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POPHP json_decode not supported, any alternatives?
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copied!<p>as the title specifies, my hosting provider does not have support for json_decode, so I need to find a way to adapt my code to achieve the same effect, but without using JSON, here is my code,</p> <p>jQuery:</p> <pre><code> var allLocations = []; $(".locations").each( function(i, location) { // for each location block location = $(location); var loc = { 'province' : $("select[data-loc*='province']", location).val(), 'town' : $("select[data-loc*='town']", location).val() }; allLocations.push( loc ); }); //POST the locations information $.ajax({ type: 'POST', url: 'locations.php', dataType: 'json', data: { locations: JSON.stringify(allLocations), uid: uid }, success: function(data){ //alert(data) } }); </code></pre> <p>PHP:</p> <pre><code>$json = $_POST['locations']; $uid = $_POST['uid']; // $json is a string $json_array = json_decode($json, true); mysql_connect('localhost','user','pass') or die(mysql_error()); mysql_select_db('eskom_products') or die(mysql_error()); //insert the locations into the database while($json_array as $key){ $query = mysql_query("INSERT INTO suppliersLocations (supplier_id, province, town) VALUES('".$uid."', '".$key['province']."', '".$key['town']."' ) ") or die(mysql_error()); } echo $text; </code></pre> <p>So as you can see, I am getting the province and town values of each location and creating a JSON object with it, which I then send off via <code>$.ajax</code> to a PHP file, but now since <code>json_decode</code> doesn't work, I need to try and find another way of fixing the problem, I was thinking of trying to pass an associative array to the php file, but I wanted to see what your guy's input would be, and if there might be a better way of achieving the desired result.</p> <p>Thanx in advance!</p>

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