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我想通過ajax函數調用一個addPost.php文件，並返回結果作爲後提交或不。如何通過ajax函數發送數據到php文件？
對於我打電話按鈕userAction功能的onClick但現在我不通過後獲取數據，當我試圖訪問addPost.php後陣列中的數據文件，它發送錯誤未定義指數類別。
任何人都可以請幫助出了什麼問題嗎？
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Post</title>
</head>
<body>
<form class="postForm" id="postForm" method="post" action="addPost.html">
<fieldset>
<legend>Please add the details below </legend>
<p>
<label for="title">Title (required, at least 2 characters)</label>
<input id="title" name="title" minlength="2" type="text" required>
</p>
<p>
<label for="url">URL (required)</label>
<input id="url" type="url" name="url">
</p>
<p>
<label for="desc">Description (required, at least 2 characters)</label>
<input id="desc" name="desc" minlength="2" type="text" required>
</p>
<p>
<label for="keywords">Keywords (eg:#facebook)(required, at least 2 characters)</label>
<input id="keywords" name="keywords" minlength="2" type="text" required>
</p>
<p>
Select Url Type :
<select name="urlType" id="urlType">
<option value="">Select Url Type...</option>
<option value="0">Server Image</option>
<option value="1">Server Video</option>
<option value="2">YouTube Video</option>
<option value="3">Vimeo Video</option>
<option value="4">Facebook Image</option>
<option value="5">Facebook Video</option>
<option value="6">Instagram Image</option>
<option value="7">Instagram Video</option>
<option value="-1">Other</option>
</select>
</p>
<p>
Select Category :
<select name="category" id="category">
</select>
</p>
<p>
<input type="button" name="Submit" id="Submit" value="Submit" onclick="userAction('add')">
</p>
</fieldset>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
getCategories();
function getCategories() {
$.ajax({
type: "POST",
url: 'getCategories.php',
dataType: 'text',
async: false,
cache: false,
success: function (result) {
$('#category').html(result);
}
});
}
function userAction(type,id){
id = (typeof id == "undefined")?'':id;
var statusArr = {add:"added",edit:"updated",delete:"deleted"};
var userData = '';
if (type == 'add') {
userData = $("#postForm").find('.form').serialize() + '&action_type=' + type + '&id=' + id;
}
$.ajax({
type: 'POST',
url: 'addPost.php',
data: userData,
success:function(report){
alert(report)
}
});
}
</script>
</form>
</body>
</html>
addPost.php
include 'Database.php';
ini_set('display_errors', 1);
error_reporting(1);
ini_set('error_reporting', E_ALL);
if(isset($_POST['action_type']) && !empty($_POST['action_type'])) {
if($_POST['action_type'] == 'add') {
/* $database = new Database(Constants::DBHOST, Constants::DBUSER, Constants::DBPASS, Constants::DBNAME);
$dbConnection = $database->getDB();
$dbConnection->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$stmt = $dbConnection->prepare("insert into keywords(keyword)
values(?)");
$stmt->execute(array($_POST['keywords']));
//insert data into posts table
$stmt = $dbConnection->prepare("insert into posts(category_id,title,url,url_type,description,keywords)
values(?,?,?,?,?,?)");
$stmt->execute(array($_POST['category'], $_POST['title'], $_POST['url'], $_POST['urlType'], $_POST['desc'], $_POST['keywords']));
$count = $stmt->rowCount();
if ($count > 0) {
//if inserted
$response = array("status" => -1, "message" => "Post Submitted");
return $response;
} else {
//if not inserted
$response = array("status" => -1, "message" => "Could not submit post.");
return $response;
}*/
echo $_POST['category'], $_POST['title'], $_POST['url'], $_POST['urlType'], $_POST['desc'], $_POST['keywords'];
}
}
編輯：
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Post</title>
</head>
<body>
<form class="postForm" id="postForm" method="post" action="addPost.php">
<fieldset>
<legend>Please add the details below </legend>
<p>
<label for="title">Title (required, at least 2 characters)</label>
<input id="title" name="title" minlength="2" type="text" required>
</p>
<p>
<label for="url">URL (required)</label>
<input id="url" type="url" name="url" required>
</p>
<p>
<label for="desc">Description (required, at least 2 characters)</label>
<input id="desc" name="desc" minlength="2" type="text" required>
</p>
<p>
<label for="keywords">Keywords (eg:#facebook)(required, at least 2 characters)</label>
<input id="keywords" name="keywords" minlength="2" type="text" required>
</p>
<p>
Select Url Type :
<select name="urlType" id="urlType">
<option value="">Select Url Type...</option>
<option value="0">Server Image</option>
<option value="1">Server Video</option>
<option value="2">YouTube Video</option>
<option value="3">Vimeo Video</option>
<option value="4">Facebook Image</option>
<option value="5">Facebook Video</option>
<option value="6">Instagram Image</option>
<option value="7">Instagram Video</option>
<option value="-1">Other</option>
</select>
</p>
<p>
Select Category :
<select name="category" id="category">
</select>
</p>
<p>
<input type="hidden" name="action_type" id="action_type_id"/>
<input type="hidden" name="id" id="p_id"/>
<input type="button" name="Submit" id="Submit" value="Submit" onclick="userAction('add')">
</p>
<p id="report"></p>
</fieldset>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
getCategories();
$("#postForm").validate();
function getCategories() {
$.ajax({
type: "POST",
url: 'getCategories.php',
dataType: 'text',
async: false,
cache: false,
success: function (result) {
$('#category').html(result);
}
});
}
function userAction(type,id){
var statusArr = {add:"added",edit:"updated",delete:"deleted"};
if (type == 'add') {
$('#action_type_id').val(type);
$('#p_id').val(id);
}
$.ajax({
type: 'POST',
url: 'addPost.php',
data: $('#postForm').serialize(),
success:function(report){
$('#report').html(result);
}
});
}
</script>
</form>
</body>
</html>
現在這個編輯代碼在數據庫中的數據越來越插入，但我想顯示的結果中添加對.html所以我已經在成功的方法給出了帕的id，但這是行不通的，也形式驗證不起作用。
請幫忙。謝謝。
您的ajax方法類型是post，所以可能'＆action_type'值和其他參數不能發送給控制器。 –
那麼應該是什麼類型？以及如何訪問這些數據？ @AtaurRahmanMunna – Sid
我的** addPost.php **文件中也沒有看到任何''標籤。 –