Question

我正在尝试创建一个更新PHP函数，它正在工作，但是我遇到了一些问题，我收到了来自表单的用户名电子邮件和密码，但是如果用户将其中一个或两个留空，我想更新只有一个不是空白的，例如“ pass”，但是当我这样做时，其他2个字段在数据库中都更新为空或为空白，我的代码有什么错误？

<?php 

    session_start();

    $id = $_SESSION['user_id'];
    $email =  $_POST['emailAlter'];
    $nome = $_POST['nameAlter'];
    $pass = $_POST['passAlter'];


    $con = mysqli_connect("localhost", "root", "", "smarttime");
    $query = mysqli_query($con,"UPDATE users SET use_name = '$nome', use_email = '$email', use_pass = '$pass' WHERE use_id = '$id'");

    $test = mysqli_query($con,"SELECT * from users");

    $row = mysqli_fetch_array($test);   

    if (!$con) {
        die('Erro de Acesso à BD' . mysqli_connect_error());
    }

    if(!isset($email) || trim($email) == '')
    {
        $email = mysqli_query($con,"SELECT use_email from users where use_id ='$id");
        $query = mysqli_query($con,"UPDATE users SET use_name = '$nome', use_email = '$email', use_pass = '$pass' WHERE use_id = '$id'");
    }

    if(!isset($nome) || trim($nome) == '')
    {
        $nome = mysqli_query($con,"SELECT use_name from users where use_id ='$id");
        $query = mysqli_query($con,"UPDATE users SET use_name = '$nome', use_email = '$email', use_pass = '$pass' WHERE use_id = '$id'");
    }

    if(!isset($pass) || trim($pass) == '')
    {
        $nome = mysqli_query($con,"SELECT use_pass from users where use_id ='$id");  
        $query = mysqli_query($con,"UPDATE users SET use_name = '$nome', use_email = '$email', use_pass = '$pass' WHERE use_id = '$id'");
    }

        $query;
        header('Location: logged.php');
        exit();



 ?>

Answer 1

如果我假设用户ID始终存在并且有效，那么您可以构建如下查询：

globals()[name]

因此，查询取决于输入的存在。

Answer 2

事情是先更新然后检查条件。更新之前必须检查所有变量

<?php 

    session_start();

    $id = $_SESSION['user_id'];
    $email =  $_POST['emailAlter'];
    $nome = $_POST['nameAlter'];
    $pass = $_POST['passAlter'];


    $con = mysqli_connect("localhost", "root", "", "smarttime");
    $query = "UPDATE users SET "; 

    if (!$con) {
        die('Erro de Acesso à BD' . mysqli_connect_error());
    }

    if(isset($email) && trim($email) != '')
    {
        $query .= "use_email = '$email' ";
    }

    if(isset($nome) && trim($nome) != '')
    {
        $query .= "use_name = '$nome' ";
    }

    if(isset($pass) && trim($pass) != '')
    {
        $query .= "use_pass = '$pass' ";
    }

        mysqli_query($con,$query." WHERE use_id = '$id'");;
        header('Location: logged.php');
        exit();



 ?>

尝试仅更新一次从表单收到的字段，而其他字段在PHP中变为空白

2 个答案: