如何从数据库中检索图像并在网页上显示图像
[英]How to retrieve image from database and display image on the web page
问题描述
I have inserted image into database and store name in the table.
我已将图像插入数据库并将名称存储在表中。 my image is saved in a folder named 'Uploads'.我的图像保存在名为“上传”的文件夹中。 Now need to retrieve image from the databse and display it.现在需要从数据库中检索图像并显示它。 when I try to display It only shows the image name which is taken from my table.but it does not show the image.当我尝试显示时，它只显示从我的表中获取的图像名称。但它不显示图像。
retrieving code is given below检索代码如下
$sql="SELECT * FROM candi_profile WHERE can_email='{$_SESSION['usr_email']}'";
$result=mysqli_query($con,$sql);
if(!$result) die(mysqli_error($con));
<div class="container">
<!-- Page Header -->
<div class="row">
<div class="col-lg-12">
<h1 class="page-header">Employer Dashboard
</h1>
</div>
</div>
<!-- /.row -->
<!-- Projects Row -->
<div class="row">
<div class="col-md-4">
<?php
while($rows=mysqli_fetch_array($result)){
$c_id = $rows['can_id'];
var_dump($c_id);
?>
<p class="lead"><?php echo $rows['can_name'] ?></p>
<div class="profile-sidebar">
<!-- SIDEBAR USERPIC -->
<div class="profile-userpic">
<p class="lead">
<?php echo $rows['pic_name'] ?></p>
</div>
<!-- END SIDEBAR USERPIC -->
<!-- SIDEBAR USER TITLE -->
<div class="profile-usertitle">
<div class="profile-usertitle-name">
Marcus Doe
</div>
<div class="profile-usertitle-job">
<?php echo $rows['can_city'] ?>
<i class="glyphicon glyphicon-map-marker">
</i>
</div>
<div class="profile-usertitle-job">
<i class="glyphicon glyphicon-envelope"></i>
<?php echo $rows['can_email'] ?>
</div>
<div class="profile-usertitle-job">
<?php echo $rows['can_country'] ?>
</div>
</div>
<!-- END SIDEBAR USER TITLE -->
<!-- SIDEBAR BUTTONS -->
<div class="profile-userbuttons">
<hr>
</div>
<!-- END SIDEBAR BUTTONS -->
<!-- SIDEBAR MENU -->
<?php
}
?>
</div>
5 个解决方案
解决方案1
1 2016-07-25 06:14:29
使用 image 标签显示图像并为其指定图像文件夹的路径
<img src="your path/<?php echo $rows['pic_name'] ?>" />
解决方案2
1 2016-07-25 06:14:30
I assume that the content of $rows['pic_name'] is string only as said on your question.我假设$rows['pic_name'] 的内容只是你的问题中所说的字符串。
Put an image attribute and call the path of the image with the corresponding filename save on the database.放置一个图像属性，并调用保存在数据库中的相应文件名的图像路径。
<img src = "<path>/<?php echo $rows['pic_name'] ?>" />
NOTE :注意：
Make sure the image is existing on your desire path.确保图像存在于您想要的路径上。
解决方案3
0 已采纳 2016-07-25 06:15:28
you can use this code to retrieve image from database您可以使用此代码从数据库中检索图像
<?php
include 'connection.php'
?>
<?php
$result = mysql_query("SELECT * FROM table") or die(mysql_error());
?>
<table border="1" cellpadding="5" cellspacing="5">
<tr> <th>Image</th></tr>
<?php
while($row = mysql_fetch_array($result)) {
$id = $row['id'];
?>
<tr>
<td><img src="uploads/<?php echo $row['pic_name'];?>" alt=" " height="75" width="75"></td>
</tr>
<?php
}
}
?>
</table>
解决方案4
0 2017-03-17 13:49:01
friend instead of making images folder you should make a new image column(ie "imageColumn ") type as blob then You need to create another php script to return the image data, eg getImage.php.朋友，而不是制作图像文件夹，您应该创建一个新的图像列（即“imageColumn”）类型为 blob 然后您需要创建另一个 php 脚本来返回图像数据，例如 getImage.php。
home.php(or display image page) code home.php（或显示图片页面）代码
<body>
<img src="getImage.php?id=1" width="175" height="200" />
</body>
Then getImage.php is然后 getImage.php 是
<?php
$id = $_GET['id'];
// do some validation here to ensure id is safe
$link = mysql_connect("localhost", "root", "");
mysql_select_db("dvddb");
$sql = "SELECT imageColumn FROM Tablename WHERE id=$id";
$result = mysql_query("$sql");
$row = mysql_fetch_assoc($result);
mysql_close($link);
header("Content-type: image/jpeg");
echo $row['imageColumn '];
?>
解决方案5
0 2020-07-14 04:13:03
- First fetch image from database using query首先使用查询从数据库中获取图像
- The imagejpeg() function is an inbuilt function in PHP which is used to display image to browser or file. imagejpeg() 函数是 PHP 中的一个内置函数，用于向浏览器或文件显示图像。
- Get data using function ob_get_contents();使用函数 ob_get_contents() 获取数据；
- Display image in page with height and width在页面中显示具有高度和宽度的图像
$id = $_GET['id'];
$sql = "select image from table where id='".$id."'";
$res = mysqli_query($sql);
$row = mysqli_fetch_assoc($res);
$image = $row['image'];
ob_start();
imagejpeg($image, null, 50);
$data = ob_get_contents();
ob_end_clean();
echo "<img src='data:image/jpg;base64,'".base64_encode($data)."' style='border:1px
black; border-radius:10px; width:100px; height:125px;'>";
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.