Моя подача форм PHP, jquery, ajax не работает
i have a php form im trying to submit using ajax and its saving data properly but its always returning the error message not the success, any help please to go about it.
Что я уже пробовал:
это файл jquery ajax.
$(document).ready(function() { var loader='<img src="https://media.giphy.com/media/11ASZtb7vdJagM/giphy.gif" />'; //if submit button is clicked $('#submit').click(function () { //show the loader $('.loading').html(loader).fadeIn(); var prog_description = $('input[name=prog_description]').val(); var type_of_contrib = $('select[name=type_of_contrib]').val(); var benefit_factor = $('input[name=benefit_factor]').val(); var budgeted_spend = $('input[name=budgeted_spend]').val(); var actual_amount = $('input[name=actual_amount]').val(); var current_status = $('select[name= current_status]').val(); var anticipated_completion = $('input[name= anticipated_completion]').val(); //organize the data properly var form_data = 'prog_description='+prog_description+ '&type_of_contrib=' +type_of_contrib+ '&benefit_factor='+benefit_factor+ '&budgeted_spend='+budgeted_spend+ '&actual_amount='+actual_amount+ '¤t_status='+current_status+ '&anticipated_completion='+anticipated_completion; //disabled all the text fields $('.text').attr('disabled','true'); //start the ajax $.ajax({ //this is the php file that processes the data and send mail url: "process.php", //POST method is used type: "POST", //pass the data data: form_data, //success success: function (html) { //if process.php returned 1/true (send mail success) if (html===1) { //hide the form $('#fupForm').fadeOut('slow'); //hide the loader $('.loading').fadeOut(); //show the success message $('.message').html('Successfully Registered ! ').fadeIn('slow'); //if process.php returned 0/false } else alert('Sorry, unexpected error. Please try again later.'); } }); //cancel the submit button default behaviours return false; }); }); </script>
process.php файл
<?php session_start(); /* Attempt MySQL server connection. Assuming you are running MySQL server with default setting (user 'root' with no password) */ $link = mysqli_connect("localhost", "root", "", "socio_economic"); // Check connection if ($link === false) { die("ERROR: Could not connect. " . mysqli_connect_error()); } // Prepare an insert statement $sql = "INSERT INTO socio_spends (prog_description,type_of_contrib,benefit_factor,budgeted_spend,actual_amount,current_status,anticipated_completion) VALUES (?, ?, ?,?,?,?,?)"; if ($stmt = mysqli_prepare($link, $sql)) { // Bind variables to the prepared statement as parameters mysqli_stmt_bind_param($stmt, "sssssss", $prog_description, $type_of_contrib, $benefit_factor, $budgeted_spend, $actual_amount, $current_status, $anticipated_completion); // Set parameters $prog_description = (isset($_POST['prog_description']) ? $_POST['prog_description'] : ''); $type_of_contrib = (isset($_POST['type_of_contrib']) ? $_POST['type_of_contrib'] : ''); $benefit_factor = (isset($_POST['benefit_factor']) ? $_POST['benefit_factor'] : ''); $budgeted_spend = (isset($_POST['budgeted_spend']) ? $_POST['budgeted_spend'] : ''); $actual_amount = (isset($_POST['actual_amount']) ? $_POST['actual_amount'] : ''); $current_status = (isset($_POST['current_status']) ? $_POST['current_status'] : ''); $anticipated_completion = (isset($_POST['anticipated_completion']) ? $_POST['anticipated_completion'] : ''); // Attempt to execute the prepared statement if (mysqli_stmt_execute($stmt)) { echo '1'; $socio_id =mysqli_insert_id($link); $_SESSION['socio_id'] =$socio_id; } else { echo "ERROR: Could not execute query: $sql. " . mysqli_error($link); } } else { echo "ERROR: Could not prepare query: $sql. " . mysqli_error($link); } header('Location:index.php'); // Close statement mysqli_stmt_close($stmt); // Close connection mysqli_close($link); exit();
F-ES Sitecore
Каково содержание" html " в вашей функции успеха? Если это не "1", то пройдите через свой php-код, чтобы проверить, какой путь принимает код и почему он не "1".
Member 14603400
спасибо, дай мне осмотреть ее