Re: newbie select statement question
Hi Jordan
Have had a look at the date input page, and the source code that it produces
looks fine. If you still haven't sorted it out could I suggest the
following:
Output the tenure date at the following points in the application:
1) When it is returned by the input form.
2) When it is returned from the database (before being input into
fixdate()).
3) Within fixdate.
See if the results are consistent throughout.
Rory McKinley
Nebula Solutions
+27 82 857 2391
[EMAIL PROTECTED]
There are 10 kinds of people in this world,
those who understand binary and those who don't (Unknown)
- Original Message -
From: Jordan Morgan [EMAIL PROTECTED]
To: Nobody [EMAIL PROTECTED]
Cc: Diana Soares [EMAIL PROTECTED]; mysql [EMAIL PROTECTED]
Sent: Friday, October 10, 2003 2:24 PM
Subject: Re: newbie select statement question
The date is stored as dates instead of timestamps. Yes, that fixDate
function is
defined by me and it worked great with other dates I have stored. I'm
really
thinking my dropdown fields for the MM, DD, and on the data entry
page has
somethng wrong with it.
do you think so?
ps: thanks for checking for me btw.
Nobody wrote:
I am not sure if this has any significance - but isn't 31 Dec 1969 the
day
before the UNIX epoch (1 Jan 1970 - i think)?
Jordan, how are the dates stored in the database - as dates or unix
timestamps?
Also, the function fixDate isn't in the online PHP manual - is it a
function
defined by you? - could that not be doing something weird?
Rory McKinley
Nebula Solutions
+27 82 857 2391
[EMAIL PROTECTED]
There are 10 kinds of people in this world,
those who understand binary and those who don't (Unknown)
- Original Message -
From: Diana Soares [EMAIL PROTECTED]
To: Jordan Morgan [EMAIL PROTECTED]
Cc: mysql [EMAIL PROTECTED]
Sent: Friday, October 10, 2003 8:58 AM
Subject: Re: newbie select statement question
Look at:
if ($tenureid=3)
You're not comparing $tenureid with 3, you're assigning 3 to
$ternureid... If you want to compare both values, you must use the
operator == (and not only =).
On Fri, 2003-10-10 at 05:43, Jordan Morgan wrote:
Hi,
I have the following statement:
echo $tenureidP;
if ($tenureid=3)
{
// get faculty employment record - award date
$sql = select TD.Award_Date from TenureDescription TD LEFT JOIN
InstitutionEmployment IE on TD.TenureDescriptionID=IE.Tenure WHERE
IE.FacultyMember='$fid';
$result = mysql_db_query($database, $sql, $connection) or die
(Error
in query: $sql. . mysql_error());
// obtain data from resultset
list($tenuredate) = mysql_fetch_row($result);
echo BDate Tenure Granted/B: ;
echo fixDate($tenuredate);
echo P;
}
else {
echo BDate Tenure Granted/B: NullP;
}
and I can't figure out why I always get the following result:
2
Date Tenure Granted: 31 Dec 1969
---
or
1
Date Tenure Granted: 31 Dec 1969
---
when 1) tenureid 3, and 2) no 31 Dec 1969 date in the database
anywhere.
The only thing I can think of is that I modified the dropdown box
for
the tenure date on the data entry page like this:
tr
tdDate Tenure Grantedbrfont size=-2(in mm-dd-
format)/font/td
td
select name=tmm
option value=0 selected='selected'/option
? for ($x=1; $x=12; $x++) { echo option value=\ .
sprintf(%02d,
$x) . \ . sprintf(%02d, $x) . /option; } ?
/select -
select name=tdd
option value=0 selected='selected'/option
? for ($x=1; $x=31; $x++) { echo option value=\ .
sprintf(%02d,
$x) . \ . sprintf(%02d, $x) . /option; } ?
/select -
select name=t
option value=0 selected='selected'/option
!-- display from 1970 to (current year) --
? for ($x=(date(Y, mktime())); $x=1970; $x--) { echo option
value=$x$x/option; } ?
/select
/td
/tr
by adding option value=0 selected='selected'/option to those 3
fields as I want null to be a default selection. but I can't imagine why
that'll mess up the if statement evaluation.
It seems that the 1st if statement just runs whatever the tenureid
is.
Can anyone help me on this? I'm using PHP 4.2.2 and MySQL 3.23.54
btw.
Thanks millions!
Jordan
--
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Jordan
Re: newbie select statement question
Look at:
if ($tenureid=3)
You're not comparing $tenureid with 3, you're assigning 3 to
$ternureid... If you want to compare both values, you must use the
operator == (and not only =).
On Fri, 2003-10-10 at 05:43, Jordan Morgan wrote:
Hi,
I have the following statement:
echo $tenureidP;
if ($tenureid=3)
{
// get faculty employment record - award date
$sql = select TD.Award_Date from TenureDescription TD LEFT JOIN
InstitutionEmployment IE on TD.TenureDescriptionID=IE.Tenure WHERE
IE.FacultyMember='$fid';
$result = mysql_db_query($database, $sql, $connection) or die (Error in query:
$sql. . mysql_error());
// obtain data from resultset
list($tenuredate) = mysql_fetch_row($result);
echo BDate Tenure Granted/B: ;
echo fixDate($tenuredate);
echo P;
}
else {
echo BDate Tenure Granted/B: NullP;
}
and I can't figure out why I always get the following result:
2
Date Tenure Granted: 31 Dec 1969
---
or
1
Date Tenure Granted: 31 Dec 1969
---
when 1) tenureid 3, and 2) no 31 Dec 1969 date in the database anywhere.
The only thing I can think of is that I modified the dropdown box for the tenure
date on the data entry page like this:
tr
tdDate Tenure Grantedbrfont size=-2(in mm-dd- format)/font/td
td
select name=tmm
option value=0 selected='selected'/option
? for ($x=1; $x=12; $x++) { echo option value=\ . sprintf(%02d, $x) . \
. sprintf(%02d, $x) . /option; } ?
/select -
select name=tdd
option value=0 selected='selected'/option
? for ($x=1; $x=31; $x++) { echo option value=\ . sprintf(%02d, $x) . \
. sprintf(%02d, $x) . /option; } ?
/select -
select name=t
option value=0 selected='selected'/option
!-- display from 1970 to (current year) --
? for ($x=(date(Y, mktime())); $x=1970; $x--) { echo option
value=$x$x/option; } ?
/select
/td
/tr
by adding option value=0 selected='selected'/option to those 3 fields as I
want null to be a default selection. but I can't imagine why that'll mess up the if
statement evaluation.
It seems that the 1st if statement just runs whatever the tenureid is.
Can anyone help me on this? I'm using PHP 4.2.2 and MySQL 3.23.54 btw.
Thanks millions!
Jordan
--
Diana Soares
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: newbie select statement question
I am not sure if this has any significance - but isn't 31 Dec 1969 the day
before the UNIX epoch (1 Jan 1970 - i think)?
Jordan, how are the dates stored in the database - as dates or unix
timestamps?
Also, the function fixDate isn't in the online PHP manual - is it a function
defined by you? - could that not be doing something weird?
Rory McKinley
Nebula Solutions
+27 82 857 2391
[EMAIL PROTECTED]
There are 10 kinds of people in this world,
those who understand binary and those who don't (Unknown)
- Original Message -
From: Diana Soares [EMAIL PROTECTED]
To: Jordan Morgan [EMAIL PROTECTED]
Cc: mysql [EMAIL PROTECTED]
Sent: Friday, October 10, 2003 8:58 AM
Subject: Re: newbie select statement question
Look at:
if ($tenureid=3)
You're not comparing $tenureid with 3, you're assigning 3 to
$ternureid... If you want to compare both values, you must use the
operator == (and not only =).
On Fri, 2003-10-10 at 05:43, Jordan Morgan wrote:
Hi,
I have the following statement:
echo $tenureidP;
if ($tenureid=3)
{
// get faculty employment record - award date
$sql = select TD.Award_Date from TenureDescription TD LEFT JOIN
InstitutionEmployment IE on TD.TenureDescriptionID=IE.Tenure WHERE
IE.FacultyMember='$fid';
$result = mysql_db_query($database, $sql, $connection) or die (Error
in query: $sql. . mysql_error());
// obtain data from resultset
list($tenuredate) = mysql_fetch_row($result);
echo BDate Tenure Granted/B: ;
echo fixDate($tenuredate);
echo P;
}
else {
echo BDate Tenure Granted/B: NullP;
}
and I can't figure out why I always get the following result:
2
Date Tenure Granted: 31 Dec 1969
---
or
1
Date Tenure Granted: 31 Dec 1969
---
when 1) tenureid 3, and 2) no 31 Dec 1969 date in the database
anywhere.
The only thing I can think of is that I modified the dropdown box for
the tenure date on the data entry page like this:
tr
tdDate Tenure Grantedbrfont size=-2(in mm-dd-
format)/font/td
td
select name=tmm
option value=0 selected='selected'/option
? for ($x=1; $x=12; $x++) { echo option value=\ . sprintf(%02d,
$x) . \ . sprintf(%02d, $x) . /option; } ?
/select -
select name=tdd
option value=0 selected='selected'/option
? for ($x=1; $x=31; $x++) { echo option value=\ . sprintf(%02d,
$x) . \ . sprintf(%02d, $x) . /option; } ?
/select -
select name=t
option value=0 selected='selected'/option
!-- display from 1970 to (current year) --
? for ($x=(date(Y, mktime())); $x=1970; $x--) { echo option
value=$x$x/option; } ?
/select
/td
/tr
by adding option value=0 selected='selected'/option to those 3
fields as I want null to be a default selection. but I can't imagine why
that'll mess up the if statement evaluation.
It seems that the 1st if statement just runs whatever the tenureid is.
Can anyone help me on this? I'm using PHP 4.2.2 and MySQL 3.23.54 btw.
Thanks millions!
Jordan
--
Diana Soares
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: newbie select statement question
oh thanks! I guess that's why I'm a newbie.
Diana Soares wrote:
Look at:
if ($tenureid=3)
You're not comparing $tenureid with 3, you're assigning 3 to
$ternureid... If you want to compare both values, you must use the
operator == (and not only =).
On Fri, 2003-10-10 at 05:43, Jordan Morgan wrote:
Hi,
I have the following statement:
echo $tenureidP;
if ($tenureid=3)
{
// get faculty employment record - award date
$sql = select TD.Award_Date from TenureDescription TD LEFT JOIN
InstitutionEmployment IE on TD.TenureDescriptionID=IE.Tenure WHERE
IE.FacultyMember='$fid';
$result = mysql_db_query($database, $sql, $connection) or die (Error in query:
$sql. . mysql_error());
// obtain data from resultset
list($tenuredate) = mysql_fetch_row($result);
echo BDate Tenure Granted/B: ;
echo fixDate($tenuredate);
echo P;
}
else {
echo BDate Tenure Granted/B: NullP;
}
and I can't figure out why I always get the following result:
2
Date Tenure Granted: 31 Dec 1969
---
or
1
Date Tenure Granted: 31 Dec 1969
---
when 1) tenureid 3, and 2) no 31 Dec 1969 date in the database anywhere.
The only thing I can think of is that I modified the dropdown box for the tenure
date on the data entry page like this:
tr
tdDate Tenure Grantedbrfont size=-2(in mm-dd- format)/font/td
td
select name=tmm
option value=0 selected='selected'/option
? for ($x=1; $x=12; $x++) { echo option value=\ . sprintf(%02d, $x) .
\ . sprintf(%02d, $x) . /option; } ?
/select -
select name=tdd
option value=0 selected='selected'/option
? for ($x=1; $x=31; $x++) { echo option value=\ . sprintf(%02d, $x) .
\ . sprintf(%02d, $x) . /option; } ?
/select -
select name=t
option value=0 selected='selected'/option
!-- display from 1970 to (current year) --
? for ($x=(date(Y, mktime())); $x=1970; $x--) { echo option
value=$x$x/option; } ?
/select
/td
/tr
by adding option value=0 selected='selected'/option to those 3 fields as I
want null to be a default selection. but I can't imagine why that'll mess up the
if statement evaluation.
It seems that the 1st if statement just runs whatever the tenureid is.
Can anyone help me on this? I'm using PHP 4.2.2 and MySQL 3.23.54 btw.
Thanks millions!
Jordan
--
Diana Soares
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
--
Jordan Morgan
Information Analyst - GeorgiaFIRST HRMS Project
Board of Regents
Office of Information and Instructional Technology
1865 West Broad Street, Athens, GA 30606-3539
Phone: (706) 369-6232
Fax: (706) 369-6429
mailto:[EMAIL PROTECTED] http://www.usg.edu
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: newbie select statement question
The date is stored as dates instead of timestamps. Yes, that fixDate function is
defined by me and it worked great with other dates I have stored. I'm really
thinking my dropdown fields for the MM, DD, and on the data entry page has
somethng wrong with it.
do you think so?
ps: thanks for checking for me btw.
Nobody wrote:
I am not sure if this has any significance - but isn't 31 Dec 1969 the day
before the UNIX epoch (1 Jan 1970 - i think)?
Jordan, how are the dates stored in the database - as dates or unix
timestamps?
Also, the function fixDate isn't in the online PHP manual - is it a function
defined by you? - could that not be doing something weird?
Rory McKinley
Nebula Solutions
+27 82 857 2391
[EMAIL PROTECTED]
There are 10 kinds of people in this world,
those who understand binary and those who don't (Unknown)
- Original Message -
From: Diana Soares [EMAIL PROTECTED]
To: Jordan Morgan [EMAIL PROTECTED]
Cc: mysql [EMAIL PROTECTED]
Sent: Friday, October 10, 2003 8:58 AM
Subject: Re: newbie select statement question
Look at:
if ($tenureid=3)
You're not comparing $tenureid with 3, you're assigning 3 to
$ternureid... If you want to compare both values, you must use the
operator == (and not only =).
On Fri, 2003-10-10 at 05:43, Jordan Morgan wrote:
Hi,
I have the following statement:
echo $tenureidP;
if ($tenureid=3)
{
// get faculty employment record - award date
$sql = select TD.Award_Date from TenureDescription TD LEFT JOIN
InstitutionEmployment IE on TD.TenureDescriptionID=IE.Tenure WHERE
IE.FacultyMember='$fid';
$result = mysql_db_query($database, $sql, $connection) or die (Error
in query: $sql. . mysql_error());
// obtain data from resultset
list($tenuredate) = mysql_fetch_row($result);
echo BDate Tenure Granted/B: ;
echo fixDate($tenuredate);
echo P;
}
else {
echo BDate Tenure Granted/B: NullP;
}
and I can't figure out why I always get the following result:
2
Date Tenure Granted: 31 Dec 1969
---
or
1
Date Tenure Granted: 31 Dec 1969
---
when 1) tenureid 3, and 2) no 31 Dec 1969 date in the database
anywhere.
The only thing I can think of is that I modified the dropdown box for
the tenure date on the data entry page like this:
tr
tdDate Tenure Grantedbrfont size=-2(in mm-dd-
format)/font/td
td
select name=tmm
option value=0 selected='selected'/option
? for ($x=1; $x=12; $x++) { echo option value=\ . sprintf(%02d,
$x) . \ . sprintf(%02d, $x) . /option; } ?
/select -
select name=tdd
option value=0 selected='selected'/option
? for ($x=1; $x=31; $x++) { echo option value=\ . sprintf(%02d,
$x) . \ . sprintf(%02d, $x) . /option; } ?
/select -
select name=t
option value=0 selected='selected'/option
!-- display from 1970 to (current year) --
? for ($x=(date(Y, mktime())); $x=1970; $x--) { echo option
value=$x$x/option; } ?
/select
/td
/tr
by adding option value=0 selected='selected'/option to those 3
fields as I want null to be a default selection. but I can't imagine why
that'll mess up the if statement evaluation.
It seems that the 1st if statement just runs whatever the tenureid is.
Can anyone help me on this? I'm using PHP 4.2.2 and MySQL 3.23.54 btw.
Thanks millions!
Jordan
--
Diana Soares
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
--
Jordan Morgan
Information Analyst - GeorgiaFIRST HRMS Project
Board of Regents
Office of Information and Instructional Technology
1865 West Broad Street, Athens, GA 30606-3539
Phone: (706) 369-6232
Fax: (706) 369-6429
mailto:[EMAIL PROTECTED] http://www.usg.edu
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
