Re: [PHP-DB] Date math functions...

2003-01-15 Thread 1LT John W. Holmes
[snip]
 if ($dateDiff == 3) {
 mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
 said='$said') or die(mysql_error());
 mysql($DBName,INSERT INTO Log VALUES('DATE_ADD($StopDate,
 INTERVAL 1 DAY','',1,2,'$CalendarDetailsID')) or die(mysql_error());

 My big question is about using the DATE_ADD MySQL function inside
 the INSERT statement.  Is this allowed?  If it is allowed, is it a bad
idea
 for some reason?  Is there a better way of doing this?  Thanks in advance.

Sure, that's allowed. You have $StopDate in PHP, so you could do it with
some math in PHP, also, but then you'd have to worry about the end of
months, years, etc, whereas DATE_ADD will do this for you.

---John Holmes...


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Re: [PHP-DB] Date math functions...

2003-01-15 Thread 1LT John W. Holmes
 [snip]
  if ($dateDiff == 3) {
  mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
  said='$said') or die(mysql_error());
  mysql($DBName,INSERT INTO Log VALUES('DATE_ADD($StopDate,
  INTERVAL 1 DAY','',1,2,'$CalendarDetailsID')) or die(mysql_error());
 
  My big question is about using the DATE_ADD MySQL function inside
  the INSERT statement.  Is this allowed?  If it is allowed, is it a bad
 idea
  for some reason?  Is there a better way of doing this?  Thanks in
advance.

 Sure, that's allowed. You have $StopDate in PHP, so you could do it with
 some math in PHP, also, but then you'd have to worry about the end of
 months, years, etc, whereas DATE_ADD will do this for you.

 ---John Holmes...

Wait... just noticed your syntax error. Don't enclose the function in single
quotes, otherwise you're trying to insert a string.

Should be:

... VALUES (DATE_ADD($StopDate,INTERVAL 1 DAY), ...

---John Holmes...


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RE: [PHP-DB] Date math functions...

2003-01-15 Thread NIPP, SCOTT V (SBCSI)
Thanks.  I was noticing that it was not working.  Let me give this a
try and see how things go.

-Original Message-
From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 15, 2003 10:32 AM
To: 1LT John W. Holmes; NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Date math functions...


 [snip]
  if ($dateDiff == 3) {
  mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
  said='$said') or die(mysql_error());
  mysql($DBName,INSERT INTO Log VALUES('DATE_ADD($StopDate,
  INTERVAL 1 DAY','',1,2,'$CalendarDetailsID')) or die(mysql_error());
 
  My big question is about using the DATE_ADD MySQL function inside
  the INSERT statement.  Is this allowed?  If it is allowed, is it a bad
 idea
  for some reason?  Is there a better way of doing this?  Thanks in
advance.

 Sure, that's allowed. You have $StopDate in PHP, so you could do it with
 some math in PHP, also, but then you'd have to worry about the end of
 months, years, etc, whereas DATE_ADD will do this for you.

 ---John Holmes...

Wait... just noticed your syntax error. Don't enclose the function in single
quotes, otherwise you're trying to insert a string.

Should be:

... VALUES (DATE_ADD($StopDate,INTERVAL 1 DAY), ...

---John Holmes...

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RE: [PHP-DB] Date math functions...

2003-01-15 Thread NIPP, SCOTT V (SBCSI)
Actually this is generating another error.  Now, without the single
quotes I am getting the following error:

Column 'StartDate' cannot be null

It looks like for some reason the DATE_ADD is returning a NULL
value.  Any more ideas?

-Original Message-
From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 15, 2003 10:32 AM
To: 1LT John W. Holmes; NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Date math functions...


 [snip]
  if ($dateDiff == 3) {
  mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
  said='$said') or die(mysql_error());
  mysql($DBName,INSERT INTO Log VALUES('DATE_ADD($StopDate,
  INTERVAL 1 DAY','',1,2,'$CalendarDetailsID')) or die(mysql_error());
 
  My big question is about using the DATE_ADD MySQL function inside
  the INSERT statement.  Is this allowed?  If it is allowed, is it a bad
 idea
  for some reason?  Is there a better way of doing this?  Thanks in
advance.

 Sure, that's allowed. You have $StopDate in PHP, so you could do it with
 some math in PHP, also, but then you'd have to worry about the end of
 months, years, etc, whereas DATE_ADD will do this for you.

 ---John Holmes...

Wait... just noticed your syntax error. Don't enclose the function in single
quotes, otherwise you're trying to insert a string.

Should be:

... VALUES (DATE_ADD($StopDate,INTERVAL 1 DAY), ...

---John Holmes...

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Re: [PHP-DB] Date math functions...

2003-01-15 Thread Jason Wong
On Thursday 16 January 2003 00:55, NIPP, SCOTT V (SBCSI) wrote:
 Actually this is generating another error.  Now, without the single
 quotes I am getting the following error:

   Column 'StartDate' cannot be null

   It looks like for some reason the DATE_ADD is returning a NULL
 value.  Any more ideas?

Which probably means your $StopDate is 0. What does it actually contain? You 
really should assign your query to some variable ($query) THEN plug it into 
your query function. If and you have any problems you can echo $query to see 
what you're passing to mysql.


-- 
Jason Wong - Gremlins Associates - www.gremlins.biz
Open Source Software Systems Integrators
* Web Design  Hosting * Internet  Intranet Applications Development *


/*
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wealthy CORPORATE LAWYERS ...
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Re: [PHP-DB] Date math functions...

2003-01-15 Thread 1LT John W. Holmes
   mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
   said='$said') or die(mysql_error());
   mysql($DBName,INSERT INTO Log VALUES(DATE_ADD($StopDate,
   INTERVAL 1 DAY,'',1,2,'$CalendarDetailsID')) or die(mysql_error());

 Actually this is generating another error.  Now, without the single
 quotes I am getting the following error:

 Column 'StartDate' cannot be null

 It looks like for some reason the DATE_ADD is returning a NULL
 value.  Any more ideas?

Are you sure $StopDate has a value and is in the right format? Echo out your
query to the screen when you get an error, so you can look for obvious
mistakes.

Is this just example code?  You're using a function called mysql(), which
isn't standard.

---John Holmes...


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