如何在PHP中創建到頁面的動態鏈接？

Question

我正在寫一個代碼來創建鏈接到一個視頻頁面的電影播放trailers.php從

原始鏈接文件cinematics.php。下面是代碼：

<?php 
$mysqli = mysqli_connect("localhost", "root", "ecncyogclone2012", "4cytedb"); 

if (mysqli_connect_errno()) { 
    printf("Connect failed: %s\n", mysqli_connect_error()); 
    exit(); 
} 
else { 
    $sql = "SELECT * FROM cinematable ORDER BY id DESC LIMIT 0,5"; 
    $res = mysqli_query($mysqli, $sql); 


    if ($res) { 
     while ($newArray = mysqli_fetch_array($res, MYSQLI_ASSOC)) { 
      $id = $newArray['id']; 
      $dateField = $newArray['cinemaDateCreat']; 
      $titleField = $newArray['cinemaTitle']; 
      $descField = $newArray['cinemaDesc']; 
      $imgField = $newArray['cinemaImg']; 
      $linkField = $newArray['cinemaLink']; 
      $display_block = "<img src=\"media/picture/".$imgField."\" width= 

\"150\" height=\"120\" alt=\"image\" />"; 

         //this next line is where I get an error 

      echo "<a href=\"cinematics-play-trailers.php?video=".$id 

['video'].">".$display_block."</a>"; 

      } 
    } 

else { 
     printf("Could not retrieve records: %s\n", mysqli_error($mysqli)); 
    } 

    mysqli_free_result($res); 
    mysqli_close($mysqli); 
} 
?> 

，這就是我對動畫播放，trailers.php頁：

<?php 

    if(isset($_GET['video'])) 
    { 
    $curr_vid_id = $_GET['video']; 

    $mysqli = mysqli_connect("localhost", "root", "ecncyogclone2012", "4cytedb"); 

    if (mysqli_connect_errno()) { 
     printf("Connect failed: %s\n", mysqli_connect_error()); 
     exit(); 
    } 
    else { 
     $sql = "SELECT * FROM cinematable ORDER BY id WHERE id = $curr_vid_id "; 
     $res = mysqli_query($mysqli, $sql); 


     if ($res) { 
      while ($newArray = mysqli_fetch_array($res, MYSQLI_ASSOC)) { 
       $id = $newArray['id']; 
       $dateField = $newArray['cinemaDateCreat']; 
       $titleField = $newArray['cinemaTitle']; 
       $descField = $newArray['cinemaDesc']; 
       $imgField = $newArray['cinemaImg']; 
       $linkField = $newArray['cinemaLink']; 

      echo 
      "<video src=\"media/$linkField\" Controls width=\"690\" ></video> 
      <br> 
      <a href=\"media/$linkField\"><img name=\"download_this_file\" 

src=\"images/download_this_file.gif\" width=\"165\" height=\"50\"></a> 
      <a href=\"http://www.youtube.com/4cyteworld\"><img src= 

\"images/play_on_youtube.fw.png\" width=\"114\" height=\"50\"></a>"; 


       } 
     } 

    else { 
      printf("Could not retrieve records: %s\n", mysqli_error($mysqli)); 
     } 

     mysqli_free_result($res); 
     mysqli_close($mysqli); 
     } 
    } 
    else { 
      echo "Cant display Video Now !"; 

      } 

    ?> 

我需要得到正確的代碼。

Answer 1

你不關閉在這裏href屬性引號：

echo "<a href=\"cinematics-play-trailers.php?video=".$id['video'].">".$display_block."</a>"; 

編輯是這樣的：

echo "<a href=\"cinematics-play-trailers.php?video=".$id['video']."\" >".$display_block."</a>";