我创建了一个带有ajax函数的页面,用于将数据插入到数据库中,所以点击提交数据就会在数据库中插入,但我想显示一条消息,说明数据是否已插入到已创建的div并显示结果在那,但我看不到文字。
<!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>
<label for="urlType">Select Url Type :(required)</label>
<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>
<label for="postType"> Select Post Type :(required)</label>
<select name="postType" id="postType">
<option value="">Select Post Type...</option>
<option value="0">Normal</option>
<option value="1">Featured</option>
<option value="2">Sponsored</option>
</select>
</p>
<p>
<label for="category"> Select Category :(required)</label>
<select name="category" id="category">
<option value="0">Select Category...</option>
</select>
</p>
<p>
<input type="hidden" name="action_type" id="action_type_id"/>
<input type="hidden" name="id" id="p_id"/>
<!-- <a href="javascript:void(0);" class="btn btn-warning" onclick="$('#postForm').slideUp();">Cancel</a>
<a href="javascript:void(0);" class="btn btn-success" onclick="userAction('add')">Add User</a>-->
<input type="button" name="Submit" id="Submit" value="Submit" onclick="userAction('add')">
</p>
</fieldset>
<div class="result"></div>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
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){
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){
$(".result").html(data);
location.reload();
}
});
}
</script>
</form>
</body>
</html>
addPost.php
<?php
include 'Database.php';
ini_set('display_errors', 1);
error_reporting(1);
ini_set('error_reporting', E_ALL);
if(isset($_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,post_type)
values(?,?,?,?,?,?,?)");
$stmt->execute(array($_POST['category'], $_POST['title'], $_POST['url'], $_POST['urlType'], $_POST['desc'], $_POST['keywords'],$_POST['postType']));
$count = $stmt->rowCount();
if ($count > 0) {
//if inserted
// echo '<p id="report">Post Submitted</p>';
echo "Post submitted.";
} else {
//if not inserted
// echo '<p id="report">Post Submitted</p>';
echo "Could not submit post.";
}
}
}
?>
请帮忙..谢谢。
答案 0 :(得分:1)
以下是您的代码的更新部分。表格重置代码已添加。
$.ajax({
type: 'POST',
url: 'addPost.php',
data: $('#postForm').serialize(),
success:function(report){
// replace data to report
$(".result").html(report);
//reset form
$(':input','#form')
.not(':button, :submit, :reset, :hidden')
.val('')
.removeAttr('checked')
.removeAttr('selected');
}
});
答案 1 :(得分:0)
请在服务器端和客户端之间制定合同
例如: 每当客户端向服务器发送请求时 服务器处理请求,为这些行构建响应 - 例如({result:“success”}或某些内容并将其发送回客户端 客户端将能够在$ .ajax的成功中访问它 客户现在将根据合同了解何时成功以及何时失败并执行下一步操作
希望这有帮助。