我敢肯定我错过了一些非常简单的东西,我已经多次运行了mysql更新脚本,但是这将无法正常工作,我也无法理解为什么。
单击更新按钮时,我正在将数据库中的数据调用到ckeditor文本框中,此脚本运行,我会收到成功消息,但数据库未更新我所缺少的内容???
session_start();
include ("mysql-connect.php");
include ("check-login.php");
if(isset($_POST['editblogbody'])){
if($_POST['editblogbody'] != "" && $_POST['editblogtitle'] != "" && $_POST['blog_id'] != ""){
$title = $_POST['editblogtitle'];
$body = $_POST['editblogbody'];
$bid = $_POST['blog_id'];
$sql = mysql_query("UPDATE blogs SET title='$title', body='$body' WHERE id='$bid'")or die (mysql_error());
echo '<img src="../_Images/round_success.png" alt="Success" width="31" height="30" />Success 1';
exit();
}
else{
echo '<img src="../_Images/round_success.png" alt="Success" width="31" height="30" />Success 2';
exit();
}
}
else{
echo '<img src="../_Images/round_error.png" alt="Error" width="31" height="30" /> Opps something went wrong. Please Try again.';
exit();
}
发布表格的脚本
$('#editblog').on('submit', function (e) {
e.preventDefault();
$('input[type=submit]', this).attr('disabled', 'disabled');
var blogTitle = $("#editblogtitle").val();
var blogText = CKEDITOR.instances['editblogbody'].getData();
var url = "../_Scripts/edit-blog.php";
if (!blogTitle) {
$('input[type=submit]', this).removeAttr('disabled');
$("#blogeditreply").html('<img src="../_Images/round_error.png" alt="Error" width="31" height="30" /> Please type a Title.').show().fadeOut(6000);
return false;
} else if (!blogText) {
$('input[type=submit]', this).removeAttr('disabled');
$("#blogeditreply").html('<img src="../_Images/round_error.png" alt="Error" width="31" height="30" /> Please type in your Blog.').show().fadeOut(6000);
return false;
} else {
$("#blogFormProcessGif").show();
for (instance in CKEDITOR.instances) {
CKEDITOR.instances['blogbody'].updateElement();
}
$.post(url, $('#editblog').serialize(), function (data) {
$("#jqueryReply").html(data).show().fadeOut(6000);
$("#blogFormProcessGif").hide();
$.modal.close();
});
}
});
连接失去了像这样
define('HOSTNAME','#');
define('DB_USERNAME','#');
define('DB_PASSWORD','#');
define('DATABASE','#');
$link = mysql_connect(constant('HOSTNAME'), constant('DB_USERNAME'), constant('DB_PASSWORD')) or die("Database connection error, please check!"); mysql_select_db(constant('DATABASE'), $link) or die("Connection to the defined database not possible, please check!");
我可以很好地回显所有变量,所以所有信息都在那里
谢谢
最佳答案
您会获得哪个成功答案?
“成功1”或“成功2”