[PHP-DB] formatting a date from mysql
I need to figure out how to format the date format: year-month-day (2005-12-06). It is a date(8) field in a mysql table that i pulled out of the table with php 5.0.5. I need it in the format: Tue December 6, 2005 (or in mysql's formatting: 6-12-2005). I am using the reformatting for display only. So it would be something like this: echo $journal['Date']; //with whatever formatting needed If possible I need php to do the formatting for me. Thanks... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] formatting a date from mysql
farily straight forward to withinn the query itself, check out http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html or for php http://uk.php.net/date RTFM Regards Adrian Eternity Records Webmaster wrote: I need to figure out how to format the date format: year-month-day (2005-12-06). It is a date(8) field in a mysql table that i pulled out of the table with php 5.0.5. I need it in the format: Tue December 6, 2005 (or in mysql's formatting: 6-12-2005). I am using the reformatting for display only. So it would be something like this: echo $journal['Date']; //with whatever formatting needed If possible I need php to do the formatting for me. Thanks... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: formatting a date from mysql
Eternity Records Webmaster wrote: I need to figure out how to format the date format: year-month-day (2005-12-06). It is a date(8) field in a mysql table that i pulled out of the table with php 5.0.5. I need it in the format: Tue December 6, 2005 (or in mysql's formatting: 6-12-2005). I am using the reformatting for display only. So it would be something like this: echo $journal['Date']; //with whatever formatting needed If possible I need php to do the formatting for me. Thanks... An easy way to do it: ?php explode(-,$date); $journal['Date'] = $date[2] . - . $date[1] . - . $date[0]; ? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Minor Change
You're getting an error, after the query, put: echo mysql_error(); to find out what's happening. On Monday 12 December 2005 11:05 am, [EMAIL PROTECTED] wrote: I made tiny changes to my php file and sql table and the table won't come up. I updated the table name (and php file name) from 109fh5 to 109fh6. In the table, I changed 6 cells, leaving a couple blank. Then I changed only the digit 5 to make it a 6 (109fh6) in the following: $get_data_query = select rep, party, state, cd, minority, afr_am, asian, am_indian, hispanic, med_hsehold_income, poverty from 109fh6 order by $sort_field $sort_order; Now I get Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in etc. I've done this many times without a problem (this is the 6th time in this sequence). What could be wrong after such a minor change? Ken -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Need a Help!
I have table and insert data using song_id, that means one artist can have
many song_ids. my question is how can I query distinct artist with his/her
song_id while I will not get duplicate data like duplicate artist?
my code is like this
$content .=form id=\form1\ method=\post\ action=\\
select name=\Quick\ onchange=\MM_jumpMenu('parent',this,0)\
option value=\#\Select Artist/option;
$result= $db-sql_query(SELECT distinct(artist), id FROM
.$prefix._lyrics order by artist asc);
if ($db-sql_numrows($result)) {
while($row = $db-sql_fetchrow($result)) {
extract($row);
$content .=option value='modules.php
?name=$module_nameamp;file=artistamp;c_id=$id'$artist/option;
}
}
$content .=/select
/form;
that code is fine except it gives me duplicate artist, so I want get rid off
that duplicate. any help
Re: [PHP-DB] Minor Change
After adding echo mysql_error(); I get the same result. I tried changing the query to include 109fh7 (a table which doesn't exist) and got the same result as with 109fh6. Changing to 109fh5 does pull up that table. The line to which the error message refers is while ($row = mysql_fetch_assoc ($data_set)) That is what always come up when there is an error in the query. ** On Mon, 12 Dec 2005 14:13:10 -0500, Micah Stevens [EMAIL PROTECTED] wrote: You're getting an error, after the query, put: echo mysql_error(); to find out what's happening. On Monday 12 December 2005 11:05 am, [EMAIL PROTECTED] wrote: I made tiny changes to my php file and sql table and the table won't come up. I updated the table name (and php file name) from 109fh5 to 109fh6. In the table, I changed 6 cells, leaving a couple blank. Then I changed only the digit 5 to make it a 6 (109fh6) in the following: $get_data_query = select rep, party, state, cd, minority, afr_am, asian, am_indian, hispanic, med_hsehold_income, poverty from 109fh6 order by $sort_field $sort_order; Now I get Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in etc. I've done this many times without a problem (this is the 6th time in this sequence). What could be wrong after such a minor change? Ken -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Minor Change
Ken: Echo out the query. Check it. Run it from the MySQL client. What does the error message say? Does the table 109fh6 actually exist in the database? David # Ken responded: # # After adding echo mysql_error(); I get the same result. I tried changing # the query to include 109fh7 (a table which doesn't exist) and got the same # result as with 109fh6. Changing to 109fh5 does pull up that table. The # line to which the error message refers is while ($row = mysql_fetch_assoc # ($data_set)) That is what always come up when there is an error in the # query. ** On Mon, 12 Dec 2005 14:13:10 -0500, Micah Stevens [EMAIL PROTECTED] wrote: You're getting an error, after the query, put: echo mysql_error(); to find out what's happening. On Monday 12 December 2005 11:05 am, [EMAIL PROTECTED] wrote: I made tiny changes to my php file and sql table and the table won't come up. I updated the table name (and php file name) from 109fh5 to 109fh6. In the table, I changed 6 cells, leaving a couple blank. Then I changed only the digit 5 to make it a 6 (109fh6) in the following: $get_data_query = select rep, party, state, cd, minority, afr_am, asian, am_indian, hispanic, med_hsehold_income, poverty from 109fh6 order by $sort_field $sort_order; Now I get Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in etc. I've done this many times without a problem (this is the 6th time in this sequence). What could be wrong after such a minor change? Ken -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Minor Change
This isn't really a MySQL error (sorta), it's a PHP error. You probably forgot
to update a variable name when you updated everything else.
Here's an example sequence for querying using PHP/MySQL:
$TestQY = SELECT * from SomeTable;
$TestRS = mysql_query($TestQY) or die(Error executing query);
while ($TestRW = mysql_fetch_assoc($TestRS)) {
$somearr[] = $TestRW; // do something with data
}
Since you're getting a Not a valid MySQL result resouce error with the
mysql_fetch_assoc() function, I'd search for all your mysql_fetch_assoc()
statements and double check their $TestRS. That error is saying that your
$TestRS isn't a valid MySQL result set. That could mean that $TestRS isn't
defined (maybe you're still using $TestOldRS and forgot to change a variable
name) or possibly that $TestQY is empty or bad somehow so mysql_query() isn't
generating a proper MySQL result set (try echoing out your $TestQY to see what
it is.. then try executing it manually on the MySQL server and see if you get
an error).
You can try the or die() syntax I use above to see if mysql_query() is
bombing out so you'll get notice before it even gets to the mysql_fetch_assoc().
Lastly.. someone recommended echoing out mysql_error(). Your response makes it
sound like you think that this fixes your problem. It's not going to fix
anything, just possibly give you some information about what failed. If you
get a MySQL error message from mysql_error(), please post it. It might help us
determine what the problem is.
It may not contain anything (under certain failing circumstances) so I'd step
through the things I've listed above. They may shed some light on where the
error is and then we can figure out how to fix it. Probably a typo in a
query, variable name or table name I'm guessing.
Let us know if you find anything else out.
-TG
= = = Original message = = =
After adding echo mysql_error(); I get the same result. I tried changing
the query to include 109fh7 (a table which doesn't exist) and got the same
result as with 109fh6. Changing to 109fh5 does pull up that table. The
line to which the error message refers is while ($row = mysql_fetch_assoc
($data_set)) That is what always come up when there is an error in the
query.
**
On Mon, 12 Dec 2005 14:13:10 -0500, Micah Stevens
[EMAIL PROTECTED] wrote:
You're getting an error, after the query, put:
echo mysql_error();
to find out what's happening.
On Monday 12 December 2005 11:05 am, [EMAIL PROTECTED] wrote:
I made tiny changes to my php file and sql table and the table won't
come
up. I updated the table name (and php file name) from 109fh5 to 109fh6.
In the table, I changed 6 cells, leaving a couple blank. Then I changed
only the digit 5 to make it a 6 (109fh6) in the following:
$get_data_query = select rep, party, state, cd, minority, afr_am,
asian,
am_indian, hispanic, med_hsehold_income, poverty from 109fh6 order by
$sort_field $sort_order;
Now I get Warning: mysql_fetch_assoc(): supplied argument is not a
valid
MySQL result resource in etc.
I've done this many times without a problem (this is the 6th time in
this
sequence). What could be wrong after such a minor change?
Ken
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Re: [PHP-DB] Need a Help!
Couple of things you can do:
1. Drop the song ID and only get the artist information SELECT
distinct(Artist) from songtable. It doesn't look like your SELECT statement
needs a song, but you include the song ID as $id anyway. Any reason for that or
can you drop it so you only get artist?
2. Pre-parse the results of your current query so you only get one artist
and/or compile a list of song ID's while you're at it. Instead of doing your
option in the database query result WHILE statement, do it outside:
while (results) {
$artistinfo[$artist][] = $id;
}
Then:
echo select name='blah';
foreach ($artistinfo as $artist = $songsarr) {
echo optgroup label='$artist';
foreach ($songsarr as $songid) {
echo option value='$id'$id or whatever/option;
}
echo /optgroup;
}
echo /select;
3. Use a different DB structure. I prefer this structure myself:
Table ARTISTS:
ArtistID
ArtistName
OtherArtistInfo
Table SONGS:
SongID
ArtistID
OtherSongInfo
Or, instead of having ArtistID, if more than one artist may be linked to a
song, you can do:
Table xrefArtistsSongs:
ArtistID
SongID
Not sure why you'd want to do that for a song unless you're counting covers of
songs and want all artists who've covered it to point to the same SongID data.
This way you can:
SELECT * from ARTISTS
for your ARTISTS select box
Then once an artist is selected, do a:
SELECT * from SONGS where ARTISTID = $artistid
Just some ideas.
Forgive the pseudo-code and mixed capitalizations.. I think you get the idea
I'm trying to convey.
-TG
= = = Original message = = =
I have table and insert data using song_id, that means one artist can have
many song_ids. my question is how can I query distinct artist with his/her
song_id while I will not get duplicate data like duplicate artist?
my code is like this
$content .=form id=\form1\ method=\post\ action=\\
select name=\Quick\ onchange=\MM_jumpMenu('parent',this,0)\
option value=\#\Select Artist/option;
$result= $db-sql_query(SELECT distinct(artist), id FROM
.$prefix._lyrics order by artist asc);
if ($db-sql_numrows($result))
while($row = $db-sql_fetchrow($result))
extract($row);
$content .=option value='modules.php
?name=$module_nameamp;file=artistamp;c_id=$id'$artist/option;
$content .=/select
/form;
that code is fine except it gives me duplicate artist, so I want get rid off
that duplicate. any help
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Re: [PHP-DB] Minor Change
did it say which line does the error Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in is referring to. if it is referring to the line which contains the function call mysql_query (), dpirago is right, there could be some mistake in the sql statement. running it through the mysql client would definitely verify whether your query is correct or not. [EMAIL PROTECTED] wrote: Ken: Echo out the query. Check it. Run it from the MySQL client. What does the error message say? Does the table 109fh6 actually exist in the database? David # Ken responded: # # After adding echo mysql_error(); I get the same result. I tried changing # the query to include 109fh7 (a table which doesn't exist) and got the same # result as with 109fh6. Changing to 109fh5 does pull up that table. The # line to which the error message refers is while ($row = mysql_fetch_assoc # ($data_set)) That is what always come up when there is an error in the # query. ** On Mon, 12 Dec 2005 14:13:10 -0500, Micah Stevens [EMAIL PROTECTED] wrote: You're getting an error, after the query, put: echo mysql_error(); to find out what's happening. On Monday 12 December 2005 11:05 am, [EMAIL PROTECTED] wrote: I made tiny changes to my php file and sql table and the table won't come up. I updated the table name (and php file name) from 109fh5 to 109fh6. In the table, I changed 6 cells, leaving a couple blank. Then I changed only the digit 5 to make it a 6 (109fh6) in the following: $get_data_query = select rep, party, state, cd, minority, afr_am, asian, am_indian, hispanic, med_hsehold_income, poverty from 109fh6 order by $sort_field $sort_order; Now I get Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in etc. I've done this many times without a problem (this is the 6th time in this sequence). What could be wrong after such a minor change? Ken -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
