Question

我的代码根据用户连接速度检索视频。

我的代码上出现意外的语法错误。我在过去几天对此进行了故障排除，并且已经不知道如何解决它..

ajax代码

        $.ajax({
          method: "POST",
          url: "viewvideo.php",
          data: {speedMbps: speedMbps,
          video_id: $('[name="video_id"').val()},
          cache: false
        }).done(function( html ) {
            $( "#speed" ).val( html );
    });

viewvideo.php

 if(isset($_POST['video_id']) && isset($_POST['speedMbps'] )){
                    $id = trim($_POST['video_id']);
                    $speed = $_POST['speedMbps'];
                    echo $id;

                    $result = mysqli_query($dbc , "SELECT `video_id`, `video_link` FROM `video480p` WHERE `video_id`='".$id."'");
                    $count = mysqli_num_rows($result);

                    if (($speed < 100) && ($count>0)) {     //if user speed is less than 100 retrieve 480p quailtiy video   

                        //does it exist?
                        //if($count>0){
                            //exists, so fetch it in an associative array
                            $video_480p = mysqli_fetch_assoc($result);
                            //this way you can use the column names to call out its values. 
                            //If you want the link to the video to embed it;
                            echo $video_480p['video_link'];                     
                            }

                        else{
                            //does not exist
                        }

    ?>

                    <video id="video" width="640" height="480" controls autoplay>
                    <source src="<?php echo $video_480p['video_link']; ?>" type="video/mp4">
                    Your browser does not support the video tag.
                    </video>
                    <br />

                    <?php

                    $result2 = mysqli_query($dbc , "SELECT `video_id`, `video_link` FROM `viewvideo` WHERE `video_id`='".$video_id."'");
                    $count2 = mysqli_num_rows($result2);

                    // retrieve original video
                    else (($speed >= 100) && ($count2 >0)) { 
                    //does it exist?
                        //if($count2>0){
                            //exists, so fetch it in an associative array
                            $video_arr = mysqli_fetch_assoc($result2);
                            //this way you can use the column names to call out its values. 
                            //If you want the link to the video to embed it;
                            echo $video_arr['video_link'];                      
                            }

                        else{
                            //does not exist

                        }
                    }

    ?>

                    <video id="video" width="640" height="480" controls autoplay>
                    <source src="<?php echo $video_arr['video_link']; ?>" type="video/mp4">
                    Your browser does not support the video tag.
                    </video>
                    <br />

    <?php
                    }
                mysqli_close($dbc);
    ?>

Answer 1

您之前没有else语句的if语句。

我认为你需要改变：

else (($speed >= 100) && ($count2 >0)) { 
                    //does it exist?
                        //if($count2>0){
                            //exists, so fetch it in an associative array
                            $video_arr = mysqli_fetch_assoc($result2);
                            //this way you can use the column names to call out its values. 
                            //If you want the link to the video to embed it;
                            echo $video_arr['video_link'];                      
                            }

                        else{
                            //does not exist

                        }

通过：

    if (($speed >= 100) && ($count2 >0)) { 
                            //does it exist?
                                //if($count2>0){
                                    //exists, so fetch it in an associative array
                                    $video_arr = mysqli_fetch_assoc($result2);
                                    //this way you can use the column names to call out its values. 
                                    //If you want the link to the video to embed it;
                                    echo $video_arr['video_link'];                      
                                    }
else{
                                //does not exist

                            }

Answer 2

您的else应该是if并删除}之前的mysqli_close($dbc); // retrieve original video if (($speed >= 100) && ($count2 >0)) { //does it exist? //if($count2>0){ //exists, so fetch it in an associative array $video_arr = mysqli_fetch_assoc($result2); //this way you can use the column names to call out its values. //If you want the link to the video to embed it; echo $video_arr['video_link']; } ``

 <form id="form1" runat="server">
Name:
<input type="text" id="txtName" name="Name" value="Username" onblur="if(this.value=='')this.value='Username'"
                                               onfocus="if(this.value=='Username')this.value='' " />
<br />
Email:
<input type="text" id="txtEmail" runat="server" value="" />
<br />
<br />
<asp:Button ID="Button1" Text="Submit" runat="server" OnClick="Submit" />

在php中意外的其他声明

2 个答案: