RE: [PHP] Query question
On Sun, 29 Oct 2006 23:04:27 -0600, Richard Lynch wrote: > On Sun, October 29, 2006 2:06 am, Beauford wrote: >> LOL, I don't know either. The format is - 01/01/2006. When I first did >> it I >> used 7, which should be right, but I ended up getting /2002 /2003, >> etc. So I >> went to 8 and all was well. Beats me. > > Dollars to donuts says your 'date' (which is really a string) has some > leading spaces which is messing up your count... > > Of course, after you fix the import process to have an actual DATE in > your DB, this will be a non-issue... > > Short-term, you should probably at least trim() the data so that your > offset of 8 is the number you would expect... Or, you use RIGHT(date, 4) as I believe MySQL always automatically removes trailing spaces in VARCHAR columns. It will save you the use of one additional function and RIGHT() may even be faster than SUBSTRING() because it doesn't need to go through the entire string. But that's just guessing and I think you probably won't even notice this in microseconds :) Ivo -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Query question
On Sun, October 29, 2006 2:06 am, Beauford wrote: > This is how I get the data and I have no control over the actual > layout of > the database. So I have to work with what I have. When you say it's "how you get the data" are they handing you a MySQL database that is this badly-constructed? Or are you importing data whose dates are in 'mm/dd/' format, and failing to realize that you CAN and SHOULD convert those to date as part of your import process. > LOL, I don't know either. The format is - 01/01/2006. When I first did > it I > used 7, which should be right, but I ended up getting /2002 /2003, > etc. So I > went to 8 and all was well. Beats me. Dollars to donuts says your 'date' (which is really a string) has some leading spaces which is messing up your count... Of course, after you fix the import process to have an actual DATE in your DB, this will be a non-issue... Short-term, you should probably at least trim() the data so that your offset of 8 is the number you would expect... -- Some people have a "gift" link here. Know what I want? I want you to buy a CD from some starving artist. http://cdbaby.com/browse/from/lynch Yeah, I get a buck. So? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Query question
You'd still benefit from converting your data into an actual date format in order to take full advantage of available features. hehe You guys are going to start thinking of me as a performance freak, but I think MySQL's date functions might provide better performance than string manipulation. On Oct 29, 2006, at 6:08 AM, Beauford wrote: I don't see this as a particular problem as the data will always be the same. Now under other circumstances in another database, I can see what you are saying. -Original Message- From: Satyam [mailto:[EMAIL PROTECTED] Sent: October 29, 2006 4:30 AM To: Beauford; 'PHP' Subject: Re: [PHP] Query question I said that you have a problem,not that you caused it, and my observation might (hopefully) help newbies in the list. Satyam - Original Message - From: "Beauford" <[EMAIL PROTECTED]> To: "'PHP'" Sent: Sunday, October 29, 2006 9:06 AM Subject: RE: [PHP] Query question This is how I get the data and I have no control over the actual layout of the database. So I have to work with what I have. Robert: LOL, I don't know either. The format is - 01/01/2006. When I first did it I used 7, which should be right, but I ended up getting /2002 /2003, etc. So I went to 8 and all was well. Beats me. B -Original Message- From: Satyam [mailto:[EMAIL PROTECTED] Sent: October 29, 2006 2:14 AM To: Beauford; 'PHP' Subject: Re: [PHP] Query question If that works then you have a problem: you are storing dates as a plain string ( varchar or whatever) instead of a native date/time datatype, which precludes the use of a large number of native SQL date/time functions, such as year() which should suit you nicely in this case. Eventually, you will bump into something more elaborate which you won't be able to do on the SQL side just with string functions and that will force you to bring whole tables into PHP to do more extensive processing. Satyam - Original Message - From: "Beauford" <[EMAIL PROTECTED]> To: "'PHP'" Sent: Sunday, October 29, 2006 7:05 AM Subject: RE: [PHP] Query question This is what I finally figured out, which works just perfectly. select count(date) as count, substring(date,8) as year from stats group by year; Thanks to all for the input. -Original Message----- From: Joe Wollard [mailto:[EMAIL PROTECTED] Sent: October 29, 2006 12:15 AM To: Beauford Cc: PHP Subject: Re: [PHP] Query question Look into the MySQL YEAR() function to extract the year from a specific date. Start here: http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html On 10/28/06, Beauford <[EMAIL PROTECTED]> wrote: I posted this here as I figured I would need to manipulate this using PHP. The code below doesn't quite work because the date is in the format of 05/05/2006. So I am getting totals for each date, not each year. I need the totals for each year, regardless of the day or month. This is why I figured I'd need to use PHP to maybe put it in an array first or something. Thanks -Original Message- From: Joe Wollard [mailto:[EMAIL PROTECTED] Sent: October 28, 2006 10:30 PM To: Ed Lazor Cc: Beauford; PHP Subject: Re: [PHP] Query question Agreed, this should go to a MySQL list. But in the spirit of helping I think the following should give you a good starting point. SELECT `year`, COUNT(`year`) AS `count` FROM `tbl` GROUP BY `year` ASC On 10/28/06, Ed Lazor <[EMAIL PROTECTED]> wrote: Use the mysql list :) On Oct 28, 2006, at 3:01 PM, Beauford wrote: Hi, I have a MySQL database with a date field and a bunch of other fields. The date field is in the format - 01/01/2006. What I want to do is query the database and create a table that shows just the year and how many instances of the year there is. I have been taxing my brain for a simple solution, but just not getting it. Any suggestions? Thanks Example output. Year Count 2002 5 2003 8 2004 9 2005 15 2006 22 ps - I get this information sent to me and I can't change any of the data. I just enter it in the db and then hopefully do the query on it. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Query question
I don't see this as a particular problem as the data will always be the same. Now under other circumstances in another database, I can see what you are saying. -Original Message- From: Satyam [mailto:[EMAIL PROTECTED] Sent: October 29, 2006 4:30 AM To: Beauford; 'PHP' Subject: Re: [PHP] Query question I said that you have a problem,not that you caused it, and my observation might (hopefully) help newbies in the list. Satyam - Original Message - From: "Beauford" <[EMAIL PROTECTED]> To: "'PHP'" Sent: Sunday, October 29, 2006 9:06 AM Subject: RE: [PHP] Query question > This is how I get the data and I have no control over the actual layout of > the database. So I have to work with what I have. > > Robert: > > LOL, I don't know either. The format is - 01/01/2006. When I first did it > I > used 7, which should be right, but I ended up getting /2002 /2003, etc. So > I > went to 8 and all was well. Beats me. > > > B > > -Original Message- > From: Satyam [mailto:[EMAIL PROTECTED] > Sent: October 29, 2006 2:14 AM > To: Beauford; 'PHP' > Subject: Re: [PHP] Query question > > If that works then you have a problem: you are storing dates as a plain > string ( varchar or whatever) instead of a native date/time datatype, > which > precludes the use of a large number of native SQL date/time functions, > such > as year() which should suit you nicely in this case. Eventually, you will > bump into something more elaborate which you won't be able to do on the > SQL > side just with string functions and that will force you to bring whole > tables into PHP to do more extensive processing. > > Satyam > > - Original Message - > From: "Beauford" <[EMAIL PROTECTED]> > To: "'PHP'" > Sent: Sunday, October 29, 2006 7:05 AM > Subject: RE: [PHP] Query question > > >> This is what I finally figured out, which works just perfectly. >> >> select count(date) as count, substring(date,8) as year from stats group >> by >> year; >> >> Thanks to all for the input. >> >> -Original Message- >> From: Joe Wollard [mailto:[EMAIL PROTECTED] >> Sent: October 29, 2006 12:15 AM >> To: Beauford >> Cc: PHP >> Subject: Re: [PHP] Query question >> >> Look into the MySQL YEAR() function to extract the year from a specific >> date. >> Start here: >> http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html >> >> >> >> >> On 10/28/06, Beauford <[EMAIL PROTECTED]> wrote: >>> >>> I posted this here as I figured I would need to manipulate this using >>> PHP. >>> The code below doesn't quite work because the date is in the format of >>> 05/05/2006. So I am getting totals for each date, not each year. I >>> need the totals for each year, regardless of the day or month. This is >>> why I figured I'd need to use PHP to maybe put it in an array first or >>> something. >>> >>> Thanks >>> >>> -Original Message- >>> From: Joe Wollard [mailto:[EMAIL PROTECTED] >>> Sent: October 28, 2006 10:30 PM >>> To: Ed Lazor >>> Cc: Beauford; PHP >>> Subject: Re: [PHP] Query question >>> >>> Agreed, this should go to a MySQL list. But in the spirit of helping I >>> think the following should give you a good starting point. >>> >>> SELECT `year`, COUNT(`year`) AS `count` FROM `tbl` GROUP BY `year` ASC >>> >>> >>> On 10/28/06, Ed Lazor <[EMAIL PROTECTED]> wrote: >>> > >>> > Use the mysql list :) >>> > >>> > >>> > On Oct 28, 2006, at 3:01 PM, Beauford wrote: >>> > >>> > > Hi, >>> > > >>> > > I have a MySQL database with a date field and a bunch of other >>> > > fields. The date field is in the format - 01/01/2006. What I want >>> > > to do is query the database and create a table that shows just the >>> > > year and how many instances of the year there is. I have been >>> > > taxing my brain for a simple solution, but just not getting it. >>> > > Any suggestions? >>> > > >>> > > Thanks >>> > > >>> > > Example output. >>> > > >>> > > Year Count >>> > > >>> > > 2002 5 >>> > > 2003 8 >>> > > 2004 9 >>> > > 2005 15 >>> > > 2006
Re: [PHP] Query question
I said that you have a problem,not that you caused it, and my observation might (hopefully) help newbies in the list. Satyam - Original Message - From: "Beauford" <[EMAIL PROTECTED]> To: "'PHP'" Sent: Sunday, October 29, 2006 9:06 AM Subject: RE: [PHP] Query question This is how I get the data and I have no control over the actual layout of the database. So I have to work with what I have. Robert: LOL, I don't know either. The format is - 01/01/2006. When I first did it I used 7, which should be right, but I ended up getting /2002 /2003, etc. So I went to 8 and all was well. Beats me. B -Original Message- From: Satyam [mailto:[EMAIL PROTECTED] Sent: October 29, 2006 2:14 AM To: Beauford; 'PHP' Subject: Re: [PHP] Query question If that works then you have a problem: you are storing dates as a plain string ( varchar or whatever) instead of a native date/time datatype, which precludes the use of a large number of native SQL date/time functions, such as year() which should suit you nicely in this case. Eventually, you will bump into something more elaborate which you won't be able to do on the SQL side just with string functions and that will force you to bring whole tables into PHP to do more extensive processing. Satyam - Original Message - From: "Beauford" <[EMAIL PROTECTED]> To: "'PHP'" Sent: Sunday, October 29, 2006 7:05 AM Subject: RE: [PHP] Query question This is what I finally figured out, which works just perfectly. select count(date) as count, substring(date,8) as year from stats group by year; Thanks to all for the input. -Original Message- From: Joe Wollard [mailto:[EMAIL PROTECTED] Sent: October 29, 2006 12:15 AM To: Beauford Cc: PHP Subject: Re: [PHP] Query question Look into the MySQL YEAR() function to extract the year from a specific date. Start here: http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html On 10/28/06, Beauford <[EMAIL PROTECTED]> wrote: I posted this here as I figured I would need to manipulate this using PHP. The code below doesn't quite work because the date is in the format of 05/05/2006. So I am getting totals for each date, not each year. I need the totals for each year, regardless of the day or month. This is why I figured I'd need to use PHP to maybe put it in an array first or something. Thanks -Original Message- From: Joe Wollard [mailto:[EMAIL PROTECTED] Sent: October 28, 2006 10:30 PM To: Ed Lazor Cc: Beauford; PHP Subject: Re: [PHP] Query question Agreed, this should go to a MySQL list. But in the spirit of helping I think the following should give you a good starting point. SELECT `year`, COUNT(`year`) AS `count` FROM `tbl` GROUP BY `year` ASC On 10/28/06, Ed Lazor <[EMAIL PROTECTED]> wrote: > > Use the mysql list :) > > > On Oct 28, 2006, at 3:01 PM, Beauford wrote: > > > Hi, > > > > I have a MySQL database with a date field and a bunch of other > > fields. The date field is in the format - 01/01/2006. What I want > > to do is query the database and create a table that shows just the > > year and how many instances of the year there is. I have been > > taxing my brain for a simple solution, but just not getting it. > > Any suggestions? > > > > Thanks > > > > Example output. > > > > Year Count > > > > 2002 5 > > 2003 8 > > 2004 9 > > 2005 15 > > 2006 22 > > > > ps - I get this information sent to me and I can't change any of > > the data. I just enter it in the db and then hopefully do the > > query on it. > > > > > > -- > PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: > http://www.php.net/unsub.php > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Query question
This is how I get the data and I have no control over the actual layout of the database. So I have to work with what I have. Robert: LOL, I don't know either. The format is - 01/01/2006. When I first did it I used 7, which should be right, but I ended up getting /2002 /2003, etc. So I went to 8 and all was well. Beats me. B -Original Message- From: Satyam [mailto:[EMAIL PROTECTED] Sent: October 29, 2006 2:14 AM To: Beauford; 'PHP' Subject: Re: [PHP] Query question If that works then you have a problem: you are storing dates as a plain string ( varchar or whatever) instead of a native date/time datatype, which precludes the use of a large number of native SQL date/time functions, such as year() which should suit you nicely in this case. Eventually, you will bump into something more elaborate which you won't be able to do on the SQL side just with string functions and that will force you to bring whole tables into PHP to do more extensive processing. Satyam - Original Message - From: "Beauford" <[EMAIL PROTECTED]> To: "'PHP'" Sent: Sunday, October 29, 2006 7:05 AM Subject: RE: [PHP] Query question > This is what I finally figured out, which works just perfectly. > > select count(date) as count, substring(date,8) as year from stats group by > year; > > Thanks to all for the input. > > -Original Message- > From: Joe Wollard [mailto:[EMAIL PROTECTED] > Sent: October 29, 2006 12:15 AM > To: Beauford > Cc: PHP > Subject: Re: [PHP] Query question > > Look into the MySQL YEAR() function to extract the year from a specific > date. > Start here: > http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html > > > > > On 10/28/06, Beauford <[EMAIL PROTECTED]> wrote: >> >> I posted this here as I figured I would need to manipulate this using >> PHP. >> The code below doesn't quite work because the date is in the format of >> 05/05/2006. So I am getting totals for each date, not each year. I >> need the totals for each year, regardless of the day or month. This is >> why I figured I'd need to use PHP to maybe put it in an array first or >> something. >> >> Thanks >> >> -Original Message- >> From: Joe Wollard [mailto:[EMAIL PROTECTED] >> Sent: October 28, 2006 10:30 PM >> To: Ed Lazor >> Cc: Beauford; PHP >> Subject: Re: [PHP] Query question >> >> Agreed, this should go to a MySQL list. But in the spirit of helping I >> think the following should give you a good starting point. >> >> SELECT `year`, COUNT(`year`) AS `count` FROM `tbl` GROUP BY `year` ASC >> >> >> On 10/28/06, Ed Lazor <[EMAIL PROTECTED]> wrote: >> > >> > Use the mysql list :) >> > >> > >> > On Oct 28, 2006, at 3:01 PM, Beauford wrote: >> > >> > > Hi, >> > > >> > > I have a MySQL database with a date field and a bunch of other >> > > fields. The date field is in the format - 01/01/2006. What I want >> > > to do is query the database and create a table that shows just the >> > > year and how many instances of the year there is. I have been >> > > taxing my brain for a simple solution, but just not getting it. >> > > Any suggestions? >> > > >> > > Thanks >> > > >> > > Example output. >> > > >> > > Year Count >> > > >> > > 2002 5 >> > > 2003 8 >> > > 2004 9 >> > > 2005 15 >> > > 2006 22 >> > > >> > > ps - I get this information sent to me and I can't change any of >> > > the data. I just enter it in the db and then hopefully do the >> > > query on it. >> > > >> > > >> > >> > -- >> > PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: >> > http://www.php.net/unsub.php >> > >> > >> >> -- >> PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: >> http://www.php.net/unsub.php >> >> > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Query question
If that works then you have a problem: you are storing dates as a plain string ( varchar or whatever) instead of a native date/time datatype, which precludes the use of a large number of native SQL date/time functions, such as year() which should suit you nicely in this case. Eventually, you will bump into something more elaborate which you won't be able to do on the SQL side just with string functions and that will force you to bring whole tables into PHP to do more extensive processing. Satyam - Original Message - From: "Beauford" <[EMAIL PROTECTED]> To: "'PHP'" Sent: Sunday, October 29, 2006 7:05 AM Subject: RE: [PHP] Query question This is what I finally figured out, which works just perfectly. select count(date) as count, substring(date,8) as year from stats group by year; Thanks to all for the input. -Original Message- From: Joe Wollard [mailto:[EMAIL PROTECTED] Sent: October 29, 2006 12:15 AM To: Beauford Cc: PHP Subject: Re: [PHP] Query question Look into the MySQL YEAR() function to extract the year from a specific date. Start here: http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html On 10/28/06, Beauford <[EMAIL PROTECTED]> wrote: I posted this here as I figured I would need to manipulate this using PHP. The code below doesn't quite work because the date is in the format of 05/05/2006. So I am getting totals for each date, not each year. I need the totals for each year, regardless of the day or month. This is why I figured I'd need to use PHP to maybe put it in an array first or something. Thanks -Original Message- From: Joe Wollard [mailto:[EMAIL PROTECTED] Sent: October 28, 2006 10:30 PM To: Ed Lazor Cc: Beauford; PHP Subject: Re: [PHP] Query question Agreed, this should go to a MySQL list. But in the spirit of helping I think the following should give you a good starting point. SELECT `year`, COUNT(`year`) AS `count` FROM `tbl` GROUP BY `year` ASC On 10/28/06, Ed Lazor <[EMAIL PROTECTED]> wrote: > > Use the mysql list :) > > > On Oct 28, 2006, at 3:01 PM, Beauford wrote: > > > Hi, > > > > I have a MySQL database with a date field and a bunch of other > > fields. The date field is in the format - 01/01/2006. What I want > > to do is query the database and create a table that shows just the > > year and how many instances of the year there is. I have been > > taxing my brain for a simple solution, but just not getting it. > > Any suggestions? > > > > Thanks > > > > Example output. > > > > Year Count > > > > 2002 5 > > 2003 8 > > 2004 9 > > 2005 15 > > 2006 22 > > > > ps - I get this information sent to me and I can't change any of > > the data. I just enter it in the db and then hopefully do the > > query on it. > > > > > > -- > PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: > http://www.php.net/unsub.php > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Query question
On Sun, 2006-10-29 at 01:05 -0500, Beauford wrote: > This is what I finally figured out, which works just perfectly. > > select count(date) as count, substring(date,8) as year from stats group by > year; > > Thanks to all for the input. So what you're saying is that you're not using a date field, you're using some kind of string concoction where the year information happens to start at the 8th character? Now that gets me wondering what kind of freaky date format you've used :) I mean I could naturally see the 5th or 7th character being the start point for the year using the following formats respectively: ddmm dd/mm/ But you're grabbing from the 8th character onwards which just seems wierd. Cheers, Rob. -- .. | InterJinn Application Framework - http://www.interjinn.com | :: | An application and templating framework for PHP. Boasting | | a powerful, scalable system for accessing system services | | such as forms, properties, sessions, and caches. InterJinn | | also provides an extremely flexible architecture for | | creating re-usable components quickly and easily. | `' -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Query question
This is what I finally figured out, which works just perfectly. select count(date) as count, substring(date,8) as year from stats group by year; Thanks to all for the input. -Original Message- From: Joe Wollard [mailto:[EMAIL PROTECTED] Sent: October 29, 2006 12:15 AM To: Beauford Cc: PHP Subject: Re: [PHP] Query question Look into the MySQL YEAR() function to extract the year from a specific date. Start here: http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html On 10/28/06, Beauford <[EMAIL PROTECTED]> wrote: > > I posted this here as I figured I would need to manipulate this using PHP. > The code below doesn't quite work because the date is in the format of > 05/05/2006. So I am getting totals for each date, not each year. I > need the totals for each year, regardless of the day or month. This is > why I figured I'd need to use PHP to maybe put it in an array first or > something. > > Thanks > > -Original Message- > From: Joe Wollard [mailto:[EMAIL PROTECTED] > Sent: October 28, 2006 10:30 PM > To: Ed Lazor > Cc: Beauford; PHP > Subject: Re: [PHP] Query question > > Agreed, this should go to a MySQL list. But in the spirit of helping I > think the following should give you a good starting point. > > SELECT `year`, COUNT(`year`) AS `count` FROM `tbl` GROUP BY `year` ASC > > > On 10/28/06, Ed Lazor <[EMAIL PROTECTED]> wrote: > > > > Use the mysql list :) > > > > > > On Oct 28, 2006, at 3:01 PM, Beauford wrote: > > > > > Hi, > > > > > > I have a MySQL database with a date field and a bunch of other > > > fields. The date field is in the format - 01/01/2006. What I want > > > to do is query the database and create a table that shows just the > > > year and how many instances of the year there is. I have been > > > taxing my brain for a simple solution, but just not getting it. > > > Any suggestions? > > > > > > Thanks > > > > > > Example output. > > > > > > Year Count > > > > > > 2002 5 > > > 2003 8 > > > 2004 9 > > > 2005 15 > > > 2006 22 > > > > > > ps - I get this information sent to me and I can't change any of > > > the data. I just enter it in the db and then hopefully do the > > > query on it. > > > > > > > > > > -- > > PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: > > http://www.php.net/unsub.php > > > > > > -- > PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: > http://www.php.net/unsub.php > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Query question
Look into the MySQL YEAR() function to extract the year from a specific date. Start here: http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html On 10/28/06, Beauford <[EMAIL PROTECTED]> wrote: I posted this here as I figured I would need to manipulate this using PHP. The code below doesn't quite work because the date is in the format of 05/05/2006. So I am getting totals for each date, not each year. I need the totals for each year, regardless of the day or month. This is why I figured I'd need to use PHP to maybe put it in an array first or something. Thanks -Original Message- From: Joe Wollard [mailto:[EMAIL PROTECTED] Sent: October 28, 2006 10:30 PM To: Ed Lazor Cc: Beauford; PHP Subject: Re: [PHP] Query question Agreed, this should go to a MySQL list. But in the spirit of helping I think the following should give you a good starting point. SELECT `year`, COUNT(`year`) AS `count` FROM `tbl` GROUP BY `year` ASC On 10/28/06, Ed Lazor <[EMAIL PROTECTED]> wrote: > > Use the mysql list :) > > > On Oct 28, 2006, at 3:01 PM, Beauford wrote: > > > Hi, > > > > I have a MySQL database with a date field and a bunch of other > > fields. The date field is in the format - 01/01/2006. What I want to > > do is query the database and create a table that shows just the year > > and how many instances of the year there is. I have been taxing my > > brain for a simple solution, but just not getting it. Any > > suggestions? > > > > Thanks > > > > Example output. > > > > Year Count > > > > 2002 5 > > 2003 8 > > 2004 9 > > 2005 15 > > 2006 22 > > > > ps - I get this information sent to me and I can't change any of the > > data. I just enter it in the db and then hopefully do the query on > > it. > > > > > > -- > PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: > http://www.php.net/unsub.php > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Query question
I posted this here as I figured I would need to manipulate this using PHP. The code below doesn't quite work because the date is in the format of 05/05/2006. So I am getting totals for each date, not each year. I need the totals for each year, regardless of the day or month. This is why I figured I'd need to use PHP to maybe put it in an array first or something. Thanks -Original Message- From: Joe Wollard [mailto:[EMAIL PROTECTED] Sent: October 28, 2006 10:30 PM To: Ed Lazor Cc: Beauford; PHP Subject: Re: [PHP] Query question Agreed, this should go to a MySQL list. But in the spirit of helping I think the following should give you a good starting point. SELECT `year`, COUNT(`year`) AS `count` FROM `tbl` GROUP BY `year` ASC On 10/28/06, Ed Lazor <[EMAIL PROTECTED]> wrote: > > Use the mysql list :) > > > On Oct 28, 2006, at 3:01 PM, Beauford wrote: > > > Hi, > > > > I have a MySQL database with a date field and a bunch of other > > fields. The date field is in the format - 01/01/2006. What I want to > > do is query the database and create a table that shows just the year > > and how many instances of the year there is. I have been taxing my > > brain for a simple solution, but just not getting it. Any > > suggestions? > > > > Thanks > > > > Example output. > > > > Year Count > > > > 2002 5 > > 2003 8 > > 2004 9 > > 2005 15 > > 2006 22 > > > > ps - I get this information sent to me and I can't change any of the > > data. I just enter it in the db and then hopefully do the query on > > it. > > > > > > -- > PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: > http://www.php.net/unsub.php > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Query question
Agreed, this should go to a MySQL list. But in the spirit of helping I think the following should give you a good starting point. SELECT `year`, COUNT(`year`) AS `count` FROM `tbl` GROUP BY `year` ASC On 10/28/06, Ed Lazor <[EMAIL PROTECTED]> wrote: Use the mysql list :) On Oct 28, 2006, at 3:01 PM, Beauford wrote: > Hi, > > I have a MySQL database with a date field and a bunch of other > fields. The > date field is in the format - 01/01/2006. What I want to do is > query the > database and create a table that shows just the year and how many > instances > of the year there is. I have been taxing my brain for a simple > solution, but > just not getting it. Any suggestions? > > Thanks > > Example output. > > Year Count > > 2002 5 > 2003 8 > 2004 9 > 2005 15 > 2006 22 > > ps - I get this information sent to me and I can't change any of > the data. I > just enter it in the db and then hopefully do the query on it. > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Query question
Use the mysql list :) On Oct 28, 2006, at 3:01 PM, Beauford wrote: Hi, I have a MySQL database with a date field and a bunch of other fields. The date field is in the format - 01/01/2006. What I want to do is query the database and create a table that shows just the year and how many instances of the year there is. I have been taxing my brain for a simple solution, but just not getting it. Any suggestions? Thanks Example output. YearCount 20025 20038 20049 200515 200622 ps - I get this information sent to me and I can't change any of the data. I just enter it in the db and then hopefully do the query on it. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Query Question
I found my problem.
$eventdetail_query = mysql_query("select DISTINCT titles, informations,
file_name from eventdetail, event where eventdetail.event = ".$id);
-Original Message-
From: Karl-Heinz Schulz [mailto:[EMAIL PROTECTED]
Sent: Wednesday, August 11, 2004 10:13 PM
To: 'Jason Davidson'
Subject: RE: [PHP] Query Question
Oops - it shows all the "eventdetails" for all the "events"
-Original Message-
From: Jason Davidson [mailto:[EMAIL PROTECTED]
Sent: Wednesday, August 11, 2004 10:10 PM
To: Karl-Heinz Schulz; [EMAIL PROTECTED]
Subject: Re: [PHP] Query Question
whats the error, are there any matching fields for the columns
eventetail.event and event.id?
Jason
"Karl-Heinz Schulz" <[EMAIL PROTECTED]> wrote:
>
> Why does this query not work?
> It should only show the eventdetails for the shown event.
>
>
> -
>
> $event = mysql_query("select id, information from event where id=".$id);
> ?>
>
> $eventdetail_query = mysql_query("select titles, informations, file_name
> from eventdetail, event where eventdetail.event = event.id");
>
> while($eventdetail = mysql_fetch_row($eventdetail_query)){
>
>
> print(" sans-serif;font-size:12px;\"> href=\"../PDF/$eventdetail[2]\"
>
class=\"decisions_links\">".strip_tags(html_decode($eventdetail[0]))." span>");
> print(" sans-serif;font-size:12px;\">".html_decode($eventdetail[1])."");
>
> }
>
> ?>
>
>
> -
>
>
> TIA
>
> Karl-Heinz
>
>
>
>
> Tracking #: B3BFF38AEF0C854C96E9B501DD35C6C783308CA5
>
>
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Re: [PHP] Query Question
whats the error, are there any matching fields for the columns
eventetail.event and event.id?
Jason
"Karl-Heinz Schulz" <[EMAIL PROTECTED]> wrote:
>
> Why does this query not work?
> It should only show the eventdetails for the shown event.
>
>
> -
>
> $event = mysql_query("select id, information from event where id=".$id);
> ?>
>
> $eventdetail_query = mysql_query("select titles, informations, file_name
> from eventdetail, event where eventdetail.event = event.id");
>
> while($eventdetail = mysql_fetch_row($eventdetail_query)){
>
>
> print(" sans-serif;font-size:12px;\"> href=\"../PDF/$eventdetail[2]\"
> class=\"decisions_links\">".strip_tags(html_decode($eventdetail[0]))." span>");
> print(" sans-serif;font-size:12px;\">".html_decode($eventdetail[1])."");
>
> }
>
> ?>
>
>
> -
>
>
> TIA
>
> Karl-Heinz
>
>
>
>
> Tracking #: B3BFF38AEF0C854C96E9B501DD35C6C783308CA5
>
>
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Re: [PHP] query question
On Wednesday 15 January 2003 01:47, Kelly Meeks wrote:
> I've got an array of values, that I need to query against a database to see
> if any records have any of the values in the array - the array is a list of
> zip codes, and the database contains a list of stores. Basically i need to
> find what stores have zip codes in the array.
>
> I know that mysql doesn't support subquerys - is there a workaround for
> this kind of scenario?
This is more of a mysql question and should be asked on the php-db list if not
the mysql list.
What you need is something like:
"SELECT doo FROM dah WHERE dee IN('val1', 'val2', ..., 'valN'"
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Open Source Software Systems Integrators
* Web Design & Hosting * Internet & Intranet Applications Development *
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Re: [PHP] query question
Not sure if I wrote and thanked you or not, but this did the trick. THANK
YOU THANK YOU THANK YOU!!!
Steve
At 08:41 PM 11/14/2002 +0100, Marek Kilimajer wrote:
This is checked:
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
} else {
list($houre,$minutee,$seconds)=explode(":",$i);
$minutee=$minutee+5;
if($minutee>=60){$minutee=$minutee-60;$houre=$houre+1;}
$tmp =
implode(':',array(str_pad($houre,2,'0',STR_PAD_LEFT)
,str_pad($minutee,2,'0',STR_PAD_LEFT),'00'));
if($tmp != $row->min) { // there is a gap
//echo $row->min ." != $tmp";
echo $i ."\n"; // print previous time
as end
echo $row->min . " - "; // print this time
as new start
}
$i = $row->min; // keep this new time for next loop
}
//echo $i." - $tmp --tmp\n";
//echo $row->min."\n";
}
echo $i ."\n"; // print last row
}
Steve Buehler wrote:
Ok. That output was my fault. I had an error in one of my other
functions. I needed to set $hours to "00" to start with in my
populatetemptable() function. Now my output is:
00:05:00 - 00:05:00
00:05:00 - 00:10:00
00:10:00 - 00:15:00
00:10:00 - 00:20:00
00:15:00 - 00:25:00
00:15:00 - 00:30:00
Problem is that it isn't giving me just the start and end times. I will
keep working on it.
Thanks
Steve
At 12:48 PM 11/14/2002 -0600, you wrote:
Thanks for the help, but that is not actually producing the desired
results either. This returns the following results:
00:00:05 - 00:00:05
00:00:05 - 5
00:00:05 - 10
00:00:05 - 15
00:00:10 - 20
00:00:10 - 25
00:00:10 - 30
00:00:10 - 35
00:00:15 - 40
00:00:15 - 45
00:00:15 - 50
00:00:15 - 55
00:00:20 - 60
00:00:20 - 65
I am not going to post all of them. :)
Here is the code that is now in the getstart_end() function. I had to
change the line that you had because the 5 minutes would give an
error. I just took out the "minutes" part.
$tmp = $i+5 minutes; // add 5 minutes to $i,
implement it yourself
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
}else{
$tmp = $i+5; // add 5 minutes to $i, implement
it yourself
if($tmp != $row->min) { // there is a gap
echo $i ."\n"; // print previous
time as end
echo $row->min . " - "; // print this
time as new start
}
$i = $tmp; // keep this new time for next loop
}
}
echo $row->min ."\n"; // print last row
}
At 07:09 PM 11/14/2002 +0100, you wrote:
Untested, but should work
Steve Buehler wrote:
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
}else{
$tmp = $i + 5 minutes // add 5 minutes to $i, implement it yourself
if($tmp != $row->min) { // there is a gap
echo $i ."\n"; // print previous time as end
echo $row->min . " - "; // print this time as new start
}
$i = $tmp; // keep this new time for next loop
}
}
echo $row->min ."\n"; // print last row
}
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Re: [PHP] query question
This is checked:
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
} else {
list($houre,$minutee,$seconds)=explode(":",$i);
$minutee=$minutee+5;
if($minutee>=60){$minutee=$minutee-60;$houre=$houre+1;}
$tmp =
implode(':',array(str_pad($houre,2,'0',STR_PAD_LEFT)
,str_pad($minutee,2,'0',STR_PAD_LEFT),'00'));
if($tmp != $row->min) { // there is a gap
//echo $row->min ." != $tmp";
echo $i ."\n"; // print previous time
as end
echo $row->min . " - "; // print this
time as new start
}
$i = $row->min; // keep this new time for next loop
}
//echo $i." - $tmp --tmp\n";
//echo $row->min."\n";
}
echo $i ."\n"; // print last row
}
Steve Buehler wrote:
Ok. That output was my fault. I had an error in one of my other
functions. I needed to set $hours to "00" to start with in my
populatetemptable() function. Now my output is:
00:05:00 - 00:05:00
00:05:00 - 00:10:00
00:10:00 - 00:15:00
00:10:00 - 00:20:00
00:15:00 - 00:25:00
00:15:00 - 00:30:00
Problem is that it isn't giving me just the start and end times. I
will keep working on it.
Thanks
Steve
At 12:48 PM 11/14/2002 -0600, you wrote:
Thanks for the help, but that is not actually producing the desired
results either. This returns the following results:
00:00:05 - 00:00:05
00:00:05 - 5
00:00:05 - 10
00:00:05 - 15
00:00:10 - 20
00:00:10 - 25
00:00:10 - 30
00:00:10 - 35
00:00:15 - 40
00:00:15 - 45
00:00:15 - 50
00:00:15 - 55
00:00:20 - 60
00:00:20 - 65
I am not going to post all of them. :)
Here is the code that is now in the getstart_end() function. I had
to change the line that you had because the 5 minutes would give an
error. I just took out the "minutes" part.
$tmp = $i+5 minutes; // add 5 minutes to $i,
implement it yourself
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY
min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
}else{
$tmp = $i+5; // add 5 minutes to $i,
implement it yourself
if($tmp != $row->min) { // there is a gap
echo $i ."\n"; // print previous
time as end
echo $row->min . " - "; // print
this time as new start
}
$i = $tmp; // keep this new time for next loop
}
}
echo $row->min ."\n"; // print last row
}
At 07:09 PM 11/14/2002 +0100, you wrote:
Untested, but should work
Steve Buehler wrote:
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
}else{
$tmp = $i + 5 minutes // add 5 minutes to $i, implement it yourself
if($tmp != $row->min) { // there is a gap
echo $i ."\n"; // print previous time as end
echo $row->min . " - "; // print this time as new start
}
$i = $tmp; // keep this new time for next loop
}
}
echo $row->min ."\n"; // print last row
}
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Re: [PHP] query question
Ok. That output was my fault. I had an error in one of my other
functions. I needed to set $hours to "00" to start with in my
populatetemptable() function. Now my output is:
00:05:00 - 00:05:00
00:05:00 - 00:10:00
00:10:00 - 00:15:00
00:10:00 - 00:20:00
00:15:00 - 00:25:00
00:15:00 - 00:30:00
Problem is that it isn't giving me just the start and end times. I will
keep working on it.
Thanks
Steve
At 12:48 PM 11/14/2002 -0600, you wrote:
Thanks for the help, but that is not actually producing the desired
results either. This returns the following results:
00:00:05 - 00:00:05
00:00:05 - 5
00:00:05 - 10
00:00:05 - 15
00:00:10 - 20
00:00:10 - 25
00:00:10 - 30
00:00:10 - 35
00:00:15 - 40
00:00:15 - 45
00:00:15 - 50
00:00:15 - 55
00:00:20 - 60
00:00:20 - 65
I am not going to post all of them. :)
Here is the code that is now in the getstart_end() function. I had to
change the line that you had because the 5 minutes would give an error. I
just took out the "minutes" part.
$tmp = $i+5 minutes; // add 5 minutes to $i,
implement it yourself
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
}else{
$tmp = $i+5; // add 5 minutes to $i, implement it
yourself
if($tmp != $row->min) { // there is a gap
echo $i ."\n"; // print previous
time as end
echo $row->min . " - "; // print this
time as new start
}
$i = $tmp; // keep this new time for next loop
}
}
echo $row->min ."\n"; // print last row
}
At 07:09 PM 11/14/2002 +0100, you wrote:
Untested, but should work
Steve Buehler wrote:
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
}else{
$tmp = $i + 5 minutes // add 5 minutes to $i, implement it yourself
if($tmp != $row->min) { // there is a gap
echo $i ."\n"; // print previous time as end
echo $row->min . " - "; // print this time as new start
}
$i = $tmp; // keep this new time for next loop
}
}
echo $row->min ."\n"; // print last row
}
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Re: [PHP] query question
Thanks for the help, but that is not actually producing the desired results
either. This returns the following results:
00:00:05 - 00:00:05
00:00:05 - 5
00:00:05 - 10
00:00:05 - 15
00:00:10 - 20
00:00:10 - 25
00:00:10 - 30
00:00:10 - 35
00:00:15 - 40
00:00:15 - 45
00:00:15 - 50
00:00:15 - 55
00:00:20 - 60
00:00:20 - 65
I am not going to post all of them. :)
Here is the code that is now in the getstart_end() function. I had to
change the line that you had because the 5 minutes would give an error. I
just took out the "minutes" part.
$tmp = $i+5 minutes; // add 5 minutes to $i,
implement it yourself
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
}else{
$tmp = $i+5; // add 5 minutes to $i, implement it
yourself
if($tmp != $row->min) { // there is a gap
echo $i ."\n"; // print previous time
as end
echo $row->min . " - "; // print this
time as new start
}
$i = $tmp; // keep this new time for next loop
}
}
echo $row->min ."\n"; // print last row
}
At 07:09 PM 11/14/2002 +0100, you wrote:
Untested, but should work
Steve Buehler wrote:
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
}else{
$tmp = $i + 5 minutes // add 5 minutes to $i, implement it yourself
if($tmp != $row->min) { // there is a gap
echo $i ."\n"; // print previous time as end
echo $row->min . " - "; // print this time as new start
}
$i = $tmp; // keep this new time for next loop
}
}
echo $row->min ."\n"; // print last row
}
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Re: [PHP] query question
One mistake
Marek Kilimajer wrote:
Untested, but should work
Steve Buehler wrote:
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
}else{
$tmp = $i + 5 minutes // add 5 minutes to $i, implement it yourself
if($tmp != $row->min) { // there is a gap
echo $i ."\n"; // print previous time as end
echo $row->min . " - "; // print this time as new start
}
// wrong: $i = $tmp; // keep this new time for next loop
$i = $row->min; // this is instead of the previos line
}
}
echo $row->min ."\n"; // print last row
}
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Re: [PHP] query question
Untested, but should work
Steve Buehler wrote:
function getstart_end(){
$result=mysql_query("SELECT * FROM aplayabletimes ORDER BY min");
while(($row=mysql_fetch_object($result))){
if(!$i){
echo $row->min . " - ";
$i=$row->min;
}else{
$tmp = $i + 5 minutes // add 5 minutes to $i, implement it yourself
if($tmp != $row->min) { // there is a gap
echo $i ."\n"; // print previous time as end
echo $row->min . " - "; // print this time as new start
}
$i = $tmp; // keep this new time for next loop
}
}
echo $row->min ."\n"; // print last row
}
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Re: [PHP] query question
On Mon, May 21, 2001 at 10:36:42AM -0500, Jacky wrote : > What do I do if i want to move all values stored in one field > to another field in the same table, from field A to field B? > Has anyone ever done that? As simple as you think it should be UPDATE table SET column_a = column_b - Markus -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP] query question
several times: UPDATE table SET B=A; Sincerely, Maxim Maletsky Founder, Chief Developer PHPBeginner.com (Where PHP Begins) [EMAIL PROTECTED] www.phpbeginner.com -Original Message- From: Jacky [mailto:[EMAIL PROTECTED]] Sent: Tuesday, May 22, 2001 12:37 AM To: [EMAIL PROTECTED] Subject: [PHP] query question Hi all What do I do if i want to move all values stored in one field to another field in the same table, from field A to field B? Has anyone ever done that? Jack [EMAIL PROTECTED] "There is nothing more rewarding than reaching the goal you set for yourself" -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
