Re: [PHP] Re: Displaying images
I suggest you use firebug https://addons.mozilla.org/en-US/firefox/addon/1843 coming back to your case.. 1. use the inspect function of firebug to check if the img src=.. is displaying correctly. 2 then use the net tab of firebug to see if the image is actually downloaded from the server. 3. open the link in new tab to see image is actually displayed. if you are struck at case 1: some problem in the echo statement of your main php case 2: firebug actually tells you what is the mistake you did case 3: i suggest one more firefox addon: live http headers see if the content-type and content-length header are being sent This page isn't working and if I try to browse this page it wants to open it with an editor, it won't view in the browser. i assume that the address in the URL bar of your browser starts with file:// rather than http:// ? simply put are you double clicking the file or clicking open with firefox ?
Re: [PHP] Re: Displaying images
Miller, Terion wrote: [..] header(Content-type: img/jpeg); [..] This page isn't working and if I try to browse this page it wants to open it with an editor, it won't view in the browser. What am I doing wrong? Is it the code or the data? Content-type: img/jpeg is undefined, the browser would not know what to do with it. Try image/jpeg which is the official MIME type. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Displaying images
Well I have tried Numerous scripts and ways and still can't get the image to display. I have echoed the file and have been able to get the gibberish image code to display but not a real image, here is my full code if anyone wants to have a look, I am going crossed eyed here. These are just a few of the ways I have been trying to get the img tag to work on my output.php page: trtdScout Photo:/td td a href=image.php?filename=?php echo$row['photoName']; ??php echo$row['photoName']; ?/Abr?php echo \nbrScout Photo : .img src=\image.php?filename= .$row['photoName']. \; ? ?php echo img src=\image.php?filename=.$row['photoName'].\\n; ? ?php echo img src=\image.php?id=.$row['ePID'].\\n; ? img src=image2.php?filename=?php echo $row['photoName']; ? Here is one way I was trying to get to work for the image.php page where the headers are supposed to work and don't include(inc/dbconn_open.php); error_reporting(E_ALL); //if (isset($_FILES['ePhoto'])){$ePhoto = $_FILES['ePhoto'];} else {$ePhoto =;} $filename = $_GET['$filename']; $image = stripslashes($_REQUEST[photoName]); $sql = SELECT ePhoto, photoName, photoType from eagleProjects WHERE photoName = $filename; $result=mysql_query($sql); $data = mysql_fetch_array ($result); $type = $data['photoType']; $name = $data['photoName']; header(Content-type: $type); header(Content-Disposition: attachment; filename=$name); echo $data[photoName]; echo $data[ePhoto]; exit; -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Displaying images
On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie nos...@mckenzies.netwrote: Miller, Terion wrote: I am trying to get an image to display but I get nothing if done like this: tr tdScout Photo:/td tdimg src=?php echo $row['ePhoto'];?/td /tr If I just echo the field I do get the file name Does the filename include the path? Does the image with said filename actually exist in that path? -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Is it an image from the db or a path to an image on the filesystem? -- Bastien Cat, the other other white meat
Re: [PHP] Re: Displaying images
On 5/27/09 12:49 PM, Bastien Koert phps...@gmail.com wrote: On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie nos...@mckenzies.netwrote: Miller, Terion wrote: I am trying to get an image to display but I get nothing if done like this: tr tdScout Photo:/td tdimg src=?php echo $row['ePhoto'];?/td /tr If I just echo the field I do get the file name Does the filename include the path? Does the image with said filename actually exist in that path? -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Is it an image from the db or a path to an image on the filesystem? -- Bastien Cat, the other other white meat Hi Bastien it is an image in the db. Thanks Terion -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Displaying images
On Wed, May 27, 2009 at 1:51 PM, Miller, Terion tmil...@springfi.gannett.com wrote: On 5/27/09 12:49 PM, Bastien Koert phps...@gmail.com wrote: On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie nos...@mckenzies.net wrote: Miller, Terion wrote: I am trying to get an image to display but I get nothing if done like this: tr tdScout Photo:/td tdimg src=?php echo $row['ePhoto'];?/td /tr If I just echo the field I do get the file name Does the filename include the path? Does the image with said filename actually exist in that path? -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Is it an image from the db or a path to an image on the filesystem? -- Bastien Cat, the other other white meat Hi Bastien it is an image in the db. Thanks Terion You can't display the image that way due to differeing headers in required for html and images. Change your image code to call a page that specializes in handling the images and the headers needed by changing to something like this: echo img src=\show_image.php?id=.$rows['id'].\\n; where the show_image page calls the image out from the db and display it as below ?php require(conn.php); //check to see if the id is passed if(isset($_GET['id'])) { $id=$_GET['id']; //query the database to get the image and the filetype $query = select bin_data, filetype from binary_data where id=$id; $result = mysql_query($query); $row = mysql_fetch_array($result); { $data = $row['bin_data']; $type = $row['filetype']; } if ($type==pjpeg) $type = jpeg; //handle the ms jpeg alternate format Header( Content-type: $type); echo $data; } ? -- Bastien Cat, the other other white meat
Re: [PHP] Re: Displaying images
Miller, Terion wrote: Does the filename include the path? Does the image with said filename actually exist in that path? -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Hmm guess I am confused because the image is being stored in the db. So I thought I could call it like any other field data. Thanks Terion Please reply all. The img tag is used to tell the browser where the image is located, it then loads the image. If you have the binary image data in the DB then you probably need to have a PHP file with a get var as the img src and then have that PHP file actually output the image data: img src=path/to/image.php?filename=?php echo $row['filename']; ? //image.php header(Content-type: img/jpeg); $filename = $_GET['filename']; //do your SQL select and fetch using filename or whatnot to get the image data, I assume from 'ePhoto' echo $row['ePhoto']; Thanks! -Shawn -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Displaying images
On Wed, 2009-05-27 at 14:03 -0400, Bastien Koert wrote: On Wed, May 27, 2009 at 1:51 PM, Miller, Terion tmil...@springfi.gannett.com wrote: On 5/27/09 12:49 PM, Bastien Koert phps...@gmail.com wrote: On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie nos...@mckenzies.net wrote: Miller, Terion wrote: I am trying to get an image to display but I get nothing if done like this: tr tdScout Photo:/td tdimg src=?php echo $row['ePhoto'];?/td /tr If I just echo the field I do get the file name Does the filename include the path? Does the image with said filename actually exist in that path? -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Is it an image from the db or a path to an image on the filesystem? -- Bastien Cat, the other other white meat Hi Bastien it is an image in the db. Thanks Terion You can't display the image that way due to differeing headers in required for html and images. Change your image code to call a page that specializes in handling the images and the headers needed by changing to something like this: echo img src=\show_image.php?id=.$rows['id'].\\n; where the show_image page calls the image out from the db and display it as below ?php require(conn.php); //check to see if the id is passed if(isset($_GET['id'])) { $id=$_GET['id']; //query the database to get the image and the filetype $query = select bin_data, filetype from binary_data where id=$id; $result = mysql_query($query); $row = mysql_fetch_array($result); { $data = $row['bin_data']; $type = $row['filetype']; } if ($type==pjpeg) $type = jpeg; //handle the ms jpeg alternate format Header( Content-type: $type); echo $data; } ? That's only true if the image is stored in the database as a BLOB or somesuch. If the database is just storing the filename of the image, then it may just need the correct directory path prefix. Go into your browser, right-click and check to see what path and filename the browser expects to find the image file. My guess is it's not the same as where the image actually is, and you need to add the path before opening the ?php tag. You can use either an absolute path (e.g. using a full http://domain) or a relative one, and relative paths can include the ../ directory if you need to move up the directory tree. Ash www.ashleysheridan.co.uk -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Displaying images
On Wed, May 27, 2009 at 3:05 PM, Ashley Sheridan a...@ashleysheridan.co.ukwrote: On Wed, 2009-05-27 at 14:03 -0400, Bastien Koert wrote: On Wed, May 27, 2009 at 1:51 PM, Miller, Terion tmil...@springfi.gannett.com wrote: On 5/27/09 12:49 PM, Bastien Koert phps...@gmail.com wrote: On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie nos...@mckenzies.net wrote: Miller, Terion wrote: I am trying to get an image to display but I get nothing if done like this: tr tdScout Photo:/td tdimg src=?php echo $row['ePhoto'];?/td /tr If I just echo the field I do get the file name Does the filename include the path? Does the image with said filename actually exist in that path? -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Is it an image from the db or a path to an image on the filesystem? -- Bastien Cat, the other other white meat Hi Bastien it is an image in the db. Thanks Terion You can't display the image that way due to differeing headers in required for html and images. Change your image code to call a page that specializes in handling the images and the headers needed by changing to something like this: echo img src=\show_image.php?id=.$rows['id'].\\n; where the show_image page calls the image out from the db and display it as below ?php require(conn.php); //check to see if the id is passed if(isset($_GET['id'])) { $id=$_GET['id']; //query the database to get the image and the filetype $query = select bin_data, filetype from binary_data where id=$id; $result = mysql_query($query); $row = mysql_fetch_array($result); { $data = $row['bin_data']; $type = $row['filetype']; } if ($type==pjpeg) $type = jpeg; //handle the ms jpeg alternate format Header( Content-type: $type); echo $data; } ? That's only true if the image is stored in the database as a BLOB or somesuch. If the database is just storing the filename of the image, then it may just need the correct directory path prefix. Go into your browser, right-click and check to see what path and filename the browser expects to find the image file. My guess is it's not the same as where the image actually is, and you need to add the path before opening the ?php tag. You can use either an absolute path (e.g. using a full http://domain) or a relative one, and relative paths can include the ../ directory if you need to move up the directory tree. Ash www.ashleysheridan.co.uk Terion stated it was in the db -- Bastien Cat, the other other white meat
Re: [PHP] Re: Displaying images
On Wed, 2009-05-27 at 15:14 -0400, Bastien Koert wrote: On Wed, May 27, 2009 at 3:05 PM, Ashley Sheridan a...@ashleysheridan.co.ukwrote: On Wed, 2009-05-27 at 14:03 -0400, Bastien Koert wrote: On Wed, May 27, 2009 at 1:51 PM, Miller, Terion tmil...@springfi.gannett.com wrote: On 5/27/09 12:49 PM, Bastien Koert phps...@gmail.com wrote: On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie nos...@mckenzies.net wrote: Miller, Terion wrote: I am trying to get an image to display but I get nothing if done like this: tr tdScout Photo:/td tdimg src=?php echo $row['ePhoto'];?/td /tr If I just echo the field I do get the file name Does the filename include the path? Does the image with said filename actually exist in that path? -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Is it an image from the db or a path to an image on the filesystem? -- Bastien Cat, the other other white meat Hi Bastien it is an image in the db. Thanks Terion You can't display the image that way due to differeing headers in required for html and images. Change your image code to call a page that specializes in handling the images and the headers needed by changing to something like this: echo img src=\show_image.php?id=.$rows['id'].\\n; where the show_image page calls the image out from the db and display it as below ?php require(conn.php); //check to see if the id is passed if(isset($_GET['id'])) { $id=$_GET['id']; //query the database to get the image and the filetype $query = select bin_data, filetype from binary_data where id=$id; $result = mysql_query($query); $row = mysql_fetch_array($result); { $data = $row['bin_data']; $type = $row['filetype']; } if ($type==pjpeg) $type = jpeg; //handle the ms jpeg alternate format Header( Content-type: $type); echo $data; } ? That's only true if the image is stored in the database as a BLOB or somesuch. If the database is just storing the filename of the image, then it may just need the correct directory path prefix. Go into your browser, right-click and check to see what path and filename the browser expects to find the image file. My guess is it's not the same as where the image actually is, and you need to add the path before opening the ?php tag. You can use either an absolute path (e.g. using a full http://domain) or a relative one, and relative paths can include the ../ directory if you need to move up the directory tree. Ash www.ashleysheridan.co.uk Terion stated it was in the db The thread didn't follow on properly, my email client wasn't able to join t by thread so I didn't see his message until after I'd replied, sorry! Ash www.ashleysheridan.co.uk -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Displaying images
Thanks for the suggestions everyone, I have this now, but still no image showing up It is stored as a blob in the database. on the output page I am calling it like this: img src=image.php?filename=?php echo $row['ePhoto']; ? Then on the image.php page I have this: ?php include(inc/dbconn_open.php); error_reporting(E_ALL); //image.php header(Content-type: img/jpeg); $filename = $_GET['filename']; $sql = SELECT ePhoto from eagleProjects WHERE $filename= ePhoto; $result=mysql_query($sql); $row = mysql_fetch_assoc ($result); ? ?php echo $row['ePhoto']; ? This page isn't working and if I try to browse this page it wants to open it with an editor, it won't view in the browser. What am I doing wrong? Is it the code or the data? Thanks Terion -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Displaying images
Miller, Terion wrote: Thanks for the suggestions everyone, I have this now, but still no image showing up It is stored as a blob in the database. on the output page I am calling it like this: img src=image.php?filename=?php echo $row['ePhoto']; ? I was assuming that ePhoto was the actual binary data? If so then you need something else here like id or something. I just used filename as an example. If ePhoto is actually the filename then you are OK here, but not below. Then on the image.php page I have this: ?php include(inc/dbconn_open.php); error_reporting(E_ALL); //image.php header(Content-type: img/jpeg); $filename = $_GET['filename']; $sql = SELECT ePhoto from eagleProjects WHERE $filename= ePhoto; Ummm... This can't work. What is ePhoto??? The data or the filename? $result=mysql_query($sql); $row = mysql_fetch_assoc ($result); ? ?php echo $row['ePhoto']; ? This page isn't working and if I try to browse this page it wants to open it with an editor, it won't view in the browser. What am I doing wrong? Is it the code or the data? Thanks Terion -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Displaying images
Shawn McKenzie wrote: Miller, Terion wrote: Thanks for the suggestions everyone, I have this now, but still no image showing up It is stored as a blob in the database. on the output page I am calling it like this: img src=image.php?filename=?php echo $row['ePhoto']; ? I was assuming that ePhoto was the actual binary data? If so then you need something else here like id or something. I just used filename as an example. If ePhoto is actually the filename then you are OK here, but not below. Then on the image.php page I have this: ?php include(inc/dbconn_open.php); error_reporting(E_ALL); //image.php header(Content-type: img/jpeg); $filename = $_GET['filename']; $sql = SELECT ePhoto from eagleProjects WHERE $filename= ePhoto; Ummm... This can't work. What is ePhoto??? The data or the filename? $result=mysql_query($sql); $row = mysql_fetch_assoc ($result); ? ?php echo $row['ePhoto']; ? This page isn't working and if I try to browse this page it wants to open it with an editor, it won't view in the browser. What am I doing wrong? Is it the code or the data? Thanks Terion I don't know what your fields are, but if ePhoto is the BLOB, and you have an id field, then try this (assuming that you have selected id in your query and it is an integer): img src=image.php?id=?php echo $row['id']; ? //image.php $id = (int)$_GET['id']; $sql = SELECT ePhoto from eagleProjects WHERE id=$id; -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php