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问如何解决回送标记URL
EN

Stack Overflow用户
提问于 2016-06-27 01:29:09
回答 2查看 58关注 0票数 1

我如何解决这个URL?

我在用回声来叫出一个标签。它运行良好,但URL包含一些我不知道如何解决它的内容?有人能跟我解释一下吗?为什么没有URL会有thoes %20%20%

下面是我回音的URL

代码语言:javascript
复制
http://localhost/pme/main/directMessageRoom.php?directMessageRoomID=1></a>%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%3Cform%20method=

这是我的密码

代码语言:javascript
复制
<div class ="direct-message-btn-container ">
        <?php 

   include '../config.php';
    if(isset($_SESSION['id'])){
    $target = $_GET['user'];
    $id = $_SESSION['id'];

    $targetusername = mysqli_query($connection,"SELECT username FROM user WHERE id='$target' ")or die(mysqli_error($connection));
    $username=  mysqli_query($connection,"SELECT username FROM user WHERE id='$id'") or die(mysqli_error($connection));

    $dm_sql="SELECT * FROM direct_message_room WHERE user_1_id ='$id' AND user_2_id='$target' OR user_1_id='$target' AND user_2_id='$id'";
    $check_direct_message_room = mysqli_query($connection, $dm_sql) or die(mysqli_error($connection));

    $check=  mysqli_num_rows($check_direct_message_room);

    if($check == 0){

       $rowUserName=mysqli_fetch_array($username);
       $rowTargetUserName=mysqli_fetch_array($targetusername);
       $room_name=$rowUserName['username']. ' and ' . $rowTargetUserName['username'];

    $direct_message_room_name ="INSERT INTO direct_message_room(`direct_message_room_name`,`user_1_id`,`user_2_id`) VALUES ('$room_name','$id','$target')";

    $newDirectMessageRoom = mysqli_query($connection,$direct_message_room_name)or die(mysqli_error($connection));
    $direct_message_room_id = mysqli_insert_id($connection); 

    echo '<a  href="directMessageRoom.php?directMessageRoomID='.$direct_message_room_id.'></a>';       
    }
    else{
        while($row3 = mysqli_fetch_array($check_direct_message_room)){
             echo '<a  href="directMessageRoom.php?directMessageRoomID='.$row3['id'].'></a>';
        }                        
    } 
?>

    <form method="post" action="" role="form">
        <div class="form-group ">
            <input type = "submit" class="btn content-btn btn-direct-message" value="message" name="chat">
        </div>
    </form>
</div>
EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2016-06-27 01:38:24

您忘记关闭href属性(双引号):

代码语言:javascript
复制
echo '<a href="directMessageRoom.php?directMessageRoomID='.$row3['id'].'"></a>';
                                                          right here ---^
票数 2
EN

Stack Overflow用户

发布于 2016-06-27 04:39:05

注意表单字段中的许多“空白”。例如,您的提交按钮编写了这个<input type = "submit" ...>。您不小心插入了空白。应该是<input type="submit" ...>。

票数 1
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/38044763

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