Question

我的代码出了什么问题，因为插入工作正常，但更新无效。我已经从phpmyadmin复制了updaet的查询，并将静态值更改为变量

<?php
if (isset($_POST['submitContactInfo'])) {
    $socityId = $_SESSION['socityid'];
    $city = $_POST['city'];
    $pin = $_POST['pin'];
    $state = $_POST['state'];
    $telephone = $_POST['telephone'];
    $mobile = $_POST['mobile'];
    $email = $_POST['email'];
    $address = $_POST['address'];

    $sql = "INSERT INTO `_abc1`.`profile` (SOCIETY_ID,ADDR,CITY,PIN,STATE,TEL,MOBILE,EMAIL) VALUES ('$socityId','$address','$city','$pin','$state','$telephone','$mobile','$email');";


    $sql = "UPDATE `_abc1`.`profile` SET `ADDR` = '$address', `CITY` = '$city', `PIN` = '$pin', `STATE` = '$state', `TEL` = '$telephone', `MOBILE` = '$mobile', `EMAIL` = '$email' WHERE `society_profile`.`SOCITY_ID` = '$socityId'; ";

    $res = mysql_query($sql);
}
?>

Answer 1

您的更新中有拼写错误。在第一个查询中，您有SOCITY_ID，它是SOCIETY_ID

试

$sql = "UPDATE `_abc1`.`profile` SET `ADDR` = '$address', `CITY` = '$city', `PIN` = '$pin', `STATE` = '$state', `TEL` = '$telephone', `MOBILE` = '$mobile', `EMAIL` = '$email' WHERE `SOCIETY_ID` = '$socityId'; ";

此外，如上所述，您应该使用mysqli_query而不是这些旧的弃用函数。

http://www.php.net/manual/en/mysqli.query.php

Answer 2

首先，您应该考虑使用mysqli或PDO而不是标准的mysql类。
其次，您还没有加入表格，那么您希望它与您尝试查询的两个表格相匹配？

Answer 3

试试这个..

$sql = "UPDATE '_abc1'.'profile' SET 'ADDR' = '".$address."', 'CITY' = '".$city."', 'PIN' = '".$pin."', 'STATE' = '".$state."', 'TEL' = '".$telephone."', 'MOBILE' = '".$mobile."', 'EMAIL' = '".$email."' WHERE 'society_profile'.'SOCITY_ID' = '".$socityId."'; ";

php和mysql - 插入工作但更新无法正常工作

3 个答案: