php - 一个数据库可以工作,而第二个数据库则不能

标签 php mysql database mysqli

我有一个数据库,其中有 4 个表,2 个用于私有(private)用户,2 个用于商业用户。由于某种原因,当我尝试使用业务用户的电子邮件登录时,它不起作用,但用户名起作用,并且在私有(private)表中它起作用,这是我的代码,如果我没有正确解释它,请告诉我,我会尽力解释再来一次

    $password = $_POST['password'];
    $emailuser = $_POST['unameemail'];
    $password = mysqli_real_escape_string($sql , $password); 
    $emailuser = mysqli_real_escape_string($sql , $emailuser); 
    $pwcheck = "
    SELECT * FROM private AS p 
    INNER JOIN user_private_data 
    AS c ON p.id = c.id 
    WHERE username='$emailuser' OR email='$emailuser'"; // part that works fine 
    $resultcheck = mysqli_query($sql , $pwcheck); // part that works fine 
    $rowcheck = mysqli_fetch_array($resultcheck , MYSQLI_ASSOC); // part that works fine 
    $hash = $rowcheck['password']; // part that works fine 
    $hash_pwd = password_verify($password , $hash);
    if ($hash_pwd != 0) {
        $_SESSION['username'] = $rowcheck['username']; // part that works fine  
        $_SESSION['logged'] = true; // part that works fine 
        header("refresh:0;url=../blablabla.php");     // part that works fine                   
    } else {
        $privateuser = "
        SELECT * FROM business AS d 
        INNER JOIN user_business_data 
        AS j ON d.id = j.id 
        WHERE username='$emailuser' OR email='$emailuser'"; // doesn't work
        $resultprivate = mysqli_query($sql , $privateuser); // doesn't work
            $rowprivate = mysqli_fetch_array($resultprivate , MYSQLI_ASSOC);
        $hashprivate = $rowprivate['password'];
        $hash_private = password_verify($password , $hashprivate);
        if ($hash_private != 0) {

            $_SESSION['username'] = $rowprivate['username'];
            $_SESSION['logged'] = true;
            $_SESSION['business'] = $rowprivate['bname'];
            $_SESSION['type'] = 'business';
} 

最佳答案

尝试一下:在获取之前您需要检查查询是否确实有结果 我假设变量 $sql 在您的连接中定义

<?php

$password = $_POST['password'];
$emailuser = $_POST['unameemail'];
$password = mysqli_real_escape_string($sql, $password);
$emailuser = mysqli_real_escape_string($sql, $emailuser);
$pwcheck = "
            SELECT * FROM private AS p 
            INNER JOIN user_private_data 
            AS c ON p.id = c.id 
            WHERE username='$emailuser' OR email='$emailuser'"; // part that works fine 
$resultcheck = mysqli_query($sql, $pwcheck); // part that works fine 
$rowcheck = mysqli_fetch_array($resultcheck, MYSQLI_ASSOC); // part that works fine 
$hash = $rowcheck['password']; // part that works fine 
$hash_pwd = password_verify($password, $hash);
if ($hash_pwd != 0) {
    $_SESSION['username'] = $rowcheck['username']; // part that works fine  
    $_SESSION['logged'] = true; // part that works fine 
    header("refresh:0;url=../blablabla.php");     // part that works fine                   
} else {
    $privateuser = "
                SELECT * FROM business AS d 
                INNER JOIN user_business_data 
                AS j ON d.id = j.id 
                WHERE username='$emailuser' OR email='$emailuser'"; // doesn't work
    $resultprivate = mysqli_query($sql, $privateuser); // doesn't work
    if ($resultprivate->num_rows > 0) {
        $rowprivate = mysqli_fetch_array($resultprivate, MYSQLI_ASSOC);
        $hashprivate = $rowprivate['password'];
        $hash_private = password_verify($password, $hashprivate);
        if ($hash_private != 0) {

            $_SESSION['username'] = $rowprivate['username'];
            $_SESSION['logged'] = true;
            $_SESSION['business'] = $rowprivate['bname'];
            $_SESSION['type'] = 'business';
        } else {
            //no record
        }
    }
?>

关于php - 一个数据库可以工作,而第二个数据库则不能,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42092561/

相关文章:

php - PHP中发生错误时如何获取所有变量

php - MOD_REWRITE 帮助!

mysql - 简单的MySQL数据库问题

database - 维护和更新IOS客户端与服务器之间的脱机更改的最佳方法是什么?

MYSQL 查询每周获取新客户

java - JDBC未插入int jdbc表

PHP windows创建隐藏文件

PHP & MySQLi - 在下拉列表中显示所选值两次

php - 有没有办法在 PHP 中执行 "forward trace"?

c++ - MySQL C++ 连接器 : Get the insert_id