problem with this code to show images
Posted: Mon Oct 08, 2012 5:17 pm
friends
i,m facing a problem with showing images stored into my database according to the following
i,m building a news website which shows news and a pic related to each news
i could make a page by which i can store news and a picture with each one
i made this page wich show these news as a title ans a pic beside to each news title every thing is ok but these pics can,t be shown all i can see is a red X mark in the place of the pic although i clicked right click on the red X and saw that the source of the pic is ok with the ip of the each opic
i,m posting the code i did so any one could tell me what is wrong or missing with it
i did a page is called getimage.php
and here is the code of it
as for the page which show these pics
the code of that part which shows the picture is as followiing
i tried alot with this code but it did not work with me as a beginner and this is my first site so i,m waiting for your point of view my dear friends
Thank you
i,m facing a problem with showing images stored into my database according to the following
i,m building a news website which shows news and a pic related to each news
i could make a page by which i can store news and a picture with each one
i made this page wich show these news as a title ans a pic beside to each news title every thing is ok but these pics can,t be shown all i can see is a red X mark in the place of the pic although i clicked right click on the red X and saw that the source of the pic is ok with the ip of the each opic
i,m posting the code i did so any one could tell me what is wrong or missing with it
i did a page is called getimage.php
and here is the code of it
Code: Select all
<?php
include ("config.php");
// this is to get the id of the required pic
$id=addslashes(@$_REQUEST['id']);
$image= mysql_query ("SELECT photo FROM tourism_news Where id=$id");
$row1 = mysql_fetch_assoc($image);
$myimage= $row1['photo'];
header("content-type:image/jpeg");
//$type ='content-type:'.$row1['imagetype'];
$type ="content-type:".$row1['imagetype'];
echo "<img src='$myimage' width='100' height='200'>";
//header($type);
//echo $row1['photo'];
echo "$myimage";
//echo <img src=\"$myimage\">;
?>the code of that part which shows the picture is as followiing
Code: Select all
echo"<img src='getimage.php?id=".$row['id']."' width=200 height=140 />"; Thank you