如何通过ajax函数发送数据到php文件？
问题描述：
我想通过ajax函数调用一个addPost.php文件，并返回结果作为后提交或不。如何通过ajax函数发送数据到php文件？
对于我打电话按钮userAction功能的onClick但现在我不通过后获取数据，当我试图访问addPost.php后阵列中的数据文件，它发送错误未定义指数类别。
任何人都可以请帮助出了什么问题吗？
<!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'];
}
}
编辑：
<!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>
现在这个编辑代码在数据库中的数据越来越插入，但我想显示的结果中添加对.html所以我已经在成功的方法给出了帕的id，但这是行不通的，也形式验证不起作用。
请帮忙。谢谢。
答
您可以使用隐藏值在服务器中发送数据。在您的表单中放置这两个字段。
<input type="hidden" name="action_type" id="action_type_id"/>
<input type="hidden" name="id" id="p_id"/>
而且你ajax方法是这样的：
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: $('#postForm').serialize(),
success:function(report){
alert(report);
}
});
}
说明：之前提交表单，您可以设置type和id值隐藏字段和发送整个形式server.now你可以通过$_POST['action_type']和$_POST['id']获得所有发布数据。
答
echo $_POST['category'], $_POST['title'], $_POST['url'], $_POST['urlType'], $_POST['desc'], $_POST['keywords']
，您尝试访问“类别”，“冠军”等，但你有没有过在阿贾克斯字段（用户数据），所以你得到了“未定义指数类别” - 这个错误
您的ajax方法类型是post，所以可能'＆action_type'值和其他参数不能发送给控制器。 –
那么应该是什么类型？以及如何访问这些数据？ @AtaurRahmanMunna – Sid
我的** addPost.php **文件中也没有看到任何''标签。 –