colap Posted January 7, 2014 Share Posted January 7, 2014 $("input.class_input_button_send").on("click",function(event) { event.preventDefault(); //var username = $("input.class_input_hidden").val(); var typed_text = $("input#id_input_text_type").val(); console.log(typed_text); rt = $.ajax({ type: "POST", url: "send.php", data: { name: "", typed_text: typed_text }, success: function(data) { console.log(data); console.log(data.length); } }).done(function() { alert( "success" ); }) .fail(function() { alert( "error" ); }); //console.log(rt); }); send.php =><?php <?php if( session_id() == '' ) { session_start(); } if($_SERVER['SERVER_PROTOCOL'] == "HTTP/1.1") { $domain_host_root = "http://" . $_SERVER['SERVER_NAME'] . "/"; } require_once 'core/model/mysql/mysqldb.php'; require_once 'core/utility/debug-app.php'; require_once 'core/utility/utility.php'; require_once 'core/view/view.php'; //$_POST['typed_text'] = "abc"; $typed_text = $_POST['typed_text']; $objDateTime = new DateTime('NOW'); $target_file = $objDateTime->format("Y-m-d"); $handle = fopen("archive/" . $target_file . ".txt", "w+"); fwrite($handle, $target_file_data); echo json_encode($target_file); ?> fopen function is working inside that send.php script if i run this as a standalone script, but fopen function returns null through the jquery $.ajax() function. Link to comment https://forums.phpfreaks.com/topic/285158-php-fopen-is-not-working-through-jquery-ajax/ Share on other sites More sharing options...
GetFreaky Posted January 7, 2014 Share Posted January 7, 2014 PHP-coder, Add dataType: "json" to your function params. $.ajax({dataType: "json", url: url, data: data, success: success }); Link to comment https://forums.phpfreaks.com/topic/285158-php-fopen-is-not-working-through-jquery-ajax/#findComment-1464204 Share on other sites More sharing options...
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