Destramic Posted May 16, 2010 Share Posted May 16, 2010 hey guys im having a problem returning the data in the json array...plus when alerting data.value it comes back at 1000+ which is truely incorrect....can anyone explain where i am going wrong please?...thank you <!DOCTYPE html> <html> <head> <script type="text/javascript" src="/ajax/jquery/libary/jquery.js"></script> </head> <body> <?php $game_type = array(); $division = array(); if ($_POST['game'] == 1) { $game_type[] = array('value' => '1', 'text' => 'TDM' ); $game_type[] = array('value' => '1', 'text' => 'CTF' ); $division[] = array('value' => '1', 'text' => 'Divison 1' ); } elseif ($_POST['game'] == 2) { $game_type[] = array('value' => '1', 'text' => 'CTF' ); $division[] = array('value' => '1', 'text' => 'Divison 2' ); } json_encode($game_type); json_encode($division); ?> <form> <script> $(document).ready(function(){ $("select#game").change(function(){ var post_string = "game=" + $(this).val(); $.ajax({ type: 'POST', data: post_string, cache: false, dataType: 'game_type', url: 'json.php', timeout: '2000', error: function() { alert("Error has occured"); }, success: function(data) { alert(data.length); alert(data[1].text); } }); }); }); //$('#game_type').html('<input type=\"checkbox\" value=\"test\" /> TDM'); //$('#division').html('<input type=\"checkbox\" value=\"test\" /> Division 1'); </script> <select name="game" id="game"> <option value=""></option> <option value="1">Counter Strike</option> <option value="2">COD</option> </select> <div id="game_type"> </div> <div id="division"> </div> <select name="sub_category" id="sub_category"> <option value="">-- Select First Value --</option> </select> </form> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/201942-json-returning-data/ Share on other sites More sharing options...
RichardRotterdam Posted May 16, 2010 Share Posted May 16, 2010 hey guys im having a problem returning the data in the json array... json is a js object not an array What does your json.php file look like? Quote Link to comment https://forums.phpfreaks.com/topic/201942-json-returning-data/#findComment-1059085 Share on other sites More sharing options...
Destramic Posted May 17, 2010 Author Share Posted May 17, 2010 here you go dj kat...i've been working on this for the last couple of days...just having a problem returning the objects in the array...maybe a second pair of eyes could notice the problem...thanks <?php ob_start(); ?> <!DOCTYPE html> <html> <head> <script type="text/javascript" src="/ajax/jquery/libary/jquery.js"></script> </head> <body> <?php if ($_POST['game'] == 1) { $array = array( 'game_type' => array( array('value' => 1, 'text' => 'CTF'), array('value' => 1, 'text' => 'TDM'), ), 'division' => array( array('value' => 1, 'text' => 'Division 1'), array('value' => 1, 'text' => 'Division 2') )); ob_clean(); header('Content-Type: application/json'); die(json_encode($array)); } elseif ($_POST['game'] == 2) { $array = array( 'game_type' => array( array('value' => 1, 'text' => 'TKOTH'), array('value' => 1, 'text' => 'DM'), ), 'division' => array( array('value' => 1, 'text' => 'Division 3'), array('value' => 1, 'text' => 'Division 4') )); ob_clean(); header('Content-Type: application/json'); die(json_encode($array)); } ?> <form> <script> $(document).ready(function(){ $("#game").change(function(){ var post_string = "game=" + $(this).val(); $.ajax({ type: 'POST', data: post_string, cache: false, dataType: 'json', url: 'json.php', timeout: '2000', error: function(data) { console.log(data); }, success: function(data) { $.each(data, function(prop, obj){ switch(prop) { case 'game_type': $.each(obj, function(i, val){ var game_type_row = val.text + '<input name=\"\" id=\"\" type=\"checkbox\" value=\"' + val.value + '\" />\n'; console.log(game_type_row); $(game_type_row).appendTo('#game_type'); }); break; case 'division': $.each(obj, function(i, val){ var division_row = val.text +'<input name=\"\" id=\"\" type=\"checkbox\" value=\"'+ val.value +'\" />\n'; console.log(division_row); $(division_row).appendTo('#division'); }); break; } }); } }); }); }); </script> <select name="game" id="game"> <option value=""></option> <option value="1">Counter Strike</option> <option value="2">COD</option> </select> <div id="game_type"> please select a game...</div> <div id="division"> please select a game...</div> </form> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/201942-json-returning-data/#findComment-1059764 Share on other sites More sharing options...
RichardRotterdam Posted May 18, 2010 Share Posted May 18, 2010 The code confuses me a bit. The relevant code seems the following: $(document).ready(function(){ $("select#game").change(function(){ var post_string = "game=" + $(this).val(); $.ajax({ type: 'POST', data: post_string, cache: false, dataType: 'game_type', url: 'json.php', timeout: '2000', error: function() { alert("Error has occured"); }, success: function(data) { alert(data.length); alert(data[1].text); } }); }); }); with the line url: 'json.php', You're calling the file json.php. Is the code you pasted in your last post indeed json.php? Offtopic why are you using ob_start, ob_clean and die if I may ask? Quote Link to comment https://forums.phpfreaks.com/topic/201942-json-returning-data/#findComment-1060026 Share on other sites More sharing options...
Destramic Posted May 18, 2010 Author Share Posted May 18, 2010 dj kat....all code posted were from json.php...just updat ed versions...although ive now got my script working as i want it...and as for the ob_clean, ob_start and die....without these functions my script wouldnt work....by all means have a look <?php ob_start(); ?> <!DOCTYPE html> <html> <head> <script type="text/javascript" src="/ajax/jquery/libary/jquery.js"></script> </head> <body> <?php if ($_POST['game'] == 1) { $array = array( 'game_type' => array( array('value' => 1, 'text' => 'CTF'), array('value' => 1, 'text' => 'TDM'), ), 'division' => array( array('value' => 1, 'text' => 'Division 1'), array('value' => 1, 'text' => 'Division 2') )); ob_clean(); header('Content-Type: application/json'); die(json_encode($array)); } elseif ($_POST['game'] == 2) { $array = array( 'game_type' => array( array('value' => 1, 'text' => 'TKOTH'), array('value' => 1, 'text' => 'DM'), ), 'division' => array( array('value' => 1, 'text' => 'Division 3'), array('value' => 1, 'text' => 'Division 4') )); ob_clean(); header('Content-Type: application/json'); die(json_encode($array)); } ?> <form> <script> $(document).ready(function(){ var game_type_value = $('#game_type').text(); var division_value = $('#division').text(); $("#game").change(function() { if ($('#game option:selected').text() == "") { $('#game_type').empty(); $('#division').empty(); $('#game_type').text(game_type_value); $('#division').text(division_value); } var post_string = "game=" + $(this).val(); $.ajax({ type: 'POST', data: post_string, cache: false, dataType: 'json', url: 'json.php', timeout: '2000', error: function(data) { console.log(data); }, success: function(data) { $.each(data, function(prop, obj){ switch(prop) { case 'game_type': $('#game_type').empty(); $.each(obj, function(i, val){ var game_type_row =$('<label>' + val.text +'</label><input name=\"\" id=\"\" type=\"checkbox\" value=\"'+ val.value +'\" />'); $(game_type_row).appendTo('#game_type'); }); break; case 'division': $('#division').empty(); $.each(obj, function(i, val){ var division_row = $('<label>' + val.text +'</label><input name=\"\" id=\"\" type=\"checkbox\" value=\"'+ val.value +'\" />'); $(division_row).appendTo('#division'); }); break; } }); } }); }); }); </script> <select name="game" id="game"> <option value=""></option> <option value="1">Counter Strike</option> <option value="2">COD</option> </select> <div id="game_type"> please select a game...</div> <div id="division"> please select a game...</div> </form> </body> </html> but there is one question i need to have two clauses in the line $("#game").change(function() { something like $("#game").change || $("#game").text(function() { but im not sure how this is done...as im still leaning...if anyknow how this is possible please let me know....thank you Quote Link to comment https://forums.phpfreaks.com/topic/201942-json-returning-data/#findComment-1060311 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.