jasonc Posted December 6, 2011 Share Posted December 6, 2011 I have searched around and found this code that suggests that I should be able to read an image file and echo it directly in to the page without hyperlinking to the file that is outside the public folder, but i get the error message that it is not there even though the file is. Warning: getimagesize() [function.getimagesize]: Unable to access "/home/****/upload/AAABBBCCC.JPG" in /home/****/public_html/client.php on line 40 $image = "AAABBBCCC.JPG"; $path= "/home/****/upload/"; $details = getimagesize($path . $image); header ('Content-Type: ' . $details['mime']); echo file_get_contents($path . $image); Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted December 6, 2011 Share Posted December 6, 2011 Is the spelling and capitalization of both the path and the file correct? Quote Link to comment Share on other sites More sharing options...
jasonc Posted December 6, 2011 Author Share Posted December 6, 2011 just in case this was the issue iI change the code so it would capitalise all the file names and the same when it is getting it. I have also tried to use permission set to 777, 666, 755 and even tried relative path ../../upload/file Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted December 6, 2011 Share Posted December 6, 2011 Without your actual code that is getting/forming the $image and triggering the error and both the error messages, it's kind of hard to help you with the half dozen possible things that could prevent a file operation from matching the actual filename. For example, the exact wording of all the error messages, less your account/domain part of them, may in fact tell someone here what is causing the problem. Quote Link to comment Share on other sites More sharing options...
jasonc Posted December 6, 2011 Author Share Posted December 6, 2011 I have checked the file and the database and both file name in file and database match. The only thing that is not working is the get image part. get image. <?PHP session_start(); $uploaddir = "/home/***/upload/"; include('connection.php'); date_default_timezone_set('Europe/London'); $format = 'Y-m-d H:i:s'; $date = date( $format ); $reportCode = ""; if(isset($_POST['viewReport']) && trim($_POST['viewReport']) == 'View Report') { $surname = mysql_real_escape_string($_POST['surname']); $code = mysql_real_escape_string($_POST['code']); $check = mysql_query("SELECT * FROM `clientinfo` WHERE `surname` = '" . strtolower($surname) . "' AND `code` = '" . strtoupper($code) . ".JPG' LIMIT 1"); if(!$surname) { $thisError = 'Please enter your Surname.'; } else if($code == "") { $thisError = 'Please enter your Security Code.'; } else if(! mysql_num_rows($check)) { $thisError = 'That report is not yet ready to view, please try again in a few days.'; } else if(mysql_num_rows($check)) { // get report file. $reports = mysql_fetch_assoc($check); $reportCode = $reports['code']; $details = getimagesize('"' . $uploaddir . $reportCode . '"'); header ('Content-Type: ' . $details['mime']); } else { // something else went wrong. } } ?> <!DOCTYPE HTML> <html lang="en"> <head> <title>Client Area</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <link rel="stylesheet" href="style.css" type="text/css"> </head> <body> <div class="headerBar"></div> <div class="content"> <div class="widthLimiter contentStyle"> <div class="formWrapper" style="width: 450px; text-align: center;"> <? if(isset($thisError)) { echo '<div class="errorDiv">',$thisError,'</div>'; } ?> <? if(isset($thisSuccess)) { echo '<div class="successDiv">',$thisSuccess,'</div>'; } ?> <span class="subHeader">Client Area</span> <? if ($reportCode) { ?> <div> <br>Your report<br><br><? echo file_get_contents('"' . $uploaddir . $reportCode . '"'); ?> </div> <? } else { ?> <form method="POST" action="client.php"> <ul style="padding: 0 10px 0 10px;"> <li class="field">Clients Surname</li> <li class="value"><input type="text" name="surname" value="<? if(isset($_POST['surname'])) { echo $surname; } ?>" style="width: 200px;"></li> <li class="field">Security Code<br><span class="small">(Please note: your Security Code may be CaSe SeNSitIvE)</span></li> <li class="value"><input type="text" name="code" value="" style="width: 200px;"></li> <li class="other"><input type="submit" name="viewReport" value=" View Report "> <input type="reset" value=" Clear "></li> </ul> </form> <? } ?> <br style="clear:both;"><br> </div> </div> </div> </body> </html> Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted December 6, 2011 Share Posted December 6, 2011 Your ACTUAL code is concatenating double quotes to the start and end of the $uploaddir . $reportCode value. Also, the response to the http request for the image can only be the content-type header followed by the image data. Quote Link to comment Share on other sites More sharing options...
jasonc Posted December 6, 2011 Author Share Posted December 6, 2011 ok i have removed the double quotes and now i get the error that the image can not be displayed as it contians errors ? I have checked the image on the server and it shows ok in the browser and when i download a copy via ftp to me pc and view it there is no problem i can see with the image, shows normally. content type ? sorry what is this ? Quote Link to comment Share on other sites More sharing options...
jasonc Posted December 6, 2011 Author Share Posted December 6, 2011 ah the content type, yes this is set by the line... header ('Content-Type: ' . $details['mime']); it first checks what file type it is and then uses that to set the type of file to show. but for some reason i think this line is reporting the problem. file_get_contents($uploaddir . $reportCode); Quote Link to comment Share on other sites More sharing options...
jasonc Posted December 6, 2011 Author Share Posted December 6, 2011 ok the only way i got the image to show was to take out all of the other html code in other words only show the image and nothing else. but I would like to have the page formatted and have a header and footer layout as well as the image showing. how can I get this to work also ? Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted December 6, 2011 Share Posted December 6, 2011 Each image on a web page requires an <img src="URL_that_results_in_the_image_being_sent_to_the_browser" alt=""> HTML tag. That's how the browser knows where to fetch the image in order to render it on the page. The URL_that_results_in_the_image_being_sent_to_the_browser that you put into the src="..." attribute is the URL of your .php script that outputs the correct content-type header followed by the image data. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.