[英]How to call php in img src tag?
This is the piece of my code from index.php: 这是我的index.php代码的一部分:
<div class="row">
<?php
$conn = mysql_connect("localhost","root","");
mysql_select_db("imdb_db");
$res = mysql_query("select * from `film` ORDER BY ID DESC");
while($row = mysql_fetch_array($res)) {
?><div class='col-xs-2'><?php
?><p style="position:relative;left:-40px"><?php echo $row["Cim"];?></p><?php
?> <img src="php/imageView.php?ID=<?php echo $row["ID"];?>" class="img-responsive" style="width:200px;height:250px;" alt="Image"> <?php
?></div><?php
}
mysql_close($conn);
?>
</div>
This is the imageView.php: 这是imageView.php:
<?php
$conn = mysql_connect("localhost","root","");
mysql_select_db("imdb_db") or die(mysql_error());
if(isset($_GET['ID'])) {
$sql = "SELECT `Boritokep` FROM `film` WHERE ID=". $_GET['ID'];
$result = mysql_query("$sql") or die("<b>Error:</b> Problem on Retrieving Image BLOB<br/>" . mysql_error());
$row = mysql_fetch_array($result);
echo $row["Boritokep"];
}
mysql_close($conn);
?>
And this is the result: 结果如下:
With php/imageView.php?ID=<?php echo $row["ID"]
I getting the path of my file: 使用php/imageView.php?ID=<?php echo $row["ID"]
我得到了文件的路径:
uploads/film/pic/5.jpg
Why it doesn't displays the image? 为什么不显示图像?
Why it doesn't displays the image? 为什么不显示图像?
Because php/imageView.php?ID=<?php echo $row["ID"];?>
returns the path of an image and doesn't return the image itself. 因为php/imageView.php?ID=<?php echo $row["ID"];?>
返回图像的路径,而不返回图像本身。
imageView.php
can take that path and redirect the user agent to it: imageView.php
可以采用该路径并将用户代理重定向到该路径:
header(sprintf('Location: %s', $row["Boritokep"]));
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