I want to display image from mysql in php page
but i have headers already sent this is what it says
will i have to change the included headers file to display images form mysql?
thank You
13 1858 Markus 6,050
Recognized Expert Expert
Post your code and we'll tell you wherre you've gone wrong.
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public function displayLogo($ID_CLIENT){
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-
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$result=mysql_query("SELECT logo,logo_type FROM ".$this->tables['clients']." WHERE ID_CLIENT='$ID_CLIENT'")or die ("SELECT ERROR:(table clients)".mysql_error());
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$type = MYSQL_RESULT($result,0,"logo_type");
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$data = MYSQL_RESULT($result,0,"logo");
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header("Content-type: $type");
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echo $data;
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}
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logo is blob
then on the site
<td>
<? echo $client->displayLogo($I D_CLIENT); ?>
</td>
but i get
Warning: Cannot modify header information - headers already sent by (output started at /home/aa24985/domains/purcon.pl/public_html/exec-jobs/config/vert_menu.php:1 5) in path_here/client.php on line 138
PNG IHDR@ŁbK GD˙˙˙ ˝ pHYs śtIME $ńIDATx ťůsG˛?}Ł KĚe{őFĽ ź6> ˇű}OUő~hF b&żĄşŞŞľ YY9RJ ż1d $$H !A $HH AB $H A $$H !A˙0Ď H!J˘BCCi S7J˘Łh`Xi˘ /;ԞqBĚԑHĐ ŔL4]Ă4͑gG CTH%ŃĐvC3ź aż="BĽěĐŹ ăą ~č#p0$ #"Ĺs˛,LŰMą eZo5)$ˇ@ ŤHJtMǶmܴ gčĆe`!ąT ń\ęeYzţđřD$P ęĐhńčahş &AŮV1ě ٢^Żl&Ť7 Čć˛ččQ%ť~(Ś ČĺsŚOŚS*0ş Tv+[uęőz<J`&ąLŇX bČřŘ8mś<v wim˝~L, ŮcĹ1&&(N iŐ[TjˇŹid˛&Ć&p <VEuŻJ Ł C,"JaZ&n `0Đt@A*ťbbl l!kĺZ'oPŇ nVXşc:( M!ť~vŤřx)Ź f.͐e0-XTv+ńřŰ >ČXľu3ĎD p˛Ĺp8^Żď
etc
hsriat 1,654
Recognized Expert Top Contributor
Similar question was asked earlier by someone. Have a look at this.
but when i add header("Content-type: $type; charset=windows-1250");
where the type is
$type=$client->getLogoType($I D_CLIENT); <--- returns image type
it tries to download it istead of displaying it
thank You
$image = imagecreatefrom string($data); doesnt work neither
hsriat 1,654
Recognized Expert Top Contributor
$image = imagecreatefrom string($data); doesnt work neither
You have to make a new PHP file for displaying images.
In that PHP file, give header of image type and print all the image data afterwards.[php]<?php
//filename: image.php
if ($_GET['img']=="")
die ('Invalid Parameter');
header("Content-type:image/jpeg");//whatever is the extension
$data = mysql_fetch_arr ay(mysql_query( "SELECT `image` FROM ...... WHERE `id`='".$_GET['img']."'"));
print($data['image']);
?>[/php]And in your HTML file, where you need to display that image, write[HTML]<img src="image.php? img=abc" alt="abc" />[/HTML]
i have -
<?
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include('../config/config.php');
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$config = new config();
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$config->mysql_server_connect();
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$result=mysql_query("SELECT logo_name,logo_size,logo_type,logo FROM ".$config->tables['clients']." WHERE ID_CLIENT=$ID_CLIENT") or die("SELECT Error(table clients): ".mysql_error());
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list($name, $type, $size, $data) = mysql_fetch_array($result);
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header("Content-length: $size");
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header("Content-type: $type");
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header("Content-Disposition: attachment; filename=$name");
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echo $data;
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?
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but how do i now pass that id_client to that file and display the image
sorry i didnt notice that you actually wrote it :)
but still i get X instead of logo
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