Re: [PHP-DB] PHP4, mssql_* and SQL Server 2005
Hi!
Have you checked that the query is escaped correctly? All single quotes
(') needs to be escaped with another single quote, but that has been for
MS-SQL all the time so I'll doubt that the problem is escaping.
Another issue I noticed is that SQL Server 2005 handles SMALLDATETIME
coulmns slightly different, and the default formatting of the retuned
date is in some collations different than in SQL Server 2000. So if you
have any queries relying on a where clause that checks a SMALLDATETIME
column you might need to adjust the date format.
Migrating to SQL 2005 might also cause some problems with character
encoding, since SQL 2005 internally works with Unicode. At least for us
who needs other characters than 7-bit ASCII... I had big trouble with
moving ISO-8859-15 XML documents to a SQL Server 2005 Unicode server and
parsing them into a native XML datatype (which is a really nice
feature!), but the problems can be overcomed with a litte extra work...
You could also check that you use the correct protocol version of
FreeTDS, if you're running PHP under Linux. I don't know how PHP
connects to MS-SQL on Windows though.
Currently I have no other thoughts of what could cause your problems, at
least without not being able to actually see the failed queries.
Regards,
Krister Karlström, Helsinki, Finland
Chris wrote:
Please always cc the mailing list so others can offer extra
advice/suggestions etc.
Weaver Hickerson wrote:
they work in query analyzer.
ok so the queries aren't the problem.
do you get an error? what is it?
Might need to play with these settings:
http://www.php.net/manual/en/function.mssql-min-error-severity.php
http://www.php.net/manual/en/function.mssql-min-message-severity.php
And they seem to work on a new server I setup for testing that has
ZendCore and on which I installed the Microsoft PHP driver and ported
the code to that driver. :(
That does seem to point to the driver, not sure what other suggestions I
can offer.
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Re: [PHP-DB] PHP4, mssql_* and SQL Server 2005
Ouch! That was like the.. longest.. query that I've ever seen! I'm surprised that it actually works and even more impressed that someone actually figured that out... It might take a long time to run, since it uses so many subqueries (if there's lots of data or even a moderate amount of data). Are you sure there's no easier way to fetch your data? Most subqueries can be re-written using a join and some more where-comparisons instead... Actually I'm more amazed that you actually can send that query and that it works.. There's a slight chance that your problems might be solved by rewriting that query... Maybe it's just to long or something, I don't know. I will however not take the time to refactor that query... :) Are you running PHP under Linux or Windows? You said that you only found two queries that breaks? Wll, if the shorter ones works, but the longer ones hangs, then maybe that's the problem...? :) Regards, Krister Karlström, Helsinki, Finland Weaver Hickerson wrote: Krister - I've included the query below - If I paste it in query analyzer on 2000, or 2005 it works fine. Just coming through PHP4 to 2005, it hangs. I actually don't get an error, and then the browser says the server is too busy to handle the request. Strange things like that. (I am using Zend Core with Apache and PHP 5). I will check for smalldatetime fields. SELECT COUNT(DISTINCT C.CHART_ID) AS TOTAL, SUM(ISNULL(FEE_SCHEDULE.FEE,0) * UNITS) AS TOTALFEE FROM HL7 H, CHART C LEFT OUTER JOIN CHART_CODE D LEFT OUTER JOIN CHART_CODE_MODIFIER CCM ON CCM.CHART_CODE_ID = D.CHART_CODE_ID AND CCM.CHART_CODE_MODIFIER_ID = (SELECT MIN(CHART_CODE_MODIFIER_ID) FROM CHART_CODE_MODIFIER WHERE CHART_CODE_ID = D.CHART_CODE_ID) LEFT OUTER JOIN FEE_SCHEDULE ON FEE_SCHEDULE.CPT_CODE = D.CHART_CODE AND FEE_SCHEDULE.START_DATE = (SELECT MSH_DTM FROM HL7, CHART, CHART_CODE WHERE CHART_CODE.CHART_ID = CHART.CHART_ID AND CHART.HL7_ID = HL7.HL7_ID AND CHART_CODE.CHART_CODE_ID = D.CHART_CODE_ID) AND FEE_SCHEDULE.END_DATE = (SELECT MSH_DTM FROM HL7, CHART, CHART_CODE WHERE CHART_CODE.CHART_ID = CHART.CHART_ID AND CHART.HL7_ID = HL7.HL7_ID AND CHART_CODE.CHART_CODE_ID = D.CHART_CODE_ID) AND FEE_SCHEDULE.SENDING_FACILITY = (SELECT SENDING_FACILITY FROM HL7, CHART, CHART_CODE WHERE CHART_CODE.CHART_ID = CHART.CHART_ID AND CHART.HL7_ID = HL7.HL7_ID AND CHART_CODE.CHART_CODE_ID = D.CHART_CODE_ID) AND FEE_SCHEDULE.CODE_TYPE = 'pro' AND FEE_SCHEDULE.MODIFIER = CCM.MODIFIER LEFT OUTER JOIN FEE_SCHEDULE FS ON FS.CPT_CODE = D.CHART_CODE AND FS.START_DATE = (SELECT MSH_DTM FROM HL7, CHART, CHART_CODE WHERE CHART_CODE.CHART_ID = CHART.CHART_ID AND CHART.HL7_ID = HL7.HL7_ID AND CHART_CODE.CHART_CODE_ID = D.CHART_CODE_ID) AND FS.END_DATE = (SELECT MSH_DTM FROM HL7
Re: [PHP-DB] mysqli_stmt_bind_param question
Hi! First of all you must connect to the server and then perform an init to get a statement. Maybe you dropped that code out here... Please have a look at the manual page: http://www.php.net/manual/en/function.mysqli-stmt-prepare.php In order to get the result you also need to bind the result to a variable and then fetch the data. I'll guess that your error comes from an unsuccessful initialization of your prepared statement. If you don't want to include your primary key column that is autogenerated you just ignore it. That means to write a proper query like: INSERT INTO legioCurrent (column_1, column_2, ...) VALUES (?,?,...) I would also like to propose for you to use the object oriented style, since it's more readable and easier to get an overview of. You also don't need to mess around with so many parameters. Greetings, Krister Karlström, Helsinki, Finland Jason Pruim wrote: Hi Everyone, I have a MySQL database that I am accessing from PHP. The table in question has a auto increment field on it and I don't want to include that in my add/edit/update query's to the database... But I can't seem to figure out how to ignore it? Everything I have done seems to fail.. I am using prepared statements so Im not sure it that is it, but I would like to keep using them since it reduces the security issues a little... Here is the code from one of my prepared statements: //Create the statement mysqli_stmt_prepare($stmt, INSERT INTO legionCurrent VALUES ( ?,?,?,?,?,?,?,?)); mysqli_stmt_bind_param($stmt, '', $FName, $LName, $Add1, $Add2, $City, $State, $Zip, $XCode)or die(mysqli_error($addlink)); //Add the record mysqli_stmt_execute($stmt) or die(mysqli_error($addlink)); Here is the error I get in my logs: [Fri Apr 4 09:35:32 2008] [error] PHP Warning: mysqli_stmt_bind_param() [a href='function.mysqli-stmt-bind-param'function.mysqli-stmt-bind-param/a]: invalid object or resource mysqli_stmt\n in /Volumes/RAIDer/webserver/Documents/dev/OLDBv2/add.php on line 91 Line 91 is the mysqli_stmt_bind_param() line above My database structure looks like this: | FName | LName | Add1| Add2| City | State | Zip | XCode | Reason | Record I know there is away to do it, but all my searching and reading hasn't told me anything... I've looked in the php manual, and mysql, plus various websites... and I just can't figure this one out... RTFM's are appreciated, as long as M is defined! :) Thanks for taking the time to look! -- Jason Pruim Raoset Inc. Technology Manager MQC Specialist 3251 132nd ave Holland, MI, 49424-9337 www.raoset.com [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] mysqli_stmt_bind_param question
Hi again Jason! Jason Pruim wrote: On Apr 4, 2008, at 2:35 PM, Krister Karlström wrote: You also don't need to mess around with so many parameters. What do you mean by this? Since you're working in object context you mostly just refer to your variable with the name of the object and the calls the method with the arrow - operator. But anyway, if you're not familiar with object orientation then I think it's no point for me to go into this discussion right now... :-) On the manual page, to which I posted a link to you in my previous post, there's the same code in both procedural style and in object oriented style. I think you'll get the point by comparing those two examples. Greetings, Krister Karlström, Helsinki, Finland -- * Ing. Krister Karlström, Zend Certified Engineer * * Systemutvecklare, IT-Centralen * * Arcada - Nylands Svenska Yrkeshögskola * * Jan-Magnus Janssons plats 1, 00550 Helsingfors, Finland * * Tel: +358(20)7699699 GSM: +358(50)5328390 * * E-mail: [EMAIL PROTECTED] * -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] mssql_connect error only on command line
Hi! That description did not help us very much... I think you're just connecting using the wrong credentials, or using no at all. From where do you grab the hostname, username and password? Maybe it's stored in your session or something, and that session is not working in your command line environment..? It could be just about anything that's your problem! :) My advice is to take a look att the parameters using to connect to the server, maybe echo them to stdout before doing your connect... It might be that they are not defined when running your script on command-line. In general, very few script are written in such a way that they work the same way in a webserver environment and on command line. You should design your scripts for either the environment. Greetings, Krister Karlström, Helsinki, Finland Molteni Davide wrote: Hi, I have a problem with all my php scripts that use MS SQL Server, but only when I start them from the command line I'm using Apache 2.2.4 with PHP 5.2.5 on a server Win2k3 with SP2, and MS SQL Server 2000 on another Win2k3 server. If I run the same script using a webbrower as an internet page it works fine, but if I run it from the command line (c:\php scipt.php) I will receive messages like this: Warning: mssql_connect(): Unable to connect to server: servername,1433 in c:\script.php on line # Can someone have an idea what could be the problem? Can someone please help me? What should I do to make it working? Tanks a lot in advance. Davide -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] mssql_connect error only on command line
Hi! Please have a look at the manual pages (if you have not allready done that). There might be some information for you in the user comments: http://www.php.net/manual/en/function.mssql-connect.php Manual page about command line scripting: http://www.php.net/manual/en/features.commandline.php Now you're using windows so I don't think that FreeTDS is used, right? But if you were on a Linux environment you should note that you need to have the server configured in the config file for FreeTDS, exactly in the same way your connecting. That means if you use fully qualified domain names the server should be registered with the same hostname in freetds.conf. But that has probably nothing to do with this now since you're using Windows... The documentation of mssql_connect() says that you can specify a portnumber like you did, but I found some examples on the net using a colon (:) instead. Try that also, or try with the default port number... I'm not sure if that is gonna help you at all, but it's worth trying... Or maybe you have some problem with name resolution through DNS... Try using the IP adress when connecting... It's really hard to tell what's wrong. However, I have seldom used PHP in a Windows environment so maybe I'm the wrong person to answer your questions! :) I do however have a command line application running on Windows XP that successfully uses MySQL on a Linux server, never had any trouble with that... Greetings, Krister Karlström, Helsinki, Finland Molteni Davide wrote: Hi, thanks for your quick answer, but in the script I don't use sessions variables to store host, user and password, I wrote something like that: $sc=mssql_connect(hostname,1433,userid,password); Can you please tell me where I can find information about how to design scripts for either the environment? Thanks a lot. Davide 2008/4/2, Krister Karlström [EMAIL PROTECTED]: Hi! That description did not help us very much... I think you're just connecting using the wrong credentials, or using no at all. From where do you grab the hostname, username and password? Maybe it's stored in your session or something, and that session is not working in your command line environment..? It could be just about anything that's your problem! :) My advice is to take a look att the parameters using to connect to the server, maybe echo them to stdout before doing your connect... It might be that they are not defined when running your script on command-line. In general, very few script are written in such a way that they work the same way in a webserver environment and on command line. You should design your scripts for either the environment. Greetings, Krister Karlström, Helsinki, Finland Molteni Davide wrote: Hi, I have a problem with all my php scripts that use MS SQL Server, but only when I start them from the command line I'm using Apache 2.2.4 with PHP 5.2.5 on a server Win2k3 with SP2, and MS SQL Server 2000 on another Win2k3 server. If I run the same script using a webbrower as an internet page it works fine, but if I run it from the command line (c:\php scipt.php) I will receive messages like this: Warning: mssql_connect(): Unable to connect to server: servername,1433 in c:\script.php on line # Can someone have an idea what could be the problem? Can someone please help me? What should I do to make it working? Tanks a lot in advance. Davide -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: Help with JOIN query
Hi! You can't use the MAX() function if you're not using a GROUP BY clause. The MAX() function can only grab the maximum value of a grouped column, as with MIN(), COUNT(), AVG() etc.. Greetings, Krister Karlström, Helsinki Jonathan Crawford wrote: I think this is what you mean. You just want the timestamp and action from B in addition to something from A (I guessed product_ref), right? The MAX() function should take care of getting the latest timestamp. explicit join: SELECT MAX(TableB.timestamp), TableB.action, TableA.product_ref FROM TableB JOIN TableA ON TableA.record_id = TableB.record_id ORDER BY TableB.action or if you want to join your tables implicitly in your WHERE clause, similar to what you had before, implicit join: SELECT MAX(TableB.timestamp), TableB.action, TableA.product_ref FROM TableB WHERE TableA.record_id = TableB.record_id ORDER BY TableB.action The problem with the implicit joins versus explicit joins is that you can't ever do OUTER JOINs, where you want many from one table and one (or many) from another table. For example if you want all sales reps and their sales, even if they don't have any. Implicit (or explicit INNER) JOINs will not show you all of the data. Jonathan Crawford [EMAIL PROTECTED] Original message Date: 6 Mar 2008 18:46:18 - From: [EMAIL PROTECTED] Subject: php-db Digest 6 Mar 2008 18:46:18 - Issue 3990 To: php-db@lists.php.net php-db Digest 6 Mar 2008 18:46:18 - Issue 3990 Topics (messages 44700 through 44700): Help with JOIN query 44700 by: Graham Cossey Administrivia: To subscribe to the digest, e-mail: [EMAIL PROTECTED] To unsubscribe from the digest, e-mail: [EMAIL PROTECTED] To post to the list, e-mail: php-db@lists.php.net -- Date: Thu, 6 Mar 2008 18:46:14 + From: Graham Cossey [EMAIL PROTECTED] Subject: Help with JOIN query To: php-db@lists.php.net I can't see how to accomplish what I need so if anyone has any suggestions they would be gratefully received... I'm using mysql 4.0.20 by the way. I have two tables : TableA record_id product_ref TableB timestamp record_id action I want to create a SELECT that joins these 2 tables where the JOIN to TableB only returns the most recent entry by timestamp. At present (using PHP) I do a SELECT on TableA then for each record returned I perform a 2nd SELECT something like : SELECT timestamp, action FROM TableB WHERE record_id = '$record_id' ORDER BY timestamp DESC LIMIT 1 I now want to do it with one query to enable sorting the results by 'action' from TableB. Any suggestions? Hopefully I've made sense, if not I'll happily try and explain further on request. -- Graham -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: Help with JOIN query
Yes, I'm totally with you Graham and I have created your tables and added some test data (you saw it in a previous post).. But I can't seem to come up with any solution for you, at least with MySQL 4.0.x. So if no one else has any brilliant solutions for this I think you'll need to stick with your solution to make the request in two steps, using PHP and an array to sort and put everyting together. Now, hopefully, you don't have hundreds of kilos of rows in your tables.. :) There might be solutions for this problem with never versions of MySQL, like using subqueries combined with joins like someone mentioned, but I'm not sure. I however, have never tested (or even needed to test) something like that, so I can't help you with that. If it is not important to keep the history of actions for each product you could simply update TableB, but I'll guess that's not the case - otherwise you wouldn't probably been asking us.. :) Another solution would be to move the previous action to an other table, like TableC for instance... Then TableC would be your history, TableB would only have your latest action.. But then again, we would have a one-to-one relation which basically means that you don't need two tables anymore. But you could at least consider this, if you are able to alter the database design a bit. Greetings, Krister Karlström, Helsinki Graham Cossey wrote: Thank you for your input Jonathan but it's not quite what I need. I need the latest action from TableB (if it exists) determined by the record_id matching TableA and where there are more than one matching record in TableB select the one with the latest timestamp. As an over-simplified example of what I'm trying to achieve : TableA record_id product_ref 1 product1 2 product2 TableB timestamp record_id action 20080301 1start 20080302 1middle 20080301 2start 20080302 2middle 20080303 2end What I need returned is : 1,product1,middle 2,product2,end --- Graham -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Help with JOIN query
Hi! Graham Cossey wrote: TableA record_id product_ref TableB timestamp record_id action I want to create a SELECT that joins these 2 tables where the JOIN to TableB only returns the most recent entry by timestamp. For instance, to select all columns: select * from TableA join TableB on TableA.record_id = TableB.record_id order by timestamp desc limit 1 So you just join everything, then order by time in descening order and then just returning the first record = the newest record in the database. If you don't want all columns, then simply replace the star with the names of the columns you want. I hope that this is what you wanted the query to do.. :) Cheers, Krister Karlström, Helsinki, Finland -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Help with JOIN query
Hi! Graham Cossey wrote: I was hoping to avoid joining everything as there can be many entries in TableB for each record in TableA. Also wouldn't your query only return one record? I need to return all records from TableA with the latest action from TableB as well. Yes, sorry - I realised that after I sent away my first reply.. :) I have a colleague who always says that people don't READ their mail - the just LOOK at their mail.. It's so true! I missed some of your points. But maybe you got some ideas from my other mail, which I seem to have posted only to you directly.. Oh well, this is a tricky one anyway... /Krister Karlström -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Help with JOIN query
Hi again! We're getting a bit of topic here, since this is pure SQL.. But anyway... I've played around with this one a bit since it seemed quite interesting... The best I can do is to get the oldest action... select TableA.record_id, product_ref, action, time_stamp from TableA join TableB on TableA.record_id = TableB.record_id group by record_id; Here's the test data: mysql select TableA.record_id, product_ref, action, time_stamp from TableA join TableB on TableA.record_id = TableB.record_id; +---+-+++ | record_id | product_ref | action | time_stamp | +---+-+++ | 1 | 100 | A | 20080306220037 | | 1 | 100 | C | 20080306220041 | | 1 | 100 | E | 20080306220045 | | 2 | 102 | A | 20080306220052 | | 3 | 110 | A | 20080306220055 | | 3 | 110 | E | 20080306220058 | | 4 | 120 | B | 20080306220105 | | 4 | 120 | C | 20080306220109 | +---+-+++ And with the query above we get the opposite of the desired behavior, the oldest action (if that's the order in the database): mysql select TableA.record_id, product_ref, action, time_stamp from TableA join TableB on TableA.record_id = TableB.record_id group by record_id; +---+-+++ | record_id | product_ref | action | time_stamp | +---+-+++ | 1 | 100 | A | 20080306220037 | | 2 | 102 | A | 20080306220052 | | 3 | 110 | A | 20080306220055 | | 4 | 120 | B | 20080306220105 | +---+-+++ 4 rows in set (0.00 sec) Now is the question: Does anyone know how to get the 'group by' clause to leave a specific row 'visible' at top? Like the last inserted or by the order of another column... Since MySQL 4.1 there are also a GROUP_CONCAT() function that can concatenate multiple 'rows' to a string in a desired order, but it does not support the limit statement... so that won't help us much I think. We can get all the actions in a string with the newest first, but then some post-stripping of the data is needed. It seems like you need to do this with two queries in PHP, if no one has an answer to the question stated above. You can always buffer your result in an array in PHP and do whatever sorting you want to before using your data... With the MAX() function we can found out when the last action was made, but we get the wrong action with the correct time: mysql select TableA.record_id, product_ref, action, max(time_stamp) from TableA join TableB on TableA.record_id = TableB.record_id group by record_id; +---+-++-+ | record_id | product_ref | action | max(time_stamp) | +---+-++-+ | 1 | 100 | A | 20080306220045 | | 2 | 102 | A | 20080306220052 | | 3 | 110 | A | 20080306220058 | | 4 | 120 | B | 20080306220109 | +---+-++-+ 4 rows in set (0.00 sec) Hmm... Now I'm stuck! :) Greetings, Krister Karlström, Helsinki, Finland Graham Cossey wrote: I can't see how to accomplish what I need so if anyone has any suggestions they would be gratefully received... I'm using mysql 4.0.20 by the way. I have two tables : TableA record_id product_ref TableB timestamp record_id action I want to create a SELECT that joins these 2 tables where the JOIN to TableB only returns the most recent entry by timestamp. At present (using PHP) I do a SELECT on TableA then for each record returned I perform a 2nd SELECT something like : SELECT timestamp, action FROM TableB WHERE record_id = '$record_id' ORDER BY timestamp DESC LIMIT 1 I now want to do it with one query to enable sorting the results by 'action' from TableB. Any suggestions? Hopefully I've made sense, if not I'll happily try and explain further on request. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Help with JOIN query
This will give you: ERROR : Invalid use of group function It seems like the use of an aggregate (or how is it spelled?) function is not allowed in a join statement... /Krister Jon L. wrote: You can try adding a quick test to the ON statement... SELECT * FROM TableA INNER JOIN TableB ON TableA.record_id = TableB.record_id AND TableB.timestamp = MAX(TableB.timestamp) Now, I haven't tested it. I can only say the theory of it is accurate. - Jon L. On Thu, Mar 6, 2008 at 12:46 PM, Graham Cossey [EMAIL PROTECTED] wrote: I can't see how to accomplish what I need so if anyone has any suggestions they would be gratefully received... I'm using mysql 4.0.20 by the way. I have two tables : TableA record_id product_ref TableB timestamp record_id action I want to create a SELECT that joins these 2 tables where the JOIN to TableB only returns the most recent entry by timestamp. At present (using PHP) I do a SELECT on TableA then for each record returned I perform a 2nd SELECT something like : SELECT timestamp, action FROM TableB WHERE record_id = '$record_id' ORDER BY timestamp DESC LIMIT 1 I now want to do it with one query to enable sorting the results by 'action' from TableB. Any suggestions? Hopefully I've made sense, if not I'll happily try and explain further on request. -- Graham -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] problem in recorset that seems temporary
Hi!
I did not quite get the point here, but maybe that's not needed to solve
the problem... :-)
First of all, you seem to have enabled the register_global setting,
since I can't see from where the parameter $id is posted. Use the $_GET
or $_POST superglobal array instead. Take big care of input filtering
also and do not put dirty data directly into your query.
Second, the function mysql_db_query() has been deprecated for a long
time now, use mysql_select_db() and mysql_query() instead.
I think your problem is in your while-loop:
You loop over the result set $result and fetch the row as an object.
Then you execute a delete statement and at the same time overwriting
your previous result set $result, thus you will break the loop with an
error that the provided parameter is not a result set.
I also guess you have a typo in your last statement where you (probably)
try to delete the gallery. I guess that you maybe want to use the
variable $galerie_id..? Well, I can't tell without any more information
about your application.
Greetings,
Krister Karlström, Helsinki, Finland
Ruprecht Helms wrote:
while($row = mysql_fetch_object($result))
{
$galerie_id=$row-ID;
$result=mysql_db_query(pferdeservice-karle,DELETE FROM
Galerie_kultur WHERE GalerieID=$id);
}
$result=mysql_db_query(pferdeservice-karle,DELETE FROM Galerie WHERE
ID=$id);
echo mysql_error();
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