尝试只更新表单中一次收到的字段，其他字段在PHP中变为空白

我正在尝试创建一个更新PHP函数，它正在工作，但我遇到了一些问题，我收到了来自表单的用户名电子邮件和密码，但如果用户将其中一个或两个留空，我想更新唯一一个没有留空的函数，例如“pass”，但当我这样做时，其他两个字段将更新为nothing或数据库中的blank，我的代码中有哪些错误

<?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();



 ?>

---

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：

$setters = array();
if ($trim($email) != '') $setters[] = "use_email = '$email'";
if ($trim($nome)  != '') $setters[] = "use_name = '$nome'";
if ($trim($pass)  != '') $setters[] = "use_pass = '$pass'";
$query  = "UPDATE users SET ".implode(", ",$setters)." WHERE use_id = '$id'";
$result = mysqli_query($con,$query);

---

因此，查询取决于输入的存在

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：

$setters = array();
if ($trim($email) != '') $setters[] = "use_email = '$email'";
if ($trim($nome)  != '') $setters[] = "use_name = '$nome'";
if ($trim($pass)  != '') $setters[] = "use_pass = '$pass'";
$query  = "UPDATE users SET ".implode(", ",$setters)." WHERE use_id = '$id'";
$result = mysqli_query($con,$query);

---

因此，查询取决于输入的存在

首先要更新，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();



 ?>

---

首先要更新，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();



 ?>

您将进行一个仅包含要更新的字段的查询。您可以检查是否输入了这样的内容：
if（trim（$pass）！=''…
。我真的希望这段代码永远不会进入公共系统，因为它对SQL注入是开放的。搜索这个词来了解更多。这是一个学校项目，老师并没有真正教我们任何东西，这不是为了学习company@KIKOSoftware它成功了，你能把它作为一个答案贴出来让我接受吗？谢谢哇，太快了。我将尝试回答。您将进行一个仅包含要更新的字段的查询。您可以检查是否输入了这样的内容：
if（trim（$pass）！=''…
。我真的希望这段代码永远不会进入公共系统，因为它对SQL注入是开放的。搜索这个词来了解更多。这是一个学校项目，老师并没有真正教我们任何东西，这不是为了学习company@KIKOSoftware它成功了，你能把它作为一个答案贴出来让我接受吗？谢谢哇，太快了。我要试着回答。