Php MYSQL更新多个表错误
我正在尝试更新我正在制作的网站上的成员详细信息,但我不断从更新事务中收到此错误。 “您的SQL语法有错误;请检查与您的MySQL服务器版本相对应的手册,以了解第1行的near'和member.memberID='6''所使用的正确语法” 我尝试了不同的编码方法,但错误仍然是一样的Php MYSQL更新多个表错误,php,mysql,sql,forms,syntax-error,Php,Mysql,Sql,Forms,Syntax Error,我正在尝试更新我正在制作的网站上的成员详细信息,但我不断从更新事务中收到此错误。 “您的SQL语法有错误;请检查与您的MySQL服务器版本相对应的手册,以了解第1行的near'和member.memberID='6''所使用的正确语法” 我尝试了不同的编码方法,但错误仍然是一样的 <?php session_start(); include "../includes/connect.php"; ?> <?php $memberID = $_POST['memberI
<?php
session_start();
include "../includes/connect.php";
?>
<?php
$memberID = $_POST['memberID']; //retrieve the memberID from the URL
$firstName = mysqli_real_escape_string($con, $_POST['firstName']);
$lastName = mysqli_real_escape_string($con, $_POST['lastName']);
$street = mysqli_real_escape_string($con, $_POST['street']);
$suburb = mysqli_real_escape_string($con, $_POST['suburb']);
$state = mysqli_real_escape_string($con, $_POST['state']);
$postcode = mysqli_real_escape_string($con, $_POST['postcode']);
$country = mysqli_real_escape_string($con, $_POST['country']);
$phone = mysqli_real_escape_string($con, $_POST['phone']);
$mobile = mysqli_real_escape_string($con, $_POST['mobile']);
$email = mysqli_real_escape_string($con, $_POST['email']);
$gender = mysqli_real_escape_string($con, $_POST['gender']);
$newsletter = mysqli_real_escape_string($con, $_POST['newsletter']);
if ($firstName == "" || $lastName == "" || $postcode == "" || $country =="" || $email == "" || $gender == "" || $newsletter == "") //check if all required fields have data
{
$_SESSION['error'] = 'All * fields are required.'; //if an error occurs intialise a session called 'error' with a msg
header("location:memberupdate.php?memberID=" . $memberID); //redirect to memberupdate.php
exit();
}
elseif (!filter_var($email, FILTER_VALIDATE_EMAIL)) //check if email is valid
{
$_SESSION['error'] = 'Please enter a valid email address.'; //if an error occurs intialise a session called 'error' with a msg
header("location:memberupdate.php?memberID=" . $memberID); //redirect to memberupdate.php
exit();
}
else
{
$sql="START TRANSACTION;
UPDATE member SET firstName='$firstName',lastName='$lastName',phone='$phone',mobile='$mobile',email='$email',gender='$gender',newsletter='$newsletter' WHERE memberID='$memberID';
UPDATE address SET street='$street',suburb='$suburb',state='$state',postcode='$postcode',country='$country' WHERE memberID='$memberID';
COMMIT;";
$result = mysqli_query($con, $sql) or die(mysqli_error($con)); //run the query
$_SESSION['success'] = 'Member updated successfully.'; //if the member is updated successfully intialise a session called 'success' with a msg
header("location:memberupdate.php?memberID=" . $memberID); //redirect to memberupdate.php
}
?>
请帮忙 表中的memberID字段是int而不是varchar吗?如果是这样,那么您的代码应该这样说(在$memberID周围没有撇号):
我认为您需要使用这个
mysqli\u多\u查询
...WHERE memberID=$memberID"