这是我将从获得图像的页面代码(完美工作)
<?php
ob_start();
session_start();
include('connect.php');
$id = $_GET['id'];
$query = mysql_query("SELECT * FROM news WHERE id=$id");
$row = mysql_fetch_assoc($query);
header("Content-type: image/jpeg");
echo $row['image'];
?>
这是我在中获取图像的页面
<?php
ob_start();
session_start();
include('includes/connect.php');
include('includes/phpCodes.php');
$id = $_GET['id'];
function showNews() {
$data = array( 'id' => $id );
$base = "includes/getImage.php";
$url = $base. "?" . "id=36";
echo $url;
echo '<img src=includes/getImage.php class="newsImage">';
echo '<h1><p class="subjecTitle">هنا العنوان</p></h1>
<div class="newsContent">
hihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihihi
</div>
';
}
?>
<!DOCTYPE html>
<html>
<head>
<title>عينٌ على الحقيقة</title>
<meta charset="utf-8">
<link rel="stylesheet" type="text/css" href="css/mainstyle.css">
<link rel="stylesheet" type="text/css" href="css/showstyle.css">
<script lang="javascript">
function logout( myFrame ) {
myFram.submit();
}
</script>
</head>
<body>
<div class="wrapper">
<?php headerCode(); ?>
<div class="content" dir="rtl">
<?php showNews(); ?>
</div>
</div>
</body>
</html>
我觉得我错在了,有人能告诉我该怎么解决吗?,抱歉我的英语不好
为您清除:
echo '<img src="includes/getImage.php?id=' . $id . '" class="newsImage">';
应该100%有效(当然,如果$id参数有值的话)。
更新以修复丢失的$id var:
<?php
ob_start();
session_start();
include('includes/connect.php');
include('includes/phpCodes.php');
$id = $_GET['id'];
function showNews(){
$id = $_GET['id'];
$base = "includes/getImage.php";
$url = $base. "?" . "id=36";
echo $url;
echo '<img src="includes/getImage.php?id=' . $id . '" class="newsImage">';
echo '
<h1><p class="subjecTitle">??? ???????</p></h1>
<div class="newsContent"></div>
';
}
?>
为什么使用两个不同的页面?把两个代码都放在一个页面上,简单地做这个
<img src=includes/<?php echo $row['image']; ?> class="newsImage">
更改:
echo ' <img src=includes/getImage.php class="newsImage">';
收件人:
echo ' <img src="includes/getImage.php?id='.$id.'" class="newsImage">';
请注意,src
具有"
,并确保includes/getImage.php
返回图像路径