Question

我有一个简单的登录，当用户登录时，他们可以上传标题和评论（以及图像，但我们可以忽略它，因为这可以正常）到数据库。我想要的是当前登录的人的用户ID也插入到数据库中。这是我目前的代码

的login.php

<?php
session_start();
if(isset($_SESSION["user"])){
    header("location:index.php");
    exit();
}

?>

<?php

if(isset($_POST["username"]) && isset($_POST["password"])){
    // Filter everything except letters and numbers
    $user = preg_replace('#^A-Za-z0-9#i','',$_POST["username"]);
    $password = preg_replace('#^A-Za-z0-9#i','',$_POST["password"]);

    // Connect to mySQL
    include"assets/scripts/sql_connect.php";

    // Query the person
    $sql = $conn->query("SELECT id FROM user_login WHERE user_name='$user' AND user_password='$password' LIMIT 1");

    // Make sure person exists
    $existCount = $sql->rowCount();

    // Evaluate the count
    if($existCount == 1){
        while($row = mysql_fetch_array($sql)){
            $id = $row["id"];
        }
        $_SESSION["id"] = $id;
        $_SESSION["user"] = $user;
        $_SESSION["password"] = $password;
        header("location:index.php");
        exit();
    }else{
        echo"Login details incorrect, try again <a href='index.php'>Click here</a>";
        exit();
    }
}
?>

upload.php的

<?php
session_start();

include"assets/scripts/sql_connect.php";

if (isset($_POST['image_title'])){

    $userid =  $_SESSION["id"];
    $image_title = $_POST['image_title'];
    $image_comment = $_POST['image_comment'];


    //Add image text to the database
    $sql = $conn->query("INSERT INTO user_image(user_id,image_title,image_comment, image_date_added) VALUES('$userid','$image_title','$image_comment', now())") or die(mysql_error());
    $id = $conn->lastInsertId();

    $image_id = $conn->lastInsertID();  
    // Places image in the images folder
    $new_name = "$image_id.jpg";
    move_uploaded_file($_FILES['app_art_image']['tmp_name'],"appArtImages/$new_name");
    header("location:gallery.php");
    exit();
}       
?>

我得到的问题是，无论用户登录了什么，都会在user_image数据库的user_id字段中插入0。

Answer 1

使用mysql_fetch_assoc()代替mysql_fetch_array() 试试

 if($existCount == 1){
            while($row = mysql_fetch_assoc($sql)){
                $id = $row["id"];
                $_SESSION["id"] = $id;
                break;
            }
            $_SESSION["user"] = $user;
            $_SESSION["password"] = $password;
            header("location:index.php");
            exit();
        }else{
            echo"Login details incorrect, try again <a href='index.php'>Click here</a>";
            exit();
        }

希望它有所帮助：）

Answer 2

有人解决它。我改变了

while($row = mysql_fetch_array($sql)){

在login.php中

while($row = $result->fetch(PDO::FETCH_ASSOC)){

使用$ _SESSION

2 个答案: