:52

You're not clear on how or where you're displaying the JSON, but all you probably need to do it use the JSON.parse method. Below is from my Chrome javascript console:

> var foo = JSON.parse('{"label":"VoIP Telephone Numbers","id":"3"}');
undefined
> foo
Object {label: "VoIP Telephone Numbers", id: "3"}

create assciative array while loop without quotes around values

2 Answers2