Question

我正在尝试创建用户个人资料图片功能。到目前为止，我能够上传和选择图像并显示它，但我无法弄清楚如何选择特定于用户的图像。我的查询工作（虽然有上述问题），直到我添加下面的代码（//添加是我更改的）。

我能够得到我的插入查询来发送用户的user_id，我只是无法弄清楚选择部分。

我的数据库很小，只有id, user_id, img。

我想选择数据库中包含用户user_id的img。我带有一个会话的user_id，这就是$user_id。

任何人都会在我的选择查询中看到我做错了什么？

function getPhoto($con,$dest)
    {
        $user_id = ( isset( $_SESSION['user'] ) ? $_SESSION['user'] : "" ); //added

       // $result = mysqli_query($con,"SELECT * FROM `profile_img` where `img` = '$dest'");
        $result = mysqli_query($con,"SELECT * FROM `profile_img` where `img` = '$user_id'"); //added

        if($row = mysqli_fetch_array($result))
            return $row;

        return 0;
    }

编辑：更多代码。

function getPhoto($con,$dest)
    {
        $user_id = ( isset( $_SESSION['user'] ) ? $_SESSION['user'] : "" ); //added

       // $result = mysqli_query($con,"SELECT * FROM `profile_img` where `img` = '$dest'");
        $result = mysqli_query($con,"SELECT * FROM `profile_img` where `user_id` = '$user_id'"); //added

        if($row = mysqli_fetch_array($result))
            return $row;

        return 0;
    }


// Make sure all functions above are include here

// Get the database connection
$con        =   Connection();
// Check for post   
if(isset($_POST['create'])) {
        // Try uploading
        $upload =   UploadFile($_FILES);
        // If upload fails
        if(!$upload['success'])
            echo '<h3>Sorry, an error occurred</h3>';
        else {
                // You could add error handling here based on the results of 
                // each function's success or failure below.

                // Try to save it
                $saveToDb   =   SaveToDb($con,$upload['file']['dest']);
                // Get the profile from image name
                $profPic    =   ($saveToDb)? getPhoto($con,$upload['file']['dest']) : false; ?>


                <?php
            }
    }
?>
<img id="profile-pic" src="<?php echo (!empty($profPic) && $profPic != 0)? $profPic['img'] : "profile_images/default.jpg"; ?>" alt="<?php echo (!empty($profPic) && $profPic != 0)? "Profile Picture" : "No Picture"; ?>" />
<form action="" method="POST" enctype="multipart/form-data">
    <input type="file" name="file" class="inputbarfile" onchange="readURL(this);">
    <img width="400px" height="300px" id="file" src="#" alt="your image">
    <input type="submit" name="create" id="signinButton" value="Upload">
</form>

Answer 1

此：

 $result = mysqli_query($con,"SELECT * FROM `profile_img` where `img` = '$user_id'"); //added

应该是：

$result = mysqli_query($con,"SELECT * FROM `profile_img` where `user_id` = '$user_id'"); //added

由于img列包含image path而您正在比较user id。

选择要与列匹配的查询

1 个答案: