我有下面的代碼。 $ last_id始終顯示爲零。我在「源」表中有一列自動遞增ID。這段代碼有什麼問題?爲什麼PHP mysql_insert_id()返回零?
的index.php:
<?php
// connect to the database
include('connect-db.php');
$last_id = mysql_insert_id($connection);
$content = "Please type your content here!";
?>
<!DOCTYPE html>
<html>
<head>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#myform").submit(function() {
$.post(
'submit.php',
$(this).serialize(),
function(data){
alert("Your ID: <?php echo $last_id;?>");
}
);
return false;
});
});
</script>
</head>
<body>
<div id="header" >
<form action="submit.php" method="post" id="myform">
<textarea id="editor" name="editor" id="editor"><?php echo $content; ?></textarea>
<br/>
<input type="submit" name="submit" value="Submit" id="submit"/>
</form>
</div>
</body>
</html>
連接-DB:
<?php
$server = 'localhost';
$user = 'mumbo';
$pass = 'mumbo123';
$db = 'jumbo';
$connection = mysql_connect($server, $user, $pass)
or die ("Could not connect to server ... \n" . mysql_error());
mysql_select_db($db)
or die ("Could not connect to database ... \n" . mysql_error());
?>
submit.php:
<?php
include('connect-db.php');
$submit_date = date("Ymd");
$content = mysql_real_escape_string(htmlspecialchars($_POST['editor']));
$ip_address = $_SERVER['REMOTE_ADDR'];
if ($content != '') {
mysql_query("INSERT INTO source (submit_date, ip, content) values('$submit_date','$ip_address','$content')") or die(mysql_error());
mysql_close($connection);
}
>
它的工作原理!謝謝! – 2013-03-10 05:34:33
@Mumbo珍寶:歡迎您 – 2013-03-10 05:52:44