[PHP-DB] php and local printing
Hello, What is the best way to print on a locally connected printer from php ? For printing we want to give the user three options : 1. Screen 2. Local Printer 3. Server Printer What you guys are using for printing on locally connected printer from php. Thanks in advance. Peace -- Rajesh : [ GNU/Linux One Stanza Tip (LOST) ]### Sub : kill-proc (killing user process by an user)LOST #285 #!/bin/sh ps -aef | grep $USER echo -en Process-No to kill : ; read P_ID kill $P_ID [EMAIL PROTECTED] : -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Help with a query
I can't figure this query out. I want to pull data from 2 tables but using a primary key from one. Here is what I think it should look like. SELECT item_id, subtotal, quantity from shopping_cart WHERE order_id = (SELECT order_id FROM orders WHERE session_id = session_id()) AND customer_id = $customer_id; -Jonathan -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Help with MySQL Logic
Hello All, This may be a bit off topic. If so, my apololgies. A client of mine, a rail car storage company, has asked that I create a PHP/MySQL application in which they will maintain and track rail cars. I am having a bit of trouble however working through one of thier requirements. They need to know in what position the rail car is on each track. For example, they might have a track called X which will hold 30 rail cars. They would enter the car information for 30 cars and associate each of them with track X. If one of the car owners decides to move car 3 out of storage, then the remaining 29 cars must be re-numbered ( ie; car 1 and 2 would remain and car 4-30 would become car 3-29 ). In the same manner, I need to be able to add a car to the track once an empty slot is available. For example, If the new car goes at the front of the track then it would become car 1 and the remaining cars, originally numbered 1-29 would become 2-30. I hope I explained thourougly. Any suggestions would be appreciated. Randy
[PHP-DB] Interbase and PHP locking (was: Interbase locking)
Hello everybody,
Some days ago I've posted a question about interbase locking. I've solved
the problem after searching at interbase developers network and php manuals.
Let me drop some line about my experiences...
Deadlock is when two (or more) transactions want update or delete a field
in the same table at the same time.
For cause a deadlock in Interbase do the following:
$tr_id1 = ibase_trans();
$tr_id2 = ibase_trans();
$result = ibase_query($tr_id1, UPDATE TEST SET FIELD_A=FIELD_C);
echo First result: $resultbr;
ibase_commit($tr_id1);
$result = ibase_query($tr_id2, UPDATE TEST SET FIELD_A=FIELD_C);
echo Second result: $resultbr;
If you run this you can see that the second result will be false and you'll
get
an error about interbase deadlock. Thats all right about interbase his do
its best when doing this. This error must be handled by the application.
So simple complete the code above with some check and rollback, like this:
$tr_id1 = ibase_trans();
$tr_id2 = ibase_trans();
$result = ibase_query($tr_id1, UPDATE TEST SET FIELD_A=FIELD_C);
echo First result: $resultbr;
ibase_commit($tr_id1);
$result = 0;
while (!$result) {
$result = ibase_query($tr_id2, UPDATE TEST SET FIELD_A=FIELD_C);
echo Second result: $resultbr;
if (!$result) {
ibase_rollback($tr_id2);
$tr_id2 = ibase_trans();
}
}
ibase_commit($tr_id2);
This will works nicely, and you will see that the first query will fail but
the second
query (in another one transaction!) will be correct. The complete solution
is when you
write a function for the queries to runs in a transaction and its check.
Regards,
Tibor
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RE: [PHP-DB] Help with a query
2) Have you tried echoing the query to see exactly what is being passed
to
the RDBMS? That is, perhaps session_id() or $customer_id is not giving
a
valid value.
Yes the values are all their.
I am using MySQL so I guess I will have to go a different route.
I'll have to run 2 queries which is what I wanted to avoid, but I guess
there's no other way
Thanks for the quick reply.
--- Jonathan
-Original Message-
From: Paul Burney [mailto:[EMAIL PROTECTED]
Sent: Monday, March 03, 2003 10:06 AM
To: [EMAIL PROTECTED]; PHP Database List
Subject: Re: [PHP-DB] Help with a query
on 3/31/03 10:41 AM, Jonathan Villa at [EMAIL PROTECTED] appended
the
following bits to my mbox:
I want to pull data from 2 tables but using a primary key from one.
Here is what I think it should look like.
SELECT item_id, subtotal, quantity from shopping_cart WHERE order_id =
(SELECT order_id FROM orders WHERE session_id = session_id()) AND
customer_id = $customer_id;
A few questions:
1) Are you sure your RDBMS supports sub-queries? Most do, but MySQL
does
not.
2) Have you tried echoing the query to see exactly what is being passed
to
the RDBMS? That is, perhaps session_id() or $customer_id is not giving
a
valid value.
3) Using the results of (2) above, have you input the query into another
RDBMS client (i.e., command line version) to see what kind of errors are
given.
HTH.
Sincerely,
Paul Burney
http://paulburney.com/
?php
while ($self != asleep) {
$sheep_count++;
}
?
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Re: [PHP-DB] Help with MySQL Logic
Heh. Sounds like a programming class homework project. I would say through the clever use of where clauses's, like: UPDATE position SET posistion = (position-1) WHERE position $idremoved; Would work, assuming $idremoved containted the position of the car removed. On Mon, 2003-03-03 at 10:59, Rankin, Randy wrote: Hello All, This may be a bit off topic. If so, my apololgies. A client of mine, a rail car storage company, has asked that I create a PHP/MySQL application in which they will maintain and track rail cars. I am having a bit of trouble however working through one of thier requirements. They need to know in what position the rail car is on each track. For example, they might have a track called X which will hold 30 rail cars. They would enter the car information for 30 cars and associate each of them with track X. If one of the car owners decides to move car 3 out of storage, then the remaining 29 cars must be re-numbered ( ie; car 1 and 2 would remain and car 4-30 would become car 3-29 ). In the same manner, I need to be able to add a car to the track once an empty slot is available. For example, If the new car goes at the front of the track then it would become car 1 and the remaining cars, originally numbered 1-29 would become 2-30. I hope I explained thourougly. Any suggestions would be appreciated. Randy -- Adam Voigt ([EMAIL PROTECTED]) The Cryptocomm Group My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc signature.asc Description: This is a digitally signed message part
Re: [PHP-DB] Help with MySQL Logic
A client of mine, a rail car storage company, has asked that I create a PHP/MySQL application in which they will maintain and track rail cars. I am having a bit of trouble however working through one of thier requirements. They need to know in what position the rail car is on each track. For example, they might have a track called X which will hold 30 rail cars. They would enter the car information for 30 cars and associate each of them with track X. If one of the car owners decides to move car 3 out of storage, then the remaining 29 cars must be re-numbered ( ie; car 1 and 2 would remain and car 4-30 would become car 3-29 ). In the same manner, I need to be able to add a car to the track once an empty slot is available. For example, If the new car goes at the front of the track then it would become car 1 and the remaining cars, originally numbered 1-29 would become 2-30. Not sure if this helps or if you already realize this... Say you have a table that identifies the track_id and car_id. Now, you delete the car_id for car #3, like you've said. So, that row is deleted and you've got a hole now. You can run a query such as: UPDATE table SET car_id = car_id - 1 WHERE car_id BETWEEN 4 AND 30 AND track_id = XX to adjust all of the other car_id numbers. Now, say you want to add a new car to the beginning. UPDATE table SET car_id = car_id + 1 WHERE track_id = XX and then insert your new car at position #1. Throw in some checks to make sure you don't go over 30 cars and you should have it. You can get a count of how many cars are on a certain track with: SELECT COUNT(*) AS c FROM table WHERE track_id = XX Hope that helps. It sounds like a fun project. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Help with MySQL Logic
What about two tables, one would have this: carID carCompany carPayload etc... the second this: position carID This table would be static and if a car was not present in certain conditions they would contain (-1) for the carID. Then finding an empty slot to place an incoming car should be doable (stepping through to find (-1), and when a car is removed from a slot, then you could update the list, just dump out the 'current' order into an array, and then you would be able to sort, order, etc... Hope this makes sense. -Brad Heh. Sounds like a programming class homework project. I would say through the clever use of where clauses's, like: UPDATE position SET posistion = (position-1) WHERE position $idremoved; Would work, assuming $idremoved containted the position of the car removed. On Mon, 2003-03-03 at 10:59, Rankin, Randy wrote: Hello All, This may be a bit off topic. If so, my apololgies. A client of mine, a rail car storage company, has asked that I create a PHP/MySQL application in which they will maintain and track rail cars. I am having a bit of trouble however working through one of thier requirements. They need to know in what position the rail car is on each track. For example, they might have a track called X which will hold 30 rail cars. They would enter the car information for 30 cars and associate each of them with track X. If one of the car owners decides to move car 3 out of storage, then the remaining 29 cars must be re-numbered ( ie; car 1 and 2 would remain and car 4-30 would become car 3-29 ). In the same manner, I need to be able to add a car to the track once an empty slot is available. For example, If the new car goes at the front of the track then it would become car 1 and the remaining cars, originally numbered 1-29 would become 2-30. I hope I explained thourougly. Any suggestions would be appreciated. Randy -- Adam Voigt ([EMAIL PROTECTED]) The Cryptocomm Group My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Help with MySQL Logic
Coming from a non-programmer... If you were to take the cars' IDs and put them in an array named for the track a specific group of cars is on, that would give you indexes, thus positions, from 0-N depending on how many cars are there. Any time you have to add or delete cars, you should be able to delete or add the ID from or to the array and the indexing would automatically tell you what position all of the cars are in. You should be able to move cars (IDs) around in the array with PHP too. So you'd have: $track1 = array(2,4,6,8); //where 2,4,6,8 are the IDs of rail cars And you'd use the serialize and unserialize functions to get the data into and out of MySQL and PHP to manipulate the order of the IDs inside the array. Like I said, I'm a non-programmer, but a track with cars on it sounds pretty similar to a real-world representation of an array structure to me. I'm sure there will be myriad other ways to get at this problem with their own good and bad points. I agree with John though, sounds like a fun project. Rich -Original Message- From: Adam Voigt [mailto:[EMAIL PROTECTED] Sent: Monday, March 03, 2003 11:25 AM To: Rankin, Randy Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Help with MySQL Logic Heh. Sounds like a programming class homework project. I would say through the clever use of where clauses's, like: UPDATE position SET posistion = (position-1) WHERE position $idremoved; Would work, assuming $idremoved containted the position of the car removed. On Mon, 2003-03-03 at 10:59, Rankin, Randy wrote: Hello All, This may be a bit off topic. If so, my apololgies. A client of mine, a rail car storage company, has asked that I create a PHP/MySQL application in which they will maintain and track rail cars. I am having a bit of trouble however working through one of thier requirements. They need to know in what position the rail car is on each track. For example, they might have a track called X which will hold 30 rail cars. They would enter the car information for 30 cars and associate each of them with track X. If one of the car owners decides to move car 3 out of storage, then the remaining 29 cars must be re-numbered ( ie; car 1 and 2 would remain and car 4-30 would become car 3-29 ). In the same manner, I need to be able to add a car to the track once an empty slot is available. For example, If the new car goes at the front of the track then it would become car 1 and the remaining cars, originally numbered 1-29 would become 2-30. I hope I explained thourougly. Any suggestions would be appreciated. Randy -- Adam Voigt ([EMAIL PROTECTED]) The Cryptocomm Group My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc
[PHP-DB] Mapping objects to tables: the actual database object
Hey everyone, I'm investigating using PHP's OOP functionality instead of the traditional model to implement a project. I've been looking at information about mapping objects to tables, and have mostly found what I needed, but am left with a question I would like to ask here. If this is the wrong list for this type of question, please be so kind to redirect me somewhere else, and please cc me on replies as I do not subscribe to this list. When mapping objects to tables, how would one usually handle the actual database object? Suppose we have a class Database derived from PostgresDatabase which implements database functionality (query(), dataQuery(), numRows(), ...) and then some actual classes: SomeClass | | SomeSubClass And SomeSubClass has a MemberClass attribute. All of these classes need access to the database to do their work. SomeClass could just have a Database attribute through which database operations are conducted, and so SubClass would have it too. The trickier part (or maybe I am just thick) seems to be MemberClass. How would this class best perform it's database operations? There are at least two possibilities: 1. Pass MemberClass a Database object when constructing it The problem with this is that in code that creates just a MemberClass object there might not be a Database class available. This could be eliminated with a factory class perhaps. 2. Make all classes that need database access derive from BaseClass and let that class set up access Has anyone else been down this road? I'd appreciate feedback on this issue. Thanks! -- Alexander Deruwe AQS-CarControl -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] ODBC Timout for MS SQL Driver
Does anyone know a reliable way to set an ODBC connection to timeout after a very short time? I've tried set_time_limit but when the query is running PHP does not drop out, I've also tried odbc_setoption on the connection before executing. Basically, I want it to just fail quicker if it is going to. 30 seconds is too long. Ryan Ryan Jameson Software Development Services Manager International Bible Society W (719) 867-2692 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] ODBC Timout for MS SQL Driver
Ok I tried this another way but using the ms sql functions along with: ini_set(mssql.timeout,2); ini_set(mssql.connect_timeout,2); ... no change, it still takes about 7-8 seconds to decide it couldn't connect. Ryan -Original Message- From: Ryan Jameson (USA) Sent: Monday, March 03, 2003 11:01 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] ODBC Timout for MS SQL Driver Does anyone know a reliable way to set an ODBC connection to timeout after a very short time? I've tried set_time_limit but when the query is running PHP does not drop out, I've also tried odbc_setoption on the connection before executing. Basically, I want it to just fail quicker if it is going to. 30 seconds is too long. Ryan Ryan Jameson Software Development Services Manager International Bible Society W (719) 867-2692 -- 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
[PHP-DB] GD support in 4.3?
According to the GD web site http://www.boutell.com/gd/, there is native GD support in v4.3.0. Anyone know how to enable this support? Thanks. Nate. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] GD support in 4.3?
Wrong mailing list. --with-gd On Mon, 3 Mar 2003, nate hurto wrote: According to the GD web site http://www.boutell.com/gd/, there is native GD support in v4.3.0. Anyone know how to enable this support? Thanks. Nate. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] How to access a MS Access DB
Hello, I'm new to this list and saw a past discussion in mid-Feb. about how to access an MS Access DB. I have Apache and PHP installed on my PC running XP Professional. I created a system DSN and got past the odbc_connect but it died in the odbc_prepare statement. Error message: Warning: odbc_prepare(): supplied argument is not a valid ODBC-Link resource in c:\perl\htdocs\test2.php on line 20 cannot prepare result Here's my code (w/username and password changed): $connection = odbc_connect(mctadb, username, password) || die(cannot open database); $sql = SELECT MeeetingTitle, MeetingDate FROM meeting;; $sql_result = odbc_prepare($connection, $sql) || die(cannot prepare result); Any help would be greatly appreciated, Beverly Steiner [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] How to access a MS Access DB
I believe I posted that thread. Try odbc_exec instead of odbc_prepare Let me know if have more problems --- Jonathan -Original Message- From: Beverly Steiner [mailto:[EMAIL PROTECTED] Sent: Monday, March 03, 2003 3:11 PM To: [EMAIL PROTECTED] Cc: Beverly Steiner Subject: RE: [PHP-DB] How to access a MS Access DB Hello, I'm new to this list and saw a past discussion in mid-Feb. about how to access an MS Access DB. I have Apache and PHP installed on my PC running XP Professional. I created a system DSN and got past the odbc_connect but it died in the odbc_prepare statement. Error message: Warning: odbc_prepare(): supplied argument is not a valid ODBC-Link resource in c:\perl\htdocs\test2.php on line 20 cannot prepare result Here's my code (w/username and password changed): $connection = odbc_connect(mctadb, username, password) || die(cannot open database); $sql = SELECT MeeetingTitle, MeetingDate FROM meeting;; $sql_result = odbc_prepare($connection, $sql) || die(cannot prepare result); Any help would be greatly appreciated, Beverly Steiner [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Confusing Date...
I have a MySQL column that was set to date. In my PHP page, I have a form and a field called 'Sch_StartDate'. I'm having difficultes understanding how to write to the date column in MySQL. If I send 03/04/02for March 4, 2002, the value that gets stored in the MySQL database is 2003-04-02 for April 2nd, 2003. The field is set to to send as a date: GetSQLValueString($HTTP_POST_VARS['Sch_StartDate'], date) So how do I send dates to MySQL so that it stores it correctly? Thank you!! Confused, Doug -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Confusing Date...
Doug, I know it works if you send the date as -mm-dd. For example yesterday, March 2, 2003 would be 2003-03-02 -- Beverly Steiner [EMAIL PROTECTED] -Original Message- From: Doug Coning [mailto:[EMAIL PROTECTED] Sent: Monday, March 03, 2003 5:29 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] Confusing Date... I have a MySQL column that was set to date. In my PHP page, I have a form and a field called 'Sch_StartDate'. I'm having difficultes understanding how to write to the date column in MySQL. If I send 03/04/02for March 4, 2002, the value that gets stored in the MySQL database is 2003-04-02 for April 2nd, 2003. The field is set to to send as a date: GetSQLValueString($HTTP_POST_VARS['Sch_StartDate'], date) So how do I send dates to MySQL so that it stores it correctly? Thank you!! Confused, Doug -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Confusing Date...
I have a MySQL column that was set to date. In my PHP page, I have a form and a field called 'Sch_StartDate'. I'm having difficultes understanding how to write to the date column in MySQL. If I send 03/04/02for March 4, 2002, the value that gets stored in the MySQL database is 2003-04-02 for April 2nd, 2003. The field is set to to send as a date: GetSQLValueString($HTTP_POST_VARS['Sch_StartDate'], date) So how do I send dates to MySQL so that it stores it correctly? MySQL expects the date in a -MM-DD or MMDD format. You can actually use other characters as the delimiter instead of the dash (-) character, as long as the year, month, and day are in the right order. As you've seen, you can also pass it a two digit year and it'll assume what century you mean. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Confusing Date...
Doug, Here's an example of how to format the date: $date = 3/2/2003; $good_date= date(Y-m-d, strtotime($date)); -- Beverly Steiner [EMAIL PROTECTED] -Original Message- From: Doug Coning [mailto:[EMAIL PROTECTED] Sent: Monday, March 03, 2003 5:38 PM To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Confusing Date... Yes, but not too many people are familiar with enter the year first, followed by the month then the date. Is there a way within PHP to take 03/02/2002 and send it as 2003-03-02. Thanks, Doug Coning - Original Message - From: Beverly Steiner [EMAIL PROTECTED] To: Doug Coning [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Sent: Monday, March 03, 2003 3:35 PM Subject: RE: [PHP-DB] Confusing Date... Doug, I know it works if you send the date as -mm-dd. For example yesterday, March 2, 2003 would be 2003-03-02 -- Beverly Steiner [EMAIL PROTECTED] -Original Message- From: Doug Coning [mailto:[EMAIL PROTECTED] Sent: Monday, March 03, 2003 5:29 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] Confusing Date... I have a MySQL column that was set to date. In my PHP page, I have a form and a field called 'Sch_StartDate'. I'm having difficultes understanding how to write to the date column in MySQL. If I send 03/04/02for March 4, 2002, the value that gets stored in the MySQL database is 2003-04-02 for April 2nd, 2003. The field is set to to send as a date: GetSQLValueString($HTTP_POST_VARS['Sch_StartDate'], date) So how do I send dates to MySQL so that it stores it correctly? Thank you!! Confused, Doug -- 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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Update MD5 field
[EMAIL PROTECTED] (Dani Matielo) writes: my problem is the following: I just imported a csv file to a MySQL database that contains name and email fields. It has about 9k lines on it and I need a new field that orinally didn't exist called code thats suposed to be MD5(name.email) I know how to do this for new data, but I don't know how to update all the old ones I already have there. If I understand you correctly, this is probably something like what you want to do: ALTER TABLE foo ADD bar VARCHAR(20); (don't know how long the field needs to be). UPDATE foo SET bar = MD5(baz); -- --Fredrik Spare no expense to save money on this one. -- Samuel Goldwyn -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Re: Update MD5 field
[snip] If I understand you correctly, this is probably something like what you want to do: ALTER TABLE foo ADD bar VARCHAR(20); (don't know how long the field needs to be). MD5() result is always 32 characters. ---John W. Holmes... PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
