hi
i'm pretty new to php. so let me ask you a pretty easy question:
the code below is supposed to import an image to a mssql database.
images are not meant to be stored in databases, i know. anyway, i'm
wondering why my code doesn't work.
can anybody give me some help?
regards
<HTML>
<HEAD>
<TITLE>php test</TITLE>
</HEAD>
<?php
$db = mssql_connect("xxx","xxx","xxx") or die("Can't connect to
mssql server.");
mssql_select_db("xxx", $db) or die("Can't select database.");
$name="51_DRAWING.jpg";
$type="image/pjpeg";
$file = fopen("C:/Temp/51_DRAWING_2.jpg", "r");
$filesize = filesize("C:/Temp/51_DRAWING_2.jpg");
$sql="INSERT INTO images(ID_IMAGE,ID_GCP,ID_LEVEL_IMG,IMAGE_BIN,IMAG E_NAME,IMAGE_SIZE,IMAGE_TYPE)";
$sql .=" VALUES('503','51','4','$img','$name','$filesize',' $type')";
$result=mssql_query($sql);
?>
</HTML> 8 2305
*** Georg Andersson wrote/escribió (3 Jun 2004 00:30:13 -0700): $name="51_DRAWING.jpg"; $type="image/pjpeg"; $file = fopen("C:/Temp/51_DRAWING_2.jpg", "r"); $filesize = filesize("C:/Temp/51_DRAWING_2.jpg");
$sql="INSERT INTO images(ID_IMAGE,ID_GCP,ID_LEVEL_IMG,IMAGE_BIN,IMAG E_NAME,IMAGE_SIZE,IMAGE_TYPE)"; $sql .=" VALUES('503','51','4','$img','$name','$filesize',' $type')";
Where is $img supposed to come from?
--
--
-- Álvaro G. Vicario - Burgos, Spain
--
Georg Andersson wrote: the code below is supposed to import an image to a mssql database. images are not meant to be stored in databases, i know. anyway, i'm wondering why my code doesn't work.
can anybody give me some help?
.... $sql="INSERT INTO images(ID_IMAGE,ID_GCP,ID_LEVEL_IMG,IMAGE_BIN,IMAG E_NAME,IMAGE_SIZE,IMAGE_TYPE)"; $sql .=" VALUES('503','51','4','$img','$name','$filesize',' $type')";
$result=mssql_query($sql);
No error-checking here?
Are you sure your $sql is correct?
You may need to use mysql_escape_string() on $img
?>
</HTML>
--
USENET would be a better place if everybody read: : mail address : http://www.catb.org/~esr/faqs/smart-questions.html : is valid for : http://www.netmeister.org/news/learn2quote2.html : "text/plain" : http://www.expita.com/nomime.html : to 10K bytes :
Hi Alvaro Where is $img supposed to come from?
--
it should actualy be $file instead of $img. sorry, i've forgot to change that...
*** Georg Andersson wrote/escribió (3 Jun 2004 06:45:29 -0700): it should actualy be $file instead of $img. sorry, i've forgot to change that...
Right, then $file is a file handler returned by fopen(). I've never
inserted files in databases but I'm pretty sure you have to provide the
file content itself, not a file handler used by a scripting language not
related to the database. Once you open the file you have to *read* it.
Check PHP manual for fread(), fgets(), etc.
--
--
-- Álvaro G. Vicario - Burgos, Spain
--
On 3 Jun 2004 00:30:13 -0700, ge*************@swisstopo.ch (Georg
Andersson) wrote: $name="51_DRAWING.jpg"; $type="image/pjpeg"; $file = fopen("C:/Temp/51_DRAWING_2.jpg", "r"); $filesize = filesize("C:/Temp/51_DRAWING_2.jpg");
$sql="INSERT INTO images(ID_IMAGE,ID_GCP,ID_LEVEL_IMG,IMAGE_BIN,IMAG E_NAME,IMAGE_SIZE,IMAGE_TYPE)"; $sql .=" VALUES('503','51','4','$img','$name','$filesize',' $type')";
From your previous post, replacing $img with $file :
$sql="INSERT INTO
images(ID_IMAGE,ID_GCP,ID_LEVEL_IMG,IMAGE_BIN,IMAG E_NAME,IMAGE_SIZE,IMAGE_TYPE)";
$sql .=" VALUES('503','51','4','$file','$name','$filesize', '$type')";
fopen() returns only a resource to the file. Use fread() to get its
contents.
Also, a quick Google search indicates that some people have had
difficulty using PHP with BLOB fields in MS SQL. If you have the same
trouble, you may have to start storing your images in the filesystem.
--
David ( @priz.co.uk )
Hi David fopen() returns only a resource to the file. Use fread() to get its contents.
i added fread() line to my code:
<?php
$db = mssql_connect("xxxx","xxx","xxx") or die("Can't connect to
mssql server.");
mssql_select_db("xxx", $db) or die("Can't select database.");
$name="51_DRAWING.jpg";
$type="image/pjpeg";
$file = fopen("C:/Temp/51_DRAWING_2.jpg", "r");
$filesize = filesize("C:/Temp/51_DRAWING_2.jpg");
$fileread = addslashes(fread($file, $filesize));
$sql="INSERT INTO images(ID_IMAGE,ID_GCP,ID_LEVEL_IMG,IMAGE_BIN,IMAG E_NAME,IMAGE_SIZE,IMAGE_TYPE)";
$sql .=" VALUES('503','51','4','$fileread','$name','$filesi ze','$type')";
$result=mssql_query($sql);
?>
i'm still getting a PHP Warning "Incorrect syntax near '\'". So, there
is no "\" in my sql-statement. i guess this means, the image is
treaten as a text and not as an image. how can this be avoided?
*** Georg Andersson wrote/escribió (4 Jun 2004 01:02:01 -0700): i'm still getting a PHP Warning "Incorrect syntax near '\'". So, there is no "\" in my sql-statement. i guess this means, the image is treaten as a text and not as an image. how can this be avoided?
If you are having problems inserting binary data into the database, you
might solve it using pure text. Check base64_encode() and base64_decode().
This makes stored info about 30% larger though.
--
--
-- Álvaro G. Vicario - Burgos, Spain
--
On 4 Jun 2004 01:02:01 -0700, ge*************@swisstopo.ch (Georg
Andersson) wrote: Hi David
fopen() returns only a resource to the file. Use fread() to get its contents.
i added fread() line to my code:
<?php $db = mssql_connect("xxxx","xxx","xxx") or die("Can't connect to mssql server."); mssql_select_db("xxx", $db) or die("Can't select database.");
$name="51_DRAWING.jpg"; $type="image/pjpeg"; $file = fopen("C:/Temp/51_DRAWING_2.jpg", "r"); $filesize = filesize("C:/Temp/51_DRAWING_2.jpg"); $fileread = addslashes(fread($file, $filesize));
$sql="INSERT INTO images(ID_IMAGE,ID_GCP,ID_LEVEL_IMG,IMAGE_BIN,IMAG E_NAME,IMAGE_SIZE,IMAGE_TYPE)"; $sql .=" VALUES('503','51','4','$fileread','$name','$filesi ze','$type')"; $result=mssql_query($sql); ?>
i'm still getting a PHP Warning "Incorrect syntax near '\'". So, there is no "\" in my sql-statement. i guess this means, the image is treaten as a text and not as an image. how can this be avoided?
SQL Server does not use \ as an escape character. You escape single
quoues by doubling them up (a ' becomes '' ). Not sure about
unprintable characters.
I used BLOB fields with SQL Server 7. I had to use special ADO
functions to INSERT and retrieve the data in chunks. I didn't use a
normal INSERT or SELECT to get it.
As suggested by Alvaro, you may have to encode it as text somehow.
--
David ( @priz.co.uk ) This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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