我如何以正确的方式显示图像,当我将其放置在显示图像而不是不同图像时,当前很难一一显示。 这是我从数据库中选择详细信息后要在其中显示图像的代码。
PHP代码
<div class="panel-body" id="panelsection">
<?php while($row1 = pg_fetch_array($result1)){ ?>
<div id="categorymoviesection">
<img src="<?php echo $row1[2]; ?>" height="300" width="200">
<h5 style="color: #FFFFFF"><?php echo $row1[1]; ?></h5>
<a href = "moviedetails.php?movieid=<?php echo $row1[0];?>" class = "btn btn-default" id="btndefault" role = "button" style="width: 200px">
More details
</a>
</div>
<?php } ?>
</div>
所有代码
<!DOCTYPE html>
<?php
session_start();
$name = "";
$userid = "";
if(array_key_exists('name', $_SESSION) && array_key_exists('userid', $_SESSION)){
$name = $_SESSION['name'];
$userid = $_SESSION['userid'];
}
?>
<html lang="en">
<head>
<?php $page_title = "All Movies" ?>
<?php include("includes/resources.php");?>
<?php include ("includes/navbar2.php");?>
<?php
require('connect.php');
$query = "SELECT M.MovieID, M.Name, rank.average
FROM (
SELECT SUM(W.Rating)/count(W.Rating) AS average, M.MovieID AS mo
FROM Movie M, Watches W
WHERE M.MovieID = W.MovieID
GROUP BY M.MovieID)rank, Movie M
WHERE M.MovieID = rank.mo
GROUP BY rank.average, rank.mo, M.Name, M.MovieID
ORDER BY M.Name";
$result = pg_query($link,$query);
$stmt = pg_prepare($link, 'ps',$query);
?>
</head>
<body style="background-color: #212121; padding-bottom: 30px;">
<div class="container">
<h1 style="color: #FFFFFF; margin-left: -45px;">
All available Products:
</h1>
<div class="row clearfix">
<!--Thumbnail for the movies-->
<div class = "row" id="thumbnailsection">
<?php while($row = pg_fetch_array($result)) { ?>
<div class = "col-sm-6 col-md-3" id="indivthumbnail" >
<div class = "thumbnail" id="indivthumbnail2" >
<img src = "<?php echo $row[3]; ?> " alt = "Generic placeholder thumbnail">
</div>
<div class = "caption" id="caption2">
<h4> <?php echo $row[1]; ?> </h4>
<p>Rating: <?php echo $row[2]; ?> </p>
<p>
<?php if (1==1){?>
<a href = "watchedlist.php?movieid=<?php echo $row[0];?>&userid=<?php echo $userid ?>" class = "btn btn-primary" id="btnprimary" role = "button">
Bought
</a>
<?php } else { ?>
<a href = "watchedlist.php?movieid=<?php echo $row[0];?>&userid=<?php echo $userid ?>" class = "btn btn-primary" id="btnprimary" role = "button">
Add to wishlist
</a>
<?php } ?>
<a href = "moviedetails.php?movieid=<?php echo $row[0];?>" class = "btn btn-default" id="btndefault" role = "button">
More details
</a>
</p>
</div>
</div>
<?php } ?>
</div>
</div>
</div>
</body>
</html>
非常感谢您的帮助
答案 0 :(得分:0)
清除空格。我认为这是个问题。
<div class = "thumbnail" id="indivthumbnail2" >
<img src = "<?php echo $row[3]; ?> " alt = "Generic placeholder thumbnail">
</div>
答案 1 :(得分:0)
您没有添加watchedlist.php
或moviedetails.php
的代码,但请确保在HTTP响应上设置正确的MIME类型,如下所示:
$filename = basename($file);
$file_extension = strtolower(substr(strrchr($filename,"."),1));
switch( $file_extension ) {
case "gif": $ctype="image/gif"; break;
case "png": $ctype="image/png"; break;
case "jpeg":
case "jpg": $ctype="image/jpeg"; break;
default:
}
header('Content-type: ' . $ctype);
readfile($file);