0
我正在創建一個小項目來解釋我的學生如何使用PHP代碼更新SQL數據庫中的值。我在MySQL中創建了所有字段爲VARCHAR的表。我寫了以下代碼，其中引發以下錯誤： 更新值時出錯您的SQL語法中有錯誤;請檢查與您的MySQL服務器版本相對應的手冊，以便在第1行的'Mbps WHERE dsl ='25610669'附近使用正確的語法，其中25610669是數據庫中的現有記錄。下面是代碼：錯誤更新值
<?php
if((isset($_POST['B2'])))
{
$server = 'localhost' ;
$un = 'root' ;
$pass = 'icsk' ;
$db = 'yusuf' ;
$conn = mysqli_connect($server, $un, $pass, $db);
$update = "UPDATE homereg SET Fname = {$_POST['First']}, Lname = {$_POST['Last']}, cid = {$_POST['cid']}, pack = {$_POST['choice']} WHERE dsl = {$_POST['dsl']}" ;
$result = mysqli_query($conn, $update);
if($result == 1)
{
echo "Successfully Updated" ;
}
else
{
echo "Error in Updting value" . mysqli_error($conn) ;
}
}
?>
<html>
<head>
<title>Update User Information</title>
</head>
<body background="HomePageMap.gif">
<CENTER><B><FONT COLOR = 'RED'>SEARCH & UPDATE THE EXISTING RECORD HERE </FONT></B></CENTER><P>
<form method="POST" action="modify.php" name = "frm">
<div align="center">
<table border="1" width="314">
<tr>
<td width="130"><b>DSL Number</b></td>
<td width="168"><input type="text" name="dsl" size="20"></td>
</tr>
<tr>
<td width="130"><b>First Name</b></td>
<td width="168"><input type="text" name="First" size="20"></td>
</tr>
<tr>
<td width="130"><b>Last Name</b></td>
<td width="168"><input type="text" name="Last" size="20"></td>
</tr>
<tr>
<td width="130"><b>Civil ID</b></td>
<td width="168"><input type="text" name="cid" size="20"></td>
</tr>
<tr>
<td width="130"><b>Net Pack</b></td>
<td width="168"><select size="1" name="choice">
<option value = "2 Mbps">2 Mbps</option>
<option value = "5 Mbps">5 Mbps</option>
<option value = "10 Mbps">10 Mbps</option>
<option value = "15 Mbps">15 Mbps</option>
</select></td>
</tr>
</table>
</div>
<p align="center"><input type="submit" value="Search" name="B1">
<p align="center"><input type="submit" value="Modify" name="B2">
<input type="reset" value="Reset" name="B2"></p>
</form>
<p align="center"> </p>
</body>
</html>
非常感謝你Luki Centuri。這工作。我忘了用單引號括住字符串值。再次感謝！ – YusufSM
不客氣:) –