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));
?
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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));
?
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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;
}
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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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