使用php从mysql数据库中的特定用户获取信息

时间:2017-06-07 01:51:06

标签: php mysql

当我输入数据库中的用户ID时,我希望它回显/打印来自该用户文件的所有数据。因此,如果我输入" user_id = 2",我希望用户"姓名","年龄"等等。我到了哪里,它给了我一切没有参数。代码:

$link = mysqli_connect("localhost", "root", "");
mysqli_select_db($link, "magicsever");

if(mysqli_connect_error()){
    die ("Database connection error");
}

$query = "SELECT * FROM classified_videos";
$result = mysqli_query($link, $query);
while($row = mysqli_fetch_array($result)){      
    print_r ($row);
}

但我希望来自特定用户,这就是我认为的代码

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

    $link = mysqli_connect("localhost", "root", "");
    mysqli_select_db($link, "magicsever");

    if(mysqli_connect_error()){     
        die ("Database connection error");
    }   

    $query = "SELECT * FROM classified_videos WHERE user_id ='".mysqli_real_escape_string($link, $_POST['userid'])."' LIMIT 1";
    $result = mysqli_query($link, $query);
    while($row = mysqli_fetch_array($result)){      
        print_r ($row['vid_category_1']);
    }
}
?>

<form method="post">
    <input type="text" name="userid" placeholder="user id...">
    <input type="submit" value="submit">
</form>

1 个答案:

答案 0 :(得分:1)

if(isset($_POST['submit']))期望提交名为submit的输入,但您目前仅发送userid

将此name="submit"添加到您的按钮:

<input type="submit" value="submit" name="submit">