retrieving images from a database.

[Axil]

I get gibberish instead of an image with this code

 

First off, keep in mind i'm pulling a binary stream from my SQL database *not* the path to some picture directory.

Secondly it's not possible that i not output anything else before the image, i know the problem has somthing to do with PHP not wanting to change the header, i also know there's supposed to be away around that by dumping the buffer before and after the header change but when i tried to do that i ran into more issues, can anyone tell me how to fix this?

 

function postBunny($name,$father,$mother,$litter,$dob,$breed,$color,$quality,$comments,$pic_name,$pic_size,$pic_type,$pic_content)

{

  echo <<<END

  <TABLE BORDER=1 height = 300px width = 550px>

  <TH colspan=3>

    <P align=center> $name  - Litter# $litter</P>

  </TH>

  <TR>

  <TD width=50%><DIV>

END;

header("Content-type: image/png");

echo ($pic_content);

header("Content-type:text/html");

echo <<<END

      </TD>

      <TD>

        <TABLE border=1 height=100% width=100%>

            <TR height=10%><TD><B>Mother: </B>$mother</TD>

            <TD><B>Father: </B>$father</TD> </TR>

            <TR height=10%><TD><B>Breed: </B>$breed</TD>

            <TD><B>Color: </B>$color</TD> </TR>

            <TR height=10%><TD><B>Quality: </B>$quality</TD>

            <TD><B> Born:</B> $dob</TD></TR>

            <TR height=10%><TD colspan=2 align=center>

            <B><U>Comments</U></B></DIV> </TD></TR>

            <TR><TD colspan=2 valign=top> $comments</TD></TR>

        </TABLE>

      </TD>

  </TR>

</TABLE>

END;

 

}

[MadTechie]

I get gibberish instead of an image with this code

 

I am not surprised..

 

Create a sctipt that only displays the image (as an image)

ie

<?php
$ID = $_GET['ID'];
//Get Bin Info from database where ID=1 and set to $img
header("Content-Type: image/jpeg");
imagejpeg($img)
exit;
?>

 

now call the above script like an image, ie <img src="image.php?id=1">

 

that should work..

[Axil]

I get gibberish instead of an image with this code

 

I am not surprised..

 

Create a sctipt that only displays the image (as an image)

ie

<?php
$ID = $_GET['ID'];
//Get Bin Info from database where ID=1 and set to $img
header("Content-Type: image/jpeg");
imagejpeg($img)
exit;
?>

 

now call the above script like an image, ie <img src="image.php?id=1">

 

that should work..

 

There's absolutely no way to do this within the function, without having to create another file to handle it?

[MadTechie]

HTML can not convert bin data into an image.. so no..

HTML will need to read a link to an image, so for a dynamic image your need to create a file that generates an image.. this is common pratice

[Axil]

HTML can not convert bin data into an image.. so no..

HTML will need to read a link to an image, so for a dynamic image your need to create a file that generates an image.. this is common pratice

well can i at least but it in the same file, but in a different function?  sorry, i'm new to PHP all my experience is in high level languages like C++ and Java, so some of the restrictions are strange to me.

[MadTechie]

[quote author=Axil link=topic=164156.msg720144#msg720144

well can i at least but it in the same file, but in a different function? 

you could.. just have an if statement from the start.. but in the long run it would be easier to do it as a independant file..

 

[quote author=Axil link=topic=164156.msg720144#msg720144

sorry, i'm new to PHP all my experience is in high level languages like C++ and Java, so some of the restrictions are strange to me.

this is a HTML restriction.. not PHP..

 

[Axil]

Well i put it in another file, but now i get a blank image.

 

When i rightclick the image and select display image i get these errors

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/axil/public_html/image.php on line 10

Warning: Cannot modify header information - headers already sent by (output started at /home/axil/public_html/image.php:10) in /home/axil/public_html/image.php on line 11

Warning: imagejpeg(): supplied argument is not a valid Image resource in /home/axil/public_html/image.php on line 12

 

my code is

<?php
   $id= $_GET['id'];
   $db = *****;
   $host = *****;
   $user = ****;
   $pass =  *****;
   $link = mysql_connect ($host, $user, $pass) or die('connection to server failed');
   mysql_select_db($db,$link) or die('connection to database failed');
   $query= ('SELECT * FROM my_db WHERE id = $id');   
   $img =   mysql_result($query,0,"my_db.pic_content");
   header("Content-Type: image/jpeg");
   imagejpeg($img);
   exit;
?>

[Axil]

whiups, made a mistake with my sql, and that's why the query didn't work right.. fixed it.... now i still get a blank image but when i rightclick it gives me the name of the page like "mydomain/images?id=1" etc

my code is

<?php
   $id= $_GET['id'];
   $db = *****;
   $host = *****;
   $user = ****;
   $pass =  *****;
   $link = mysql_connect ($host, $user, $pass) or die('connection to server failed');
   mysql_select_db($db,$link) or die('connection to database failed');
   $query= ('SELECT * FROM test WHERE bunny_id = ' . $id);
   $result = mysql_query($query,$link)or die('query failed');    
   $img =   mysql_result($result,0,"my_db.pic_content");
   header("Content-Type: image/jpeg");
   imagejpeg($img);
   exit;
?>

[Axil]

i fixed it, the problem was with the imagejpg($img) function, for whatever reason it was giving me the problem i mentioned before but when i just used echo($img) it worked, i had thought that i'd read somewhere you could echo() binary if you set the headers appropriately, which leaves me wondering what the use of imagejpg() is... is there a way i could use imagejpg to output images *without* having to change the header, and thus avoid all these problems?