在您標記我的問題之前重複這是我的最後手段。我已嘗試過每種可能的語法沒有錯誤,但我不能更新表,雖然我可以插入表,但不能更新已經存在的內容。 config.php用於與db進行連接。不能更新表使用PHP嘗試每種可能的組合語法
<html>
<head>
<title>
Thank You FOR Entering Data !!
</title>
</head>
<style>
div {
width:100%;
position: absolute;
bottom:0;
}
img {
position: relative;
display: block;
float: right;
}
</style>
<body background="wall.jpg">
<font face="WenQuanYi Micro Hei Mono" color="white"><b>
<h1>Thank You For Entering Data</h1>
<?php
include("config.php");
// Data coming into form via POST
if ($_POST) {
$empide = $_POST["empid"];
$name = $_POST["name"];
$address = $_POST["address"];
$pnumber = $_POST["pnumber"];
$email = $_POST["email"];
$gender = $_POST["gender"];
if ($con) {
echo nl2br("empid= $empide \nNAME = $name\n Address = $address \n Phone Number = $pnumber \n Email = $email \n Gender = $gender ");
echo $empide;
//$abc=mysqli_query($con,"INSERT INTO `employee` (`name`, `address`, `pnumber`, `email`,`gender`) VALUES ('$name', '$address', '$pnumber', '$email','$gender') ;");
$query = mysqli_query($con, "UPDATE 'employee' SET 'name' = '$name', 'address' = '$address', 'email' = '$email', 'pnumber' = '$pnumber', 'gender' = '$gender' WHERE 'employee'.'empid' = '$empide';");
} else {
echo "Unable to make connection with your data base";
}
}
?>
<right>
<div>
<a href="index.php"><img src="home.png" width="110" height="110"></a>
</div>
</right>
</body>
</html>
打印您的$ CON什麼,它會給你? – RK12
您應該在開發模式下開啓您的error_reporting。 –
不要在表名或列中使用簡單的引號,只能在值中使用! – rray