RE: [PHP-DB] date functions (generates parse error)
This is very weird. What are you using to create this code?
If you remove all of the spaces (?) before $query, the code doesn't give
a parse error. It shouldn't give one either way, though, if those are
spaces or a tab before your $query = ... line.
Even if you remove all of the text and only leave , and those
"spaces", PHP will spit something like:
Notice: Use of undefined constant - assumed ' ' in
c:\inetpub\wwwroot\test.php on line 3
So, the fix is to remove those characters and replace them with spaces
or tabs.
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/
> -Original Message-
> From: David Rice [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, March 05, 2003 12:34 PM
> To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] date functions (generates parse error)
>
>
> Sorry, I forgot to add the part at the bottom where I am calling the
> function
>
> =
>
> function tips($weekstart){
>
> $start = date('Ymd',strtotime($weekstart));
>
> $query = "SELECT * FROM Rota WHERE date >= $start and date <=
> ($start +
> INTERVAL 6 DAY) ORDER BY staffid";
> $result = mysql_query($query);
> while ($row = mysql_fetch_array($result)){
>
> if ( isset ( $tips ) ){
>
> if (isset ( $tips[$row[staffid]] ) ){
>
> $hours = $row[finish] - $row[start];
> $tips[$row[staffid]] =
$tips[$row[staffid]] +
> $hours;
>
> }
>
> else{
>
> $tips[$row[staffid]] = $row[finish] -
$row[start];
>
> }
> }
>
> else{
>
> $tips = array('$row[staffid]' =>( $row[finish] -
> $row[start] ) );
>
> }
>
> }
>
> return $tips;
>
> }
>
> function dbconnect(){
> mysql_connect("localhost", "filterseveuk", "godisadj");
> mysql_select_db("filterseveuk");
> }
>
> dbconnect();
> $date = "2003-03-02";
>
> var_dump(tips($date));
> ?>
>
> _
> Express yourself with cool emoticons http://messenger.msn.co.uk
>
>
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Re: [PHP-DB] date functions (generates parse error)
Sorry, I forgot to add the part at the bottom where I am calling the
function
=
$start = date('Ymd',strtotime($weekstart));
$query = "SELECT * FROM Rota WHERE date >= $start and date <= ($start +
INTERVAL 6 DAY) ORDER BY staffid";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)){
if ( isset ( $tips ) ){
if (isset ( $tips[$row[staffid]] ) ){
$hours = $row[finish] - $row[start];
$tips[$row[staffid]] = $tips[$row[staffid]] + $hours;
}
else{
$tips[$row[staffid]] = $row[finish] - $row[start];
}
}
else{
$tips = array('$row[staffid]' =>( $row[finish] - $row[start] ) );
}
}
return $tips;
}
function dbconnect(){
mysql_connect("localhost", "filterseveuk", "godisadj");
mysql_select_db("filterseveuk");
}
dbconnect();
$date = "2003-03-02";
var_dump(tips($date));
?>
_
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Re: [PHP-DB] date functions (generates parse error)
> Here is the whole code of my function
>
> Whenever i run it, it say's there is a parse error on line 6, can't see
what
> is the problem
> the format of $weekstart (as it is stored in the Database) is -MM-DD
>
> =
> function tips($weekstart){
>
> $start = date('Ymd',strtotime($weekstart));
>
> $query = "SELECT * FROM Rota WHERE date >= $start and date <= ($start +
> INTERVAL 6 DAY) ORDER BY staffid";
> $result = mysql_query($query);
> while ($row = mysql_fetch_array($result)){
>
> if ( isset ( $tips ) ){
>
> if (isset ( $tips[$row[staffid]] ) ){
>
> $hours = $row[finish] - $row[start];
> $tips[$row[staffid]] = $tips[$row[staffid]] + $hours;
>
> }
>
> else{
>
> $tips[$row[staffid]] = $row[finish] - $row[start];
>
> }
> }
>
> else{
>
> $tips = array('$row[staffid]' =>( $row[finish] - $row[start] ) );
>
> }
>
> }
>
> return $tips;
>
> }
I cut and pasted your exact code here and didn't get a parse error.
---John Holmes...
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Re: [PHP-DB] date functions (generates parse error)
Here is the whole code of my function
Whenever i run it, it say's there is a parse error on line 6, can't see what
is the problem
the format of $weekstart (as it is stored in the Database) is -MM-DD
=
$start = date('Ymd',strtotime($weekstart));
$query = "SELECT * FROM Rota WHERE date >= $start and date <= ($start +
INTERVAL 6 DAY) ORDER BY staffid";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)){
if ( isset ( $tips ) ){
if (isset ( $tips[$row[staffid]] ) ){
$hours = $row[finish] - $row[start];
$tips[$row[staffid]] = $tips[$row[staffid]] + $hours;
}
else{
$tips[$row[staffid]] = $row[finish] - $row[start];
}
}
else{
$tips = array('$row[staffid]' =>( $row[finish] - $row[start] ) );
}
}
return $tips;
}
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Re: [PHP-DB] date functions
> I want to use it in this function that i am creating (it's for a
resteraunt
> automated tips system, to work out how much tips each staff member is
> entitled to.
>
>
> function tips($weekstart){
> /* JUST BELOW HERE IS WHERE I NEED TO GET THE DATE OF 6
> DAYS AFTER THE
> SPECIFIED DATE OF THE START OF THE WEEK */
> $weekend = date($weekstart +6);
>
> $query = "SELECT * FROM Rota WHERE date => $weekstart
> AND date <= $weekend ORDER BY staffid";
>
> etc...
What format is $weekstart in? Maybe you said already, I can't remember.
You could try
$weekend = strtotime($weekstart . ' +6 days');
Since I think your 'date' column is a TIMESTAMP, you'll need to format
$weekstart and $weekend as MMDD in order to get what you have above to
work.
Maybe try this:
function tips($weekstart){
$start = date('Ymd',strtotime($weekstart));
$query = "SELECT * FROM Rota WHERE date >= $start and date <= ($start +
INTERVAL 6 DAY) ORDER BY staffid";
etc...
---John Holmes...
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RE: [PHP-DB] date functions
Good stuff here. http://www.mysql.com/doc/en/Date_and_time_functions.html Check out the SELECT DATE_ADD section. That might be what you're looking for. HTH, Rich > -Original Message- > From: David Rice [mailto:[EMAIL PROTECTED] > Sent: Tuesday, March 04, 2003 9:36 AM > To: [EMAIL PROTECTED] > Subject: [PHP-DB] date functions > > > > > I am looking for a way to take a date stored in a mysql > database... and find > out the date seven days later. > > how would i do this?! > > > cheers, dave > > > > _ > Chat online in real time with MSN Messenger http://messenger.msn.co.uk > > > -- > 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] date functions
> I am looking for a way to take a date stored in a mysql database... and find > out the date seven days later. > > how would i do this?! too easy... SELECT date_column + INTERVAL 7 DAY FROM your_table ... ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date functions
On Mon, 1 Jul 2002, Achilleas Maroulis wrote: > Hi. > I want to know what is the right way to make some comparisons with date variables. > I have used only the date() function to get a variable in this format=> 01-07-2002 > Is there a way to add 20 days in order to have 21-07-2002 or 2 months to have >01-09-2002? Hi! Try to read PHP documentation: http://www.php.net/manual/en/function.mktime.php Regards Urosh -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
