我正在尝试提交表单而不使用 json 离开页面。
但是,下面的方法忽略了我在CKEditor中输入的数据。
任何想法(并随时更正我的术语)?
<script type="text/javascript">
$(document).ready( function() {
$("#addStory input[type=submit]").click(function(e) {
e.preventDefault();
$.post('_posteddata.php', $("#addStory").serialize(), function(result) {
alert(result.adminList);
}, "json");
});
});
</script>
<form name="addStory" action="" method="post" id="addStory">
<label for="story_story">Story: </label><textarea id="story_story" name="story_story"><p></p></textarea>
<?php
// Include the CKEditor class.
include("ckeditor/ckeditor.php");
// Create a class instance.
$CKEditor = new CKEditor();
$CKEditor->basePath = '/ckeditor/'
$CKEditor->replace("story_story");
?>
<input type="submit" value="Submit" />
</form>
_posteddata.php:
include 'connection.php';
function check_input($value, $quoteIt)
{
// Stripslashes
if (get_magic_quotes_gpc())
{
$value = stripslashes($value);
}
// Quote if not a number
if (is_null($value) || $value=="") {
$value = 'NULL';
} else if (!is_numeric($value) && $quoteIt == 1) {
$value = "'" . mysql_real_escape_string($value) . "'";
}
return $value;
}
// CKEDITOR STUFF FOR STORY_STORY
if (isset($_POST)) {
$postArray = &$_POST;
}
foreach ( $postArray as $sForm => $value )
{
if($sForm == "story_story") {
$story_story = check_input($value, 1);
}
}
$query = "INSERT INTO story_table (story) VALUES ($story_story)";
mysql_query($query) or die(mysql_error() . $query);
$return = array();
$return['adminList'] = "New story added with ID: " . mysql_insert_id();
header('application/json');
echo json_encode($return);
mysql_close();