Re: [PHP-DB] Php if statement in a form
On Fri, 6 Aug 2004 08:53:53 +1000, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Hello, I wonder if someone could point me in the right direction here.
I have a table that is displayed that is also a form, and allowed a
person to select a record to update using a radio button. With one of
the fields of the form/table however, I would like it to display the
value in the db (if there is one). If there is no value then I want the
field to display the current time and then submit it to the db as a
variable.
I can get it to display the current time and submit it to the db, but I
don't know how to do the IF bit of the form/table.
Anyway, the code is below.
Thanks
?
$dbcnx = @mysql_connect( localhost, user, password);
mysql_select_db(movements);
$result = mysql_query(SELECT id, name, location, timein, timeout FROM
workalone WHERE date=CURRENT_DATE()) or die (mysql_error());
// Display the results of the query in a table
print bCurrent Staff Working Alone/b;
//below is the table/form with the id, name, location, timein and
timeout taken from the db
print table border=\1\ cellpadding=\3\ cellspacing=\0\\n;
print
trtd/tdtdbName/b/tdtdbLocation/b/tdtdbTime
in/b/tdtdbTime Out/b/td/tr;
while ($row = mysql_fetch_array($result, MYSQL_BOTH))
{
print trtd;
print INPUT TYPE='RADIO' NAME='id' VALUE='.$row[id].';
print /tdtd;
print $row[name];
print /tdtd;
print $row[location];
print /tdtd;
print $row[timein];
print /tdtd;
//below is the IF thing I am having problems with
?input name=timeout type=text value=if {$timeout=null ?php
echo date('H:i');?} else {?print $row[timeout];??
print /td/tr\n;
}
print /table\n;
?input type=submit name=submit2 value=Select your name and click
here to record your timeout/form ?
if ($submit2)
{
$dbcnx = @mysql_connect( localhost, user, password);
mysql_select_db(movements);
$result = mysql_query(UPDATE workalone SET timeout='$timeout' WHERE
id='$id') or die (mysql_error());
echo bThank you, your Time Out has been recorded./b;
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You seem to have your php tags not around your if statement and you
are checking to see if $timeout rather than $row[timeout] has a
value. I also think the proper way to check for NULL is to say IS
NULL or NOT NULL, not $foo = NULL. Could be wrong about that one.
?php
if ($row[timeout] IS NULL) {
echo date('H:i');
} else {
print $row[timeout];
}
?
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Re: [PHP-DB] Php if statement in a form
On Fri, 6 Aug 2004 10:21:19 +1000, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Thanks Eric,
I have changed it somewhat but am just getting a parse error,
unexpected T_STRING on that line.
My revised code is below:
print bCurrent Staff Working Alone/b;
print /td/tr;
print /table\n;
print table border=\1\ cellpadding=\3\ cellspacing=\0\\n;
print
trtd/tdtdbName/b/tdtdbLocation/b/tdtdbTime
in/b/tdtdbTime Out/b/td/tr;
while ($row = mysql_fetch_array($result, MYSQL_BOTH))
{
print trtd;
print INPUT TYPE='RADIO' NAME='id' VALUE='.$row[id].';
print /tdtd;
print $row[name];
print /tdtd;
print $row[location];
print /tdtd;
print $row[timein];
print /tdtd;
if ($row[timeout] IS NULL);
{
print input name=timeout type=text value=echo date('H:i');
} else {
print $row[timeout];
}
print /td/tr\n;
}
print /table\n;
input type=submit name=submit2 value=Select your name and click
here to record your timeout/form
-Original Message-
From: Eric Schwartz [mailto:[EMAIL PROTECTED]
Sent: Friday, 6 August 2004 9:31 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Php if statement in a form
On Fri, 6 Aug 2004 08:53:53 +1000, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Hello, I wonder if someone could point me in the right direction here.
I have a table that is displayed that is also a form, and allowed a
person to select a record to update using a radio button. With one of
the fields of the form/table however, I would like it to display the
value in the db (if there is one). If there is no value then I want
the
field to display the current time and then submit it to the db as a
variable.
I can get it to display the current time and submit it to the db, but
I
don't know how to do the IF bit of the form/table.
Anyway, the code is below.
Thanks
?
$dbcnx = @mysql_connect( localhost, user, password);
mysql_select_db(movements);
$result = mysql_query(SELECT id, name, location, timein, timeout FROM
workalone WHERE date=CURRENT_DATE()) or die (mysql_error());
// Display the results of the query in a table
print bCurrent Staff Working Alone/b;
//below is the table/form with the id, name, location, timein and
timeout taken from the db
print table border=\1\ cellpadding=\3\ cellspacing=\0\\n;
print
trtd/tdtdbName/b/tdtdbLocation/b/tdtdbTime
in/b/tdtdbTime Out/b/td/tr;
while ($row = mysql_fetch_array($result, MYSQL_BOTH))
{
print trtd;
print INPUT TYPE='RADIO' NAME='id' VALUE='.$row[id].';
print /tdtd;
print $row[name];
print /tdtd;
print $row[location];
print /tdtd;
print $row[timein];
print /tdtd;
//below is the IF thing I am having problems with
?input name=timeout type=text value=if {$timeout=null ?php
echo date('H:i');?} else {?print $row[timeout];??
print /td/tr\n;
}
print /table\n;
?input type=submit name=submit2 value=Select your name and
click
here to record your timeout/form ?
if ($submit2)
{
$dbcnx = @mysql_connect( localhost, user, password);
mysql_select_db(movements);
$result = mysql_query(UPDATE workalone SET timeout='$timeout' WHERE
id='$id') or die (mysql_error());
echo bThank you, your Time Out has been recorded./b;
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You seem to have your php tags not around your if statement and you
are checking to see if $timeout rather than $row[timeout] has a
value. I also think the proper way to check for NULL is to say IS
NULL or NOT NULL, not $foo = NULL. Could be wrong about that one.
?php
if ($row[timeout] IS NULL) {
echo date('H:i');
} else {
print $row[timeout];
}
?
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Try adding a quote on the line and escaping the other quotes, like this:
print input name=\timeout\ type=\text\ value=\.echo date('H:i').\;
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Re: [PHP-DB] Php if statement in a form
Sorry. Forgot to remove the word echo from that line.
On Thu, 5 Aug 2004 21:29:01 -0400, Eric Schwartz
[EMAIL PROTECTED] wrote:
On Fri, 6 Aug 2004 10:21:19 +1000, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Thanks Eric,
I have changed it somewhat but am just getting a parse error,
unexpected T_STRING on that line.
My revised code is below:
print bCurrent Staff Working Alone/b;
print /td/tr;
print /table\n;
print table border=\1\ cellpadding=\3\ cellspacing=\0\\n;
print
trtd/tdtdbName/b/tdtdbLocation/b/tdtdbTime
in/b/tdtdbTime Out/b/td/tr;
while ($row = mysql_fetch_array($result, MYSQL_BOTH))
{
print trtd;
print INPUT TYPE='RADIO' NAME='id' VALUE='.$row[id].';
print /tdtd;
print $row[name];
print /tdtd;
print $row[location];
print /tdtd;
print $row[timein];
print /tdtd;
if ($row[timeout] IS NULL);
{
print input name=timeout type=text value=echo date('H:i');
} else {
print $row[timeout];
}
print /td/tr\n;
}
print /table\n;
input type=submit name=submit2 value=Select your name and click
here to record your timeout/form
-Original Message-
From: Eric Schwartz [mailto:[EMAIL PROTECTED]
Sent: Friday, 6 August 2004 9:31 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Php if statement in a form
On Fri, 6 Aug 2004 08:53:53 +1000, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Hello, I wonder if someone could point me in the right direction here.
I have a table that is displayed that is also a form, and allowed a
person to select a record to update using a radio button. With one of
the fields of the form/table however, I would like it to display the
value in the db (if there is one). If there is no value then I want
the
field to display the current time and then submit it to the db as a
variable.
I can get it to display the current time and submit it to the db, but
I
don't know how to do the IF bit of the form/table.
Anyway, the code is below.
Thanks
?
$dbcnx = @mysql_connect( localhost, user, password);
mysql_select_db(movements);
$result = mysql_query(SELECT id, name, location, timein, timeout FROM
workalone WHERE date=CURRENT_DATE()) or die (mysql_error());
// Display the results of the query in a table
print bCurrent Staff Working Alone/b;
//below is the table/form with the id, name, location, timein and
timeout taken from the db
print table border=\1\ cellpadding=\3\ cellspacing=\0\\n;
print
trtd/tdtdbName/b/tdtdbLocation/b/tdtdbTime
in/b/tdtdbTime Out/b/td/tr;
while ($row = mysql_fetch_array($result, MYSQL_BOTH))
{
print trtd;
print INPUT TYPE='RADIO' NAME='id' VALUE='.$row[id].';
print /tdtd;
print $row[name];
print /tdtd;
print $row[location];
print /tdtd;
print $row[timein];
print /tdtd;
//below is the IF thing I am having problems with
?input name=timeout type=text value=if {$timeout=null ?php
echo date('H:i');?} else {?print $row[timeout];??
print /td/tr\n;
}
print /table\n;
?input type=submit name=submit2 value=Select your name and
click
here to record your timeout/form ?
if ($submit2)
{
$dbcnx = @mysql_connect( localhost, user, password);
mysql_select_db(movements);
$result = mysql_query(UPDATE workalone SET timeout='$timeout' WHERE
id='$id') or die (mysql_error());
echo bThank you, your Time Out has been recorded./b;
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
You seem to have your php tags not around your if statement and you
are checking to see if $timeout rather than $row[timeout] has a
value. I also think the proper way to check for NULL is to say IS
NULL or NOT NULL, not $foo = NULL. Could be wrong about that one.
?php
if ($row[timeout] IS NULL) {
echo date('H:i');
} else {
print $row[timeout];
}
?
--
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
Try adding a quote on the line and escaping the other quotes, like this:
print input name=\timeout\ type=\text\ value=\.echo date('H:i').\;
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
