gpt4 book ai didi

php - 记录不会用 PHP 更新 MySQL

转载 作者:行者123 更新时间:2023-11-29 01:16:36 25 4
gpt4 key购买 nike

我这辈子都想不通为什么下面的记录不会更新。它给了我一条成功消息,但实际上并没有更新记录。我经历了大量的修改并尝试了我能想到的一切。其他人能看出问题出在哪里吗?

<?php
$id = $_POST['id'];
$username = $_POST['username'];
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$email = $_POST['email'];
$company = $_POST['company'];
$rep1 = $_POST['rep1'];
$rep2 = $_POST['rep2'];
$rep3 = $_POST['rep3'];
$phone = $_POST['phone'];
$fax = $_POST['fax'];
$cell = $_POST['cell'];
$service = $_POST['service'];
$license = $_POST['license'];
$expdate = $_POST['expdate'];
$active = $_POST['active'];
$userlevel = $_POST['userlevel'];

$host="XXXXXXX"; // Host name
$username="XXXXXX"; // Mysql username
$password="XXXXXXX"; // Mysql password
$db_name="XXXXXXX"; // Database name
$tbl_name="users"; // Table name

// Connect to server and select database.
$con = mysqli_connect("$host", "$username", "$password","$db_name")or die("cannot connect");


// update data in mysql database
$sql="UPDATE $tbl_name SET id='".$id."', fname='".$fname."', lname='".$lname."', email='".$email."', company='".$company."', rep1='".$rep1."', rep2='".$rep2."', rep3='".$rep3."', phone='".$phone."', fax='".$fax."', cell='".$cell."', service='".$service."', license='".$license."', expdate='".$expdate."', active='".$active."', userlevel='".$userlevel."'".
"WHERE id = '".$id."'";
$result=mysqli_query($con,$sql);

// if successfully updated.
if($result){
$emailID = "info@domain.com";
$subject = "Registration notification from. $fname . through website";
$body = <<<EOD

<table cellspacing="0" cellpadding="1" border="1">
<tbody>
<tr>
<td style="padding: 5px 10px;" width="150">Name: </td>
<td style="padding: 5px 10px;">$fname $lname</td>
</tr>
<tr>
<td style="padding: 5px 10px;" width="150">Mobile: </td>
<td style="padding: 5px 10px;">$cell</td>
</tr>
<tr>
<td style="padding: 5px 10px;" width="150">Email: </td>
<td style="padding: 5px 10px;">$email</td>
</tr>
<tr>
<td style="padding: 5px 10px;" width="150">License Number: </td>
<td style="padding: 5px 10px;">$license</td>
</tr>
</tbody>
</table>

EOD;

$headers = "From: info@domain.com\r\n";
$headers .= "MIME-Version: 1.0\r\n";
$headers .= "Content-Type: text/html; charset=\"iso-8859-1\"\n";
$headers .= "X-Priority: 1\r\n";
$headers .= "X-MSMail-Priority: High\n";
$headers .= "X-Mailer: PHP". phpversion() ."\r\n";

mail($emailID, $subject, $body, $headers );
echo "<h4>Thank you for updating your account info. We will authorize your account and notify you once we have verified your license number and expiration.</h4>";
}
else {
echo "ERROR";
}

?>

最佳答案

您的最后一个值后面没有空格,因此 where 关键字“卡在”了它。只需在 where 之前添加一个空格,就可以了:

$sql="UPDATE $tbl_name SET id='".$id."', fname='".$fname."', lname='".$lname."', email='".$email."', company='".$company."', rep1='".$rep1."', rep2='".$rep2."', rep3='".$rep3."', phone='".$phone."', fax='".$fax."', cell='".$cell."', service='".$service."', license='".$license."', expdate='".$expdate."', active='".$active."', userlevel='".$userlevel."'".
" WHERE id = '".$id."'"; // Note the additional space here

关于php - 记录不会用 PHP 更新 MySQL,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25652096/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com