我正在運行一個wamp服務器並創建一系列html網頁,用戶放入的數據然後通過PHP腳本提交給SQL數據庫。我用Id. set to INT auto increment Crime. and Suspect. set as VARCHAR(40)
創建了一個數據庫表。我可以連接到數據庫,但現在當我嘗試插入我的錯誤:PHP腳本,SQL插入錯誤
'Problem with insert: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'case (Id, Case, Suspect) VALUES('NULL', 'test', 'test12')' at line 1'
所以我知道它與插入功能的問題,但是不能修復它!
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$db = 'fid';
$case = (isset($_POST['case'])? $_POST['case']:"None");
$suspect= (isset($_POST['suspect'])? $_POST['suspect']:"None");
$conn = mysql_connect($dbhost,$dbuser,$dbpass);
if (!$conn)
die('Could not connect: ' . mysql_error());
mysql_select_db($db);
mysql_query("INSERT INTO case
(Id, Case, Suspect) VALUES('NULL', '$case', '$suspect') ")
or die(' Problem with insert: ' . mysql_error());
echo 'Connected successfully';
mysql_close($conn);
?>
感謝您在過去一天試圖解決的問題,我一直在圈子裏尋找幫助,只是看着PDO感謝您的建議! – James 2013-03-17 18:05:46