自學的同時SQL和PHP,並正在此錯誤:MySQL錯誤:插入表單數據到數據庫
錯誤:你在你的SQL語法錯誤; VALUES('duke','john','doe','cpsc','teacher'at line)檢查與您的MySQL服務器版本相對應的手冊,以找到正確的語法,以便在'姓名,姓氏,Course_id,Professor' 1與查詢
我的代碼應該從表格中的信息,並將其插入到一個數據庫,我已經cread我已經看了它幾個小時,並已經搜索了一段時間的互聯網,但仍然可以而她不知道什麼造成這種誰能幫我出
下面是與形式我test.php的文件:?
<html>
<head>
<title>Insert data into database</title>
</head>
<body>
<form method="post" action="handler.php">
<p>What is your University:</p>
<input type="text" name="uni"><br>
<p>What is your First Name:</p>
<input type="text" name="f_name"><br>
<p>What is your Last Name:</p>
<input type="text" name="l_name"><br>
<p>What is your Course:</p>
<input type="text" name="cou_id"><br>
<p>Who was your Professor for that course?:</p>
<input type="text" name="pro_id"><br>
<input type="submit" value="Submit" />
</form>
</body>
</html>
,這裏是我的^ h存儲從表格中的信息到數據庫andler.php文件:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>
<body>
<?php
$uni = $_POST['uni'];
$uni= mysql_real_escape_string($uni);
$f_name = $_POST['f_name'];
$f_name = mysql_real_escape_string($f_name);
$l_name = $_POST['l_name'];
$l_name = mysql_real_escape_string($l_name);
$cou_id = $_POST['cou_id'];
$cou_id = mysql_real_escape_string($cou_id);
$pro_id = $_POST['pro_id'];
$pro_id = mysql_real_escape_string($pro_id);
$con = mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("tester") or die(mysql_error());
if (!mysql_query("INSERT INTO my_hw (University, First Name, Last Name, Course_id, Professor) VALUES ('$uni','$f_name','$l_name','$cou_id','$pro_id')"))
{
die('Error: ' . mysql_error() . " with query ". $sql);
}
echo "1 record added";
mysql_close($con);
?>
</body>
</html>
感謝
之間謝謝空格分隔,這是一個愚蠢的錯誤。你應該有反駁保留字以及?他們還有其他用途嗎? – spentpat 2013-02-21 09:31:43
@spentpat - 在你的情況下,它們可以用於列名中的空格。我不會推薦使用保留字作爲列名。 – 2013-02-21 11:19:28