VineetDubey Posted June 30, 2011 Share Posted June 30, 2011 Hello All, I tried and search in google so many times but i not getting what i exactly want. As i was trying a bit typical task as i think though i am not able to display image using fopen and fread function, Please help me to get a code else tell me whether we can display image using these function or not. please check the below code is correct? $location='C:/xampp/htdocs/dog.jpg'; $size = getimagesize($location); $img_type = $size['mime']; $data = file_get_contents($location); //$img_data = addslashes($data); header("Content-type: ".$img_type); print $data; Link to comment https://forums.phpfreaks.com/topic/240793-how-to-display-image-in-php/ Share on other sites More sharing options...
Alex Posted June 30, 2011 Share Posted June 30, 2011 Did you try it? Link to comment https://forums.phpfreaks.com/topic/240793-how-to-display-image-in-php/#findComment-1236771 Share on other sites More sharing options...
VineetDubey Posted June 30, 2011 Author Share Posted June 30, 2011 yes Link to comment https://forums.phpfreaks.com/topic/240793-how-to-display-image-in-php/#findComment-1236871 Share on other sites More sharing options...
VineetDubey Posted June 30, 2011 Author Share Posted June 30, 2011 i tried the above code in xampp and that was not working but rt now i tried in wamp and it is working .Please let me know what was the problem. i am new in this technology. Link to comment https://forums.phpfreaks.com/topic/240793-how-to-display-image-in-php/#findComment-1236892 Share on other sites More sharing options...
VineetDubey Posted June 30, 2011 Author Share Posted June 30, 2011 slq: query is attached; Please let me know why this code is not displaying image. <?php $cid=mysql_connect('localhost','root',''); mysql_select_db('filesystem',$cid); $fp=fopen('DSC00194 copy.jpg','r'); $fs=filesize('DSC00194 copy.jpg'); //$data=fread($fp,$fs); //$data=addslashes($data); //mysql_query("insert into file2db(name,size,type,data) values('DSC00194 copy.jpg',$fs,'image/jpeg','$data')"); $rs=mysql_query("select * from file2db"); $row=mysql_fetch_array($rs); $data=file_get_contents($row['data']); $data=stripslashes($data); header("Content-type:".$row['type']); print $data; ?> [attachment deleted by admin] Link to comment https://forums.phpfreaks.com/topic/240793-how-to-display-image-in-php/#findComment-1236910 Share on other sites More sharing options...
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