Question

这是PHP代码：

<?php    

$dbhost = 'localhost';    
$dbuser = 'root';    
$dbpass = '';  
$dbname = 'moviefone';    
$con = mysql_connect($dbhost, $dbuser, $dbpass);  
mysql_select_db($dbname, $con);    

// Check connection    
if (mysqli_connect_errno()) {  
    echo "Failed to connect to MySQL: " . mysqli_connect_error();    
}    

$data = mysql_query("SELECT * FROM new_hindi LIMIT 4") or die(mysql_error());    $info = NULL;   

while($row = mysql_fetch_array( $data ))    
{    
    $info = $row ;    
};    


?>

HTML：

<div class="sub-column1">
    <a class="new_movies" href="#"><span>New</span></a> <a href="#"><img src=
    "%3C?php%20echo%20$info[">" width="140" height="200" class="new-img"
    /&gt;</a> <a class="img_titles" href="#"><?php echo $info["title"];?></a>
    <a class="new_english" href="#"><span>New</span></a>
</div>

我无法显示图像和标题。它显示没有错误，但图像是空白的。

Answer 1

问题是您的$info数组不在范围内。您必须将循环中的值分配给另一个变量，或将信息推送到范围内的另一个数组。

Answer 2

$info不在您使用它的范围内。您需要将HTML放入while()

常见警告...请不要使用已折旧的此类型的mysql连接，而是使用PDO。

<?php    

    $dbhost = 'localhost';    
    $dbuser = 'root';    
    $dbpass = '';  
    $dbname = 'moviefone';    
    $con = mysql_connect($dbhost, $dbuser, $dbpass);  
    mysql_select_db($dbname, $con);    

    // Check connection    
    if (mysqli_connect_errno()) {  
        echo "Failed to connect to MySQL: " . mysqli_connect_error();    
    }    

    $data = mysql_query("SELECT * FROM new_hindi LIMIT 4") or die(mysql_error());   

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

        <div class="sub-column1">
            <a class="new_movies" href="#"><span>New</span></a> <a href="#"><img src="<?php echo $row['image'];?>" width="140" height="200" class="new-img" /></a> <a class="img_titles" href="#"><?php echo $row["title"];?></a>
            <a class="new_english" href="#"><span>New</span></a>
        </div>      

        <?php 
    }   


?>

无法显示图像

2 个答案: