2013-12-09 37 views
0

本質上,我創建了一個有7種電影流派的電影訂購系統,並且我希望每個電影標籤都根據流派進行填充。此時我甚至不能讓桌子出現。誰能幫我嗎?使用MySql來使用特定列填充html表格

<div id="content"> 
    <table> 
    <tr><th>Name</th> <th>Genre</th> <th>Year</th> <th>Rating</th></tr> 
<?php 
//connect to database 
$dbc = mysql_connect('localhost' , 'username' , 'password'); 
$test = mysql_select_db('movies', $dbc); 

if ($test = mysql_select_db('movies', $dbc))//test 
{ 
    $query = "SELECT * FROM 'movies' WHERE 'genre' = 'Action'"; 

    //call query 
    $result = mysql_query($query, $dbc); 

    while ($row = mysql_fetch_array($result)) 
    { 
     ?> 

     <tr> 
     <td><?php print $row['name']; ?></td> 
     <td><?php print $row['genre']; ?></td> 
     <td><?php print $row['year']; ?></td> 
     <td><?php print $row['rating'];?></td> 
     </tr> 
<table> 

     <?php 

    }//close the while loop 

}//close the if statement 

mysql_close($dbc); 

?> 
+1

你可能得到一個MySQL錯誤。添加或死亡(mysql_error())到你的mysql語句的結尾,如下所示:$ result = mysql_query($ query,$ dbc)或die(mysql_error()); – Jade

回答

0

首先,不要使用mysql,因爲存在安全問題,它將不會在PHP後期支持。閱讀關於http://php.net/manual/en/function.mysql-connect.php

嘗試使用PDO或mysqli的,例如:

$con=mysqli_connect("example.com","peter","abc123","my_db"); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

$result = mysqli_query($con,"SELECT * FROM 'movies' WHERE 'genre' = 'Action'"); 

while($row = mysqli_fetch_array($result)) 
    { 
    echo $row['name']; 
    echo "<br>"; 
    } 

mysqli_close($con); 
+0

謝謝!我很感謝你的迴應! – user3084249