[PHP] PHP displaying images

2005-01-28 Thread Ian Johnson
Hello
I am trying to use GD to create and manipulate images but the statement:
header (Content-type: image/jpg);
generates the error message:
	The image http://localhost/gdtst2.php; cannot be displayed, 		because 
it contains errors.

There is no error message in the ../httpd/error_log file.
I am using PHP 5.0.3 with GD 2.0.28.  The report from phpinfo() 
indicates gd present and enabled. Other nongraphics content-types do not 
gererate errors.

Code used was First Program from tutorial at 
http://www.phpfreaks.com/tutorials/105/1.php.  Similar code from php 
manual generates the same error.

I suspect the solution is simple but I'm just not seeing it!
Help
Ian Johnson
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Re: [PHP] PHP displaying images

2005-01-28 Thread Richard Lynch
Ian Johnson wrote:
 I am trying to use GD to create and manipulate images but the statement:

   header (Content-type: image/jpg);

 generates the error message:

   The image http://localhost/gdtst2.php; cannot be displayed,
 because
 it contains errors.

contains errors here means It's not a valid JPEG

So comment out the header() and see if you see any error messages then.

After it looks like it's a JPEG (starts with yoya) put the header() line
back in.

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Re: [PHP] PHP displaying images

2005-01-28 Thread Jason Wong
On Saturday 29 January 2005 10:13, Ian Johnson wrote:

 There is no error message in the ../httpd/error_log file.

That looks like the Apache error log file, which is most likely not what you 
want to be looking at. You want the PHP error log, see settings in php.ini, 
and check phpinfo().

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Re: [PHP] PHP displaying images

2005-01-28 Thread Ian Johnson
This does the same for png as well.  No errors are generated when header
() is commented out or other content-types are specified

Ian Johnson



On Fri, 2005-01-28 at 20:45 -0800, Richard Lynch wrote:
 Ian Johnson wrote:
  I am trying to use GD to create and manipulate images but the statement:
 
  header (Content-type: image/jpg);
 
  generates the error message:
 
  The image http://localhost/gdtst2.php; cannot be displayed,
  because
  it contains errors.
 
 contains errors here means It's not a valid JPEG
 
 So comment out the header() and see if you see any error messages then.
 
 After it looks like it's a JPEG (starts with yoya) put the header() line
 back in.
 

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Re: [PHP] PHP displaying images

2005-01-28 Thread Ian Johnson
Jason Wong wrote:
On Saturday 29 January 2005 10:13, Ian Johnson wrote:

There is no error message in the ../httpd/error_log file.

That looks like the Apache error log file, which is most likely not what you 
want to be looking at. You want the PHP error log, see settings in php.ini, 
and check phpinfo().

My php is configured to write error messages to the http error_log and 
not to display errors.

In this case the error is displayed and not logged.
Note that there is no problem with the code from phpfreaks.  It seems to 
run for  everyone else on their computers.

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Re: [PHP] PHP displaying images

2005-01-28 Thread Jason Wong
On Saturday 29 January 2005 13:42, Ian Johnson wrote:

 My php is configured to write error messages to the http error_log and
 not to display errors.

OK. 

 In this case the error is displayed and not logged.

If you're referring to this ...

  The image http://localhost/gdtst2.php; cannot be displayed,
    because it contains errors.

... then I'm pretty sure that is generated by your browser and not by the 
server.

When using the image functions I always have them save to file initially, then 
when I know that my code is 100% correct and the saved images are viewable 
*then* I output them directly. 

 Note that there is no problem with the code from phpfreaks.  It seems to
 run for  everyone else on their computers.

In that case you're doing something different to everyone else, find out what 
that difference is.

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[PHP] php and images

2004-07-23 Thread Roman Duriancik
I have problem with jpg pictures in php pages. Path and picture is 
correct but when i want  see picture in php script  picture don't show. 
I use PHP4.3.4 on apache server 2.0.44 on Windopws 2000 Server.

Thanks for help.
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Re: [PHP] php and images

2004-07-23 Thread Miroslav Hudak (php/ml)
I have a certain difficulties in case of processing large images thru GD2 
library. GD2 has some problems with image processing and it results in some 
exception and GD just crashes. I have to mention, that the crash causes whole 
php interpreter to crash and execution of the script is terminated without any 
warning or handlable event :(...

I don't know for certain now, which command causes it, but it seems that 
imagecopyresampled is the origin of the problems.

So if you are using GD2 with large images, this may be the issue.
regards,
m.
Roman Duriancik wrote:
I have problem with jpg pictures in php pages. Path and picture is 
correct but when i want  see picture in php script  picture don't show. 
I use PHP4.3.4 on apache server 2.0.44 on Windopws 2000 Server.

Thanks for help.
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Re: [PHP] PHP MSSQL Images

2003-08-14 Thread Curt Zirzow
* Thus wrote [EMAIL PROTECTED] ([EMAIL PROTECTED]):
 ... 
 
 The following is the tag inserted to my webpage:
 
 img src=showpic.php?picid=1

View the source  of this image to see if any php errors are being
displayed.

Curt
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[PHP] PHP MSSQL Images

2003-08-12 Thread Chris . Jones
Hey ..

I am having a problem when i try to pull an image that has been stored in 
MS SQL Server from displaying properly.  I know, its a waste of space to 
store the actual image in SQL, but this is the way it needs to be done to 
be compatible with some of the other systems.  Anyway, if you have any 
input at all as to what could be wrong, please let me know.  I will 
provide my system information and code below for your review.  Thanks.

Platform: Linux Redhat 9
PHP Version  4.3.1
FreeTDS 0.61 (used to connect to Microsoft SQL Server)
SQL Server 2000

Here is my code (except for connection information, of course)

showpic.php

?
$eventid=$_GET[picid];

$dsource = SQL IP;
$sqluser = Username;
$sqlpass = Password;
$catalog = Database;

[EMAIL PROTECTED]($dsource,$sqluser,$sqlpass) or die(Could not connect 
to the database);
@mssql_select_db($catalog) or die (invalid);

$getpic = mssql_query(select Picture from GetPics where Id='$eventid') 
or die(Query is not achievable by SQL Server) or die(invalid);

while($rs=mssql_fetch_array($getpic))
{

header(Content-type: image/jpeg);
$image=$rs[Picture];
print $image;
}

?


The following is the tag inserted to my webpage:

img src=showpic.php?picid=1

Please let me know if you have any ideas, and if you dont, thanks for your 
help anyway. 

Chris
[EMAIL PROTECTED]

[PHP] PHP, DB, images, ODBC

2002-03-26 Thread David JURAS

Hi,
I'm working with PHP and ODBC to get images stored in a database (Blob fields). But I 
experiment little problems to display these pics. Has anyone any information about 
PHP/ODBC/images ?

Thanks,

Bilbo




Re: [PHP] PHP and Images stored in mysql

2002-01-10 Thread Rodrigo Peres

How can I echo an image from other location than the mysql using this kind
of script?? since I'm using php to output the image from database, there's a
way to change it to a ftp link for example if image doesn't exists??

this is my script:

?
$sql = SELECT Imagem_data,Imagem_type FROM imagens WHERE
CelebID='$celebID';
$query = new Query($conexao);
$query-executa($sql);
$resultado = $query-dados();
mysql_close();

$imagem_banco = $resultado['Imagem_data'];
$type = $resultado['Imagem_type'];
if($imagem_banco != ) {
HEADER(Content-type: $type);
echo($imagem_banco);
}
}
?

and I call it from my page with img src=lib_image.php?celebID='1'.

Thank's

Rodrigo


on 1/9/02 9:16 PM, Bogdan Stancescu at [EMAIL PROTECTED] wrote:

 My way around this is using two fallback stages: for one you test
 mysql_num_rows() after performing the query. If that's 0, use a default image
 placed in the database beforehand. If that also returns a null
 mysql_num_rows()
 then echo an image (probably the same default) from a known location in the
 filesystem.
 
 HTH
 
 Bogdan
 
 Rodrigo Peres wrote:
 
 Hi list,
 
 I have some images stored in a blob in mysql. I've made this code to output
 them to html, but the problem is if there's no image at a given id the page
 tooks a long time to load and display the broken image, there's a way to
 avoid this, I mean, there's a way to print a message or another image if
 there's none in mysql ???
 

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[PHP] PHP and Images stored in mysql

2002-01-09 Thread Rodrigo Peres

Hi list,

I have some images stored in a blob in mysql. I've made this code to output
them to html, but the problem is if there's no image at a given id the page
tooks a long time to load and display the broken image, there's a way to
avoid this, I mean, there's a way to print a message or another image if
there's none in mysql ???

?
$sql = SELECT Imagem_data,Imagem_type FROM imagens WHERE
CelebID='$celebID';
$query = new Query($conexao);
$query-executa($sql);
$resultado = $query-dados();
mysql_close();

$imagem_banco = $resultado['Imagem_data'];
$type = $resultado['Imagem_type'];
if($imagem_banco != ) {
HEADER(Content-type: $type);
echo($imagem_banco);
}
}
?

Thank's in advance

Rodrigo
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Re: [PHP] PHP and Images stored in mysql

2002-01-09 Thread Bogdan Stancescu

My way around this is using two fallback stages: for one you test
mysql_num_rows() after performing the query. If that's 0, use a default image
placed in the database beforehand. If that also returns a null mysql_num_rows()
then echo an image (probably the same default) from a known location in the
filesystem.

HTH

Bogdan

Rodrigo Peres wrote:

 Hi list,

 I have some images stored in a blob in mysql. I've made this code to output
 them to html, but the problem is if there's no image at a given id the page
 tooks a long time to load and display the broken image, there's a way to
 avoid this, I mean, there's a way to print a message or another image if
 there's none in mysql ???


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