doufu9521 2017-08-07 16:55
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如何从数据库中显示登录用户的图像?

Say if a user uploaded multiple images to their profile page, how would I display that specific user's images? I tried I did something like this:

$db = mysqli_connect("localhost", "root", "", "testdb");
        $sql = "SELECT * FROM users ORDER BY id DESC";
        $result = mysqli_query($db, $sql);
        while ($row = mysqli_fetch_array($result)) {
            echo "<a href='profiles/uploads/".$row['image']."> ";
                echo "<img  id='img_div' title='".$row['image']."' alt='".$row['image']."'  src='profiles/uploads/".$row['image']."'/>";
                //echo "<p id='img_div'>".$row['desc']."</p>";
                echo "</a>";

But I feel like this is terribly wrong because it is showing everyone's images and not the user's images. I tried looking up answers, but can't seem to find one.

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  • duanao2688 2017-08-07 16:57
    关注

    Try this:

    You need to set user in sql query like id = 10.

    And also missing ' last of this line: echo "<a href='profiles/uploads/".$row['image'].">";

    <?php
    $user_id = 10;
    $db = mysqli_connect("localhost", "root", "", "testdb");
    $sql = "SELECT * FROM users WHERE id = ". $user_id ." LIMIT 1";
    $result = mysqli_query($db, $sql);
    
    while ($row = mysqli_fetch_array($result)) {
        echo "<a href='profiles/uploads/". $row['image'] ."'>";
        echo "<img id='img_div' title='".$row['image']."' alt='".$row['image']."' src='profiles/uploads/".$row['quotes']."'/>";
        //echo "<p id='img_div'>".$row['desc']."</p>";
        echo "</a>";
    }
    ?>
    
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