[PHP-DB] select from two tables
I have the following code:
$query = SELECT * FROM xoops_album,xoops_artist WHERE
xoops_album.artist_id = xoops_artist.artist_id;
$albumby = mysql_query($query) or die(Select Failed!);
echotdcenter;
if (mysql_num_rows($albumr)) {
while ($album = mysql_fetch_array($albumr))
{
echo ($album[album]);
echoi by ; echo ($albumby[xoops_artist.artist]); echoibr;
}
echo /center/td;
}
What I'm trying to do is select the album title from one table ( this
works ok ) and get the artist name from another table ( this is the part
I can't get to work ). I believe it has something to do with the code
above as the rest works ok. The result I end up with is below:
Waterloo by ( this is where I want to put the artist name)
Thanks for in help in advance.
RE: [PHP-DB] select from two tables
Ok I tired this but it did not help, but thank you. -Original Message- From: Herman Verkade [mailto:[EMAIL PROTECTED]] Sent: Sunday, 28 July 2002 1:18 a.m. To: 'Barry Rumsey' Cc: [EMAIL PROTECTED] Subject: RE: [PHP-DB] select from two tables Well, I'm just a beginner myself, but I would say that: echoi by ; echo ($albumby[xoops_artist.artist]); echoibr; should at least be: echoi by ; echo ($album[xoops_artist.artist]); echoibr; or even: echoi by ; echo ($album[artist]); echoibr; Hope this helps, Herman -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] multiple inserts
I have the following insert : mysql_connect( localhost, , ); mysql_select_db( ); mysql_query(INSERT INTO music_album VALUES (NULL, '$artist_id' ,'$album' ,NULL ,NULL)); mysql_query(INSERT INTO music_songs VALUES (NULL ,NULL ,NULL ,NULL ,'$songname' ,'$lyrics')); When I run this, the first insert works alright but the second insert does nothing. What am I doing wrong ? How do I do a mysql_insert_id()? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] drop list inserts
Yes I see what I have done wrong. I named it music_artist instead of nameing it music_artist.id Sorry about that but thanks for the help. I think I'll need more help as I'm still learning. -Original Message- From: Christian Schneider [EMAIL PROTECTED] To: Barry Rumsey [EMAIL PROTECTED] Date: Thu, 18 Apr 2002 12:17:34 +0200 Subject: Re: [PHP-DB] drop list inserts Hi Barry, Barry Rumsey wrote: $query_id = mysql_query(INSERT INTO music_album VALUES (NULL, '$music_artist.id' ,'$album' ,NULL ,NULL)); echo $artist_name and $album has been added to the Could someone please point out what I'm doing wrong? is the value of $music_artist.id correct? Also, your select has the name of music_artist and the value of the respective id, isn't that what you want? Regards, Christian -- 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
[PHP-DB] drop list inserts
I have the following to pages( just testing them at the moment ):
? mysql_connect(host,,);
mysql_select_db(music);
echo form name='add_album' method='post' action='test-album-add.php';
$getlist = mysql_query(SELECT * FROM music_artist ORDER BY
artist_name ASC);
echo Artist Name : select name=\artist_name\\n;
while ($row = mysql_fetch_array($getlist)) {
echo 'option value='.$row[id].''.$row[artist_name]./option\n;
}
echo /select\n;
echo brAlbum Name : input type='text' name='album' value='$album';
echo input type='submit' name='Submit' value='Submit';
echo /form;
?
and
?
include(../mainfile.php);
include(../header.php);
OpenTable();
mysql_connect( host, , );
mysql_select_db( xoops );
$query_id = mysql_query(INSERT INTO music_album VALUES
(NULL, '$music_artist.id' ,'$album' ,NULL ,NULL));
echo $artist_name and $album has been added to the
database.;
CloseTable();
include(../footer.php);
?
What I am trying to do is insert the id of the artist they selected in
the first page into a second table. At the moment all I get is 0
inserted instead of the artist id from page 1.
Could someone please point out what I'm doing wrong?
Thanks in advance.
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RE: [PHP-DB] drop list inserts
The $music_artist.id is the id from the first page. ( database = music ,
table = music_artist id ). This is what I need, a drop down list of the
artists in table 'music_artist'( I've got this), I want the id of the
artist they selected in the drop list to be inserted into the
table 'music_album' as $artist_id.
-Original Message-
From: Gurhan Ozen [EMAIL PROTECTED]
To: Barry Rumsey [EMAIL PROTECTED], php-db list php-
[EMAIL PROTECTED]
Date: Wed, 17 Apr 2002 19:02:52 -0400
Subject: RE: [PHP-DB] drop list inserts
Hi Barry,
First of all,
$query_id = mysql_query(INSERT INTO...); is wrong. That line will
just
assign the resultset of the whatever mysql_query() function returns to
the
variable $query_id .. Get rid of $query_id and just have
mysql_query(INSERt
INTO ); See: http://www.php.net/manual/en/function.mysql-query.php
for
this..
Second of all, in your INSERT INTO query you are trying to insert the
value
of a variable called $music_artist.id which doesn't exist anywhere. I
think
you meant to insert $artist_name instead???
Gurhan
-Original Message-
From: Barry Rumsey [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, April 17, 2002 6:23 PM
To: php-db list
Subject: [PHP-DB] drop list inserts
I have the following to pages( just testing them at the moment ):
? mysql_connect(host,,);
mysql_select_db(music);
echo form name='add_album' method='post'
action='test-album-add.php';
$getlist = mysql_query(SELECT * FROM music_artist ORDER BY
artist_name ASC);
echo Artist Name : select name=\artist_name\\n;
while ($row = mysql_fetch_array($getlist)) {
echo 'option
value='.$row[id].''.$row[artist_name]./option\n;
}
echo /select\n;
echo brAlbum Name : input type='text' name='album'
value='$album';
echo input type='submit' name='Submit' value='Submit';
echo /form;
?
and
?
include(../mainfile.php);
include(../header.php);
OpenTable();
mysql_connect( host, , );
mysql_select_db( xoops );
$query_id = mysql_query(INSERT INTO music_album VALUES
(NULL, '$music_artist.id' ,'$album' ,NULL ,NULL));
echo $artist_name and $album has been added to the
database.;
CloseTable();
include(../footer.php);
?
What I am trying to do is insert the id of the artist they selected in
the first page into a second table. At the moment all I get is 0
inserted instead of the artist id from page 1.
Could someone please point out what I'm doing wrong?
Thanks in advance.
--
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To unsubscribe, visit: http://www.php.net/unsub.php
--
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[PHP-DB] Tutorials
Are there any good tutorials for beginners on forms and inserts into mulit-tables ? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] database table layout help
I am setting up a music info site and have decided on the follow tables ( not to sure weather to add the mp3 part). The one main question I have before starting to make the tables is weather mysql_insert_id() would be able to handle this with the way I have it ? :Artist id artist_name :Album id - auto-inc artist_id album_name rec_id image_id :Songs id - auto-inc album_id genre_id mp3_id lyrics :Genre id - auto-inc genre :Recording id - auto-inc recording_studio date :Album Image id - auto-inc image :Artist Info id - auto-inc info :Mp3 id - auto-inc mp3_location Any other suggestions are welcome. B.J.Rumsey -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Simple Display
I am trying to do a simple fetch of the lastest add name. I have the following code : $query = SELECT name FROM name ORDER BY name DESC LIMIT 1; $latename = mysql_query($query) or die(Select Failed!); $latename = mysql_fetch_array($latename); echo bLastest Name Added:/b $latename I know I am doing something wrong as being new to php it get be confused. It returns : Lastest Name Added: Array Could someone please point out where i'm going wrong Thank you -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] option in forms
I know how to do the option call e.g. This is the list of artists already in our database : SELECT name=artist ?php echo $option_block; ? /SELECT But I want to do a form where they select the option then fill out rest of form. The option above returns the artist name, but I want to insert the artistid. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Count(*)
Thank you everyone. The "COUNT(*) AS c" worked great. ---Original Message--- From: Kai Voigt Date: Friday, 25 January 2002 11:11:06 a. To: Barry Rumsey Cc: [EMAIL PROTECTED] Subject: Re: Count(*) Barry Rumsey wrote: I have this small query on a page:emit source="sql" host=xoops query= SELECT COUNT(*) FROM xp_topics WHERE artist='artist' ORDER BY topicid DESC LIMIT 1"sql.artist;/emitUse "SELECT COUNT(*) AS C FROM ..." instead. Then you can access thecount value as sql.c;Kai-- dreiecksplatz 8, d-24105 kiel, +49-431-22199869, http://k.123.org/. IncrediMail - Email has finally evolved - Click Here
Re: [PHP-DB] Books and what nots
Just got back from Newmarket. Yes they had 1 copy of the book and got
her to put it aside for me. Now it's down to some hard reading in my spare
time or even take it to work.
Thanks
---Original Message---
From: DL Neil
Date: Thursday, 24
January 2002 9:25:06 a.
To: Barry Rumsey
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB]
Books and what nots
and as they say "if we haven't got it in stock, we'll get it in for
you..."
At Newmarket one of the staff is reasonably 'up' on computer books -
the others tend to take deep breaths and roll their eyes. Just in case,
the ISBN is 0-672-31784-2 (but that may have changed if a new edition has
been published).
Regards,
=dn
- Original Message -
From:
Barry Rumsey
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: 23 January 2002 18:31
Subject: Re: [PHP-DB] Books and
what nots
No haven't tried Amazon because I haven't got a credit card,
used to ,but it got me into a lot of trouble. No I'll head down to
tech books in a couple of of hours when they open.
---Original
Message---
From: DL Neil
Date: Thursday,
24 January 2002 7:23:17 a.
To: Barry Rumsey
Cc: [EMAIL PROTECTED]
Subject: Re:
[PHP-DB] Books and what nots
Barry,
You can't buy a decent computer book for less than
NZD100!
That's a fair price against what I paid.
The USD price looks even worse.
Tried Amazon etc?
=dn
snip
IncrediMail - Email has finally evolved -
Click
Here
RE: [PHP-DB] # of Records in Table
Zach wrote: "I do generally find that the more brief I am in describing an issue, themore likely I am to get a reply. Then details can be given if needed, infurther correspondences." I would like to disagree with this. I am only a newbie to php and have found that if I can describe to my best ability the problem I'm having I can get a reply that helps me in the best way in 1 or 2 replies. There is none of this back and forward and end up still getting nowhere. You have towait for some replies, but they are worth it. ---Original Message--- From: Zach Curtis Date: Thursday, 24 January 2002 1:02:07 p. To: DL Neil Cc: [EMAIL PROTECTED] Subject: RE: [PHP-DB] # of Records in Table There are many columns of data in the table. The only column that I wouldlike to extract a value from is the password field (this is the key aswell). I do not have an AUTO_INCREMENT field. Although, I see how addingthat field could be of use by using it with the SELECT MAX() as youmentioned.The last record is whatever the last record was inserted using INSERT INTO.To give you the overall picture of what I am trying to accomplish:1) I am opening up a db table and searching for the password of the lastrecord in the table2) I then open a flat file and search for that password and grab the nextrecord after that (as records are appended to the end of the flat file) andany other subsequent records added to the flat file3) Write those new records to the db tableIn this scenario, the last record would be the last record from the flatfile added to the db table.Another suggestion from [EMAIL PROTECTED] was to:Select your password field, and whatever other fields you need toUse-mysql_num_rows() to capture the number of rows returned-mysql_data_seek() to position pointer on last row (remember indexingstarts at 0)-mysql_fetch_row() or mysql_fetch_array(), your choice, to return thedata.This is done with one select statement, however all the passwords from thetable have to be loaded into array (perhaps not the most efficient way? Butbetter than what I had come up with.).I do generally find that the more brief I am in describing an issue, themore likely I am to get a reply. Then details can be given if needed, infurther correspondences.I will give that AUTO_INCREMENT and SELECT MAX() a try as well.Zach-Original Message-From: DL Neil [mailto:[EMAIL PROTECTED]]Sent: Wednesday, January 23, 2002 3:27 PMTo: Zach CurtisCc: [EMAIL PROTECTED]Subject: Re: [PHP-DB] # of Records in TableZach,We're going back and forth on this and getting no where...So far all you have informed us is that the table has one column, whichcontains a bunch of passwords (I alsosuggest that this is not really the case and there'll at least be some sortof userId field - but you haven'tbothered to tell me/us that)The concept of "last" in relational terminology is ambiguous. Do you meanlast entered, or do you mean the rowwith the field containing the highest value in the column - for example.Some people are used to the idea that there is also some 'phantom' rowIdthat counts/labels each row. In otherfile systems this might have been the case, but the physically last recordin an RDBMS table-file may notconform to either of the "last" definitions mentioned above. There is nosuch 'highest' in an RDBMS unless youput it there - as mentioned by another correspondent, it's a good idea touse some sort of 'id' field in everytable. This can be generated for you with the AUTO_INCREMENT feature. Withappropriate design, eg first recordhas id=1 and the succeeding records have id-s with ascending values, thenyou can use SELECT MAX() against the'id' or possibly against the AUTO_INCREMENT feature.These commands are well documented - RTFM. Many tutorials with plenty ofexamples of authentication techniquesexist on the PHP/MySQL web sites.Next time, please don't be so (repeatedly) parsimonious in the provision ofinformation, if you expect/hope thatsomeone is going to give their (free) time to help you out. (grumble)=dn- Original Message -From: "Zach Curtis" lt;[EMAIL PROTECTED]To: "DL Neil" lt;[EMAIL PROTECTED]Cc: lt;[EMAIL PROTECTED]Sent: 23 January 2002 21:52Subject: RE: [PHP-DB] # of Records in Table The key field is the password field, which is also the field that I would like to retrieve for that last record. Zach -Original Message- From: DL Neil [mailto:[EMAIL PROTECTED]] Sent: Wednesday, January 23,
[PHP-DB] Counter view total
In one of my tables I have a field called count_view, How do I add all the totals up to come up with one grand total of views. IncrediMail - Email has finally evolved - Click Here
[PHP-DB] distinct - how to do
I have the following query: $query = SELECT * FROM xp_topics, xp_stories WHERE xp_topics.topicid = xp_stories.topicid AND xp_topics.artistname='Faith Hill' ORDER BY topictext DESC LIMIT 0,20; It returns the lot but a lot of doubles. I want to put a DISTINCT call the query, but if I put it after the select ( select distinct * from...) it returns nothing. How do you do a DISTINCT?
[PHP-DB] very simple answer
I can't figure out where I have gone wrong. Maybe not enough sleep, anyway this is the simple query: $query = SELECT * FROM xp_topics WHERE artist='artist'ORDER BY topicid DESC LIMIT 0,1; $result = mysql_db_query($query) or die(Select Failed!); echo(Latest Artists : $topictext br); It says query failed. I know I've forgoten something but don't know what
Re: [PHP-DB] very simple answer
I ended up rewriting it again and somewhere I must have done something wrong as it now works. Thanks guys. Maybe I should think about more sleep. - Original Message - From: Miles Thompson [EMAIL PROTECTED] To: Barry Rumsey [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Friday, January 18, 2002 9:48 AM Subject: Re: [PHP-DB] very simple answer Barry, Add or die( mysql_errno() . : . mysql_error() ) It could be you don't have any data, also there's no space between /artist' and ORDER BY. Also trim off everything after 'artist' and see what you get. If it still doesn't work there's something wrong with the field names. Then start adding stuff back. BTW you don't need ORDER BY if you are LIMITing to 1. g Miles At 08:52 AM 1/18/2002 +1300, Barry Rumsey wrote: I can't figure out where I have gone wrong. Maybe not enough sleep, anyway this is the simple query: $query = SELECT * FROM xp_topics WHERE artist='artist'ORDER BY topicid DESC LIMIT 0,1; $result = mysql_db_query($query) or die(Select Failed!); echo(Latest Artists : $topictext br); It says query failed. I know I've forgoten something but don't know what -- PHP Database 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]
[PHP-DB] Count
How do you add a count feature to a link. I can get stuff from the db but want to know how many times it's been read or click on?
[PHP-DB] How are site layout blocks made
I'll ask this question again but in a different way. In these php news portals like phpnuke, myphpnuke, they lay their site out in these so called blocks.I'm looking for information on how these blocks are made, are they done in php or html and where can I find more information on this.
Re: [PHP-DB] PHP blocks
ones like phpnuke, myphpnuke, postnuke etc - Original Message - From: olinux [EMAIL PROTECTED] To: Barry Rumsey [EMAIL PROTECTED] Sent: Friday, January 11, 2002 7:26 PM Subject: Re: [PHP-DB] PHP blocks What portals? olinux --- Barry Rumsey [EMAIL PROTECTED] wrote: On these web portals they have things call blocks that arrange the layout of the site. Where can one go to learn how to do this sort of layout. __ Do You Yahoo!? Send FREE video emails in Yahoo! Mail! http://promo.yahoo.com/videomail/ -- PHP Database 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-DB] multiple tables insert
Thanks for that. But it has brought up another question. I'll list the tables so you'll know what i'm on about. TABLE 1 : Album alb_id int(11) NOT NULL auto_increment, art_id int(11) NOT NULL default '0', alb_name varchar(255) NOT NULL default '', alb_image varchar(255) NOT NULL default '', alb_year date NOT NULL default '-00-00', alb_genre varchar(100) NOT NULL default '', PRIMARY KEY (alb_id) TABLE 2 : Artist art_id int(11) NOT NULL auto_increment, art_name varchar(255) NOT NULL default '', art_details mediumtext NOT NULL, PRIMARY KEY (art_id) TABLE 3 : Songs song_id int(11) NOT NULL auto_increment, song_name varchar(255) NOT NULL default '', song_lyrics mediumtext NOT NULL, song_info mediumtext NOT NULL, PRIMARY KEY (song_id) TABLE 4 : Tracks track_id int(11) NOT NULL auto_increment, alb_id int(11) NOT NULL default '0', art_id int(11) NOT NULL default '0', song_id int(11) NOT NULL default '0', PRIMARY KEY (track_id) Ok I'll be able to work out the simple part of the inserts, but what about the auto_increment parts thatt relie on the information in the other files. eg: 'artist.art_id' should be the same as 'album.art_id' the problem is that 'artist.art_id' is auto_increment and 'album.art_id' is not. So if I added an artist in the artist table and the auto_increment gave it the value of 5, how would I update the 'album.art_id' with the same value? - Original Message - From: Miles Thompson [EMAIL PROTECTED] To: Barry Rumsey [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Saturday, January 12, 2002 9:50 AM Subject: Re: [PHP-DB] multiple tables insert Barry See below ... At 09:09 AM 1/12/2002 +1300, Barry Rumsey wrote: Two questions: 1) If I have a url in the db that points to a image , how do I get php to get this image and display it ? Check the IMG tag in an HTML reference 2) I have 4 tables in the db and would like to know of a good tutorial on inserting to multiple tables form a single form. There are many tutorials, the one I most frequently recommend is by Julie Meloni at http://www.thickbook.com pick the one on custom error messages as she develops it very nicely. As for the inserts, in the processing part of the script, you'll know what I mean when you examine the tutorial, execute an INSERT for the data you want to insert in each of the tables. INSERT acts on one table at a time, so you'll have four of them. That should get you going - Miles Thompson -- PHP Database 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] -- PHP Database 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-DB] multiple tables insert
I had been looking for a pre made script for a lyric site like phpnuke but the only one I could find was myphplyrics which is quite hopeless ,so I've had no choice but to try and make my own any way I would like to thank you for all your help so far. B.J.Rumsey. - Original Message - From: Miles Thompson [EMAIL PROTECTED] To: Barry Rumsey [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Saturday, January 12, 2002 12:51 PM Subject: Re: [PHP-DB] multiple tables insert Barry, Good question, and you have discovered one of MySQL's limitations, it does not enforce functional constraints. To put it another way, you can design parent-child relationships, but the db will never say You can't add this because the parent record doesn't exist. So you have to do it in code, not a big deal,and it's been done that way for years with other flat file, pseudo-relational database -- FoxPro, dBase, etc. You work around it by capturing the auto_increment value using the mysql_insert_id() function. Hence you can insert the values for the artist table, call mysql_insert_id() and assign it to a variable, say $art_id_key, and use it in the insert for the album table. Then repeat, calling mysql_insert_id() and assigning it to $album_id_key after the insert into the album table so that you will have it for the insert into the tracks table. And so forth. You will be doing all these inserts on the same thread, so if someone else inserts into the same table you should be safe, according to the documentation. If A inserts into artist, B inserts into artist, then A calls mysql_insert_id() the value returned is for A's insert, not B's. MySQL has a separate thread for each of A and B. Now, don't trust me. Read up on the last_insert_id() and mysql_insert_id() functions in both the MySQL and PHP docs. I'll toss this out - what if you decide to capture other information, such as the name of the producer. Can your design handle that? What changes might you have to make to accommodate the change? Regards - Miles Thompson At 10:17 AM 1/12/2002 +1300, Barry Rumsey wrote: Thanks for that. But it has brought up another question. I'll list the tables so you'll know what i'm on about. TABLE 1 : Album alb_id int(11) NOT NULL auto_increment, art_id int(11) NOT NULL default '0', alb_name varchar(255) NOT NULL default '', alb_image varchar(255) NOT NULL default '', alb_year date NOT NULL default '-00-00', alb_genre varchar(100) NOT NULL default '', PRIMARY KEY (alb_id) TABLE 2 : Artist art_id int(11) NOT NULL auto_increment, art_name varchar(255) NOT NULL default '', art_details mediumtext NOT NULL, PRIMARY KEY (art_id) TABLE 3 : Songs song_id int(11) NOT NULL auto_increment, song_name varchar(255) NOT NULL default '', song_lyrics mediumtext NOT NULL, song_info mediumtext NOT NULL, PRIMARY KEY (song_id) TABLE 4 : Tracks track_id int(11) NOT NULL auto_increment, alb_id int(11) NOT NULL default '0', art_id int(11) NOT NULL default '0', song_id int(11) NOT NULL default '0', PRIMARY KEY (track_id) Ok I'll be able to work out the simple part of the inserts, but what about the auto_increment parts thatt relie on the information in the other files. eg: 'artist.art_id' should be the same as 'album.art_id' the problem is that 'artist.art_id' is auto_increment and 'album.art_id' is not. So if I added an artist in the artist table and the auto_increment gave it the value of 5, how would I update the 'album.art_id' with the same value? - Original Message - From: Miles Thompson [EMAIL PROTECTED] To: Barry Rumsey [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Saturday, January 12, 2002 9:50 AM Subject: Re: [PHP-DB] multiple tables insert Barry See below ... At 09:09 AM 1/12/2002 +1300, Barry Rumsey wrote: Two questions: 1) If I have a url in the db that points to a image , how do I get php to get this image and display it ? Check the IMG tag in an HTML reference 2) I have 4 tables in the db and would like to know of a good tutorial on inserting to multiple tables form a single form. There are many tutorials, the one I most frequently recommend is by Julie Meloni at http://www.thickbook.com pick the one on custom error messages as she develops it very nicely. As for the inserts, in the processing part of the script, you'll know what I mean when you examine the tutorial, execute an INSERT for the data you want to insert in each of the tables. INSERT acts on one table at a time, so you'll have four of them. That should get you going - Miles Thompson -- PHP Database 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] -- PHP Database Mailing
[PHP-DB] simple question
I have a colum with the numbers between 0-9 ( table is a list of bands )eg. 2 live crew , 2pac , 1 maniacs also in this table Ihave normal names like : abba , queen , police etc.. How would I query the table to list only the ones starting with numbers.
[PHP-DB] count
I'm trying to do the following query:
$db_connect = mysql_connect($sqlhostname,$login,$password);
$base_selection = mysql_select_db($base,$db_connect);
$query = SELECT DISTINCT * FROM xp_topics WHERE artist_count='artist' AND topictext
LIKE
'A%' ORDER BY artist_count ASC limit 0,5; $req = mysql_query($query);
$res = mysql_num_rows($req); if ($res == 0)
{ echo /p
pbN/A/b;} else { while($row = mysql_fetch_array($req))
$query2 = SELECT SUM(1) FROM xp_stories,xp_topics WHERE xp_topics.topicid =
xp_stories_topicid AND xp_topics.topictext = '$topictest';
$numtopic = mysql_query($query2) or die(Select Failed!);
$numtopic = mysql_fetch_array($numtopic);
{ extract($row); echo ($topictext $numtopic[0]br
); } }
?
Which should return a name and then count how many topics by that person.
If I leave out the query2 statement I do get a list of name, but I can't get query2 to
work.
[PHP-DB] PHP blocks
On these web portals they have things call blocks that arrange the layout of the site. Where can one go to learn how to do this sort of layout.
[PHP-DB] query and count
I have two tables, the first one I query like :
$query = SELECT * FROM xp_topics WHERE artist_count='artist' AND topictext LIKE
'B%' ORDER BY artist_count DESC limit 0,5; $req = mysql_query($query);
$res = mysql_num_rows($req); if ($res == 0) { echo /p
pbSorry there is no result./b;} else { while($row =
mysql_fetch_array($req))
{ extract($row); echo ($topictextbr
); } }
The second one I want to do a count for each one that was returned above. eg Billie
(2) , Bill (4) Bret (1)
How would I set up the query to do both?
[PHP-DB] count query
Is it posible to do a count(*) on a tabe where id=2 and count just those that id = 2
[PHP-DB] Formated Results
Hi I have a database that has lyrics in it. I can get them to list but they are not aligned the same as whats in the database. How can I get the results aligned as whats in the database ?
Re: [PHP-DB] Formated Results
Yes thank you that was the answer. While I'm asking questions , I've got one good one : Whats the best book to buy on php mysql ? - Original Message - From: Andrey Hristov [EMAIL PROTECTED] To: Barry Rumsey [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Sent: Tuesday, January 08, 2002 1:27 AM Subject: Re: [PHP-DB] Formated Results Try printing pre before and /pre after them Regards, Andrey Hristov - Original Message - From: Barry Rumsey [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Monday, January 07, 2002 2:19 PM Subject: [PHP-DB] Formated Results Hi I have a database that has lyrics in it. I can get them to list but they are not aligned the same as whats in the database. How can I get the results aligned as whats in the database ? -- PHP Database 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]
[PHP-DB] ABC listings
One last question for tonight as I have been given alot of help from this group to keep me going for awhile. I would like to set up one of those A B C D E F page listings. How do i query the database to grab the first letter from each entrey.
[PHP-DB] Blocks
Hi In these web portals they have these things called blocks, is there a tutorial on how to do these anywhere
[PHP-DB] DESC LIMIT
$query = SELECT * FROM xp_sings,xp_artist,xp_songs WHERE xp_artist.artist_id =
xp_sings.artist_id AND xp_sings.song_id = xp_songs.song_id AND song_name LIKE 'b%'
DESC LIMIT 5;
$req = mysql_query($query);
$res = mysql_num_rows($req);
if ($res == 0)
{ echo centerbSorry there is no result./b/center;}
else
{ while($row = mysql_fetch_array($req))
{
extract($row);
It returns as Sorry there is no result but if I leave of the DESC and limit , it
works. Can someone tell me where I'm going wrong.
[PHP-DB] help on mulit query
Hi I have three tables set out below: xp_artist: artist_id , artist _name xp_sings: artist_id , songs_id xp_songs: songs_id , song_name , lyrics I can not figure out how to query and return the results for a query like: select song_name FROM xp_artist,xp_sings,xp_songs WHERE xp_artist.artist_id = xp_sings.artist_id AND xp_sings.song_id = xp_songs.songs_id AND artist_name LIKE 'b%';
Re: [PHP-DB] help on mulit query
I did not see that ( I must be blind hehe ). One other question , I have the lyrics also stored in the database, how can I create a link so that when the query is returned they can click on song name and the lyrics will be viewed? - Original Message - From: Bogdan Stancescu [EMAIL PROTECTED] To: Barry Rumsey [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Sent: Monday, January 07, 2002 1:51 PM Subject: Re: [PHP-DB] help on mulit query You seem to be doing it fine as far as I can see, except for song_id which should be songs_id in the query... Bogdan Barry Rumsey wrote: Hi I have three tables set out below: xp_artist: artist_id , artist _name xp_sings: artist_id , songs_id xp_songs: songs_id , song_name , lyrics I can not figure out how to query and return the results for a query like: select song_name FROM xp_artist,xp_sings,xp_songs WHERE xp_artist.artist_id = xp_sings.artist_id AND xp_sings.song_id = xp_songs.songs_id AND artist_name LIKE 'b%'; -- PHP Database 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] -- PHP Database 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]
