bubblegum.anarchy

Yeah, I never tested the SELECT @id := in php - does not appear to work. Use SQL_CALC_FOUND_ROWS in the LIMITed query then SELECT found_rows() and some math to calculate the rank in php.

The query looks fine syntactically, maybe the tbl prefix is an issue... try again with the table name and column names wrapped in backticks: SELECT * FROM `tbl` WHERE CHAR_LENGTH(`col`) > 5;

I recently learnt from fenway that mysql 5.1 has an inbuilt event scheduler: http://dev.mysql.com/doc/refman/5.1/en/events.html

or mysql_insert_id() in php.

Also consider wrapping the strings with html content in single quotes so that the following: $departments .= "<select name=\"user_department\" size=\"5\" id=\"user_department\">$lst_departments</option>"; Would be more readable in this format: $departments .= '<select name="user_department" size="5" id="user_department">$lst_departments</option>';

yeah... event_rsvp would certainly make a good double for an invitation record with date set to null as default making the NOT rsvp'd query a great deal simpler: event_rsvp.id event_rsvp.event_id event_rsvp.member_id <= forgot this member earlier event_rsvp.date SELECT member.id , member.name , event.name FROM event_rsvp INNER JOIN event ON event_rsvp.event_id = event.id INNER JOIN member ON event_rsvp.member_id = member.id WHERE event_rsvp.event_id = $event_id AND event_rsvp.date IS NULL much better!

To get a list of all members that have NOT rsvp'd an event: SELECT member.id , member.name , event.name , event_rsvp.date FROM member LEFT JOIN event_rsvp ON event_rsvp.member_id = member.id LEFT JOIN event ON event.id = event_rsvp.event_id AND event.id = $event_id WHERE event_rsvp.member_id IS NULL; something like that anyway.

The three core tables: member.id member.name event.id event.name group.id group.name A table to track events rsvp: event_rsvp.id event_rsvp.event_id event_rsvp.date A table to track members in groups: member_in_group.member_id member_in_group.group_id To get a list of all members and their associated groups: SELECT member.id , member.name , group_concat(group.name SEPARATOR '\n') AS groups FROM member LEFT JOIN member_in_group ON member.id = member_in_group.member_id LEFT JOIN group ON member_in_group.group_id = group.id GROUP BY member.id ORDER BY member.name To get a list of all members that have rsvp'd an event: SELECT member.id , member.name , event.name , event_rsvp.date FROM event INNER JOIN event_rsvp ON event.id = event_rsvp.event_id INNER JOIN member ON event_rsvp.member_id = member.id WHERE event.id = $event_id;

Only one php function would be required.. with using a newline separator in the group_concat... $caption = explode("\n", $record['caption']);

Columns can not be created dynamically. A group_concat can group all the captions into one column and add a seperator as well... have a look at the group_concat information here: http://dev.mysql.com/doc/refman/4.1/en/group-by-functions.html

I get conflicting results. The following table definitions and record inserts: DROP TABLE IF EXISTS property; CREATE TABLE property ( idINT NOT NULL AUTO_INCREMENT primary key, address CHAR(50) ); INSERT INTO property (address) VALUES ('123 Fake Street, Faketown'); DROP TABLE IF EXISTS deals; CREATE TABLE deals ( property_id INT, deal_type_id INT, sales_pitch TEXT, price INT, INDEX (sales_pitch(10), price), PRIMARY KEY (property_id, deal_type_id) ); INSERT INTO deals (property_id, deal_type_id, sales_pitch, price) VALUES (1, 1, 'buy buy buy', 100000), (1, 2, 'hurry hurry hurry', 100000), (1, 3, 'make an offer', 100000), (1, 4, 'hurry last days', 100000), (1, 5, 'once in a lifetime', 100000), (1, 6, 'golf course nearby', 100000), (1, 7, 'beach side property', 100000); The following explain: EXPLAIN SELECT deals.property_id , deals.price # , deals.sales_pitch # , property.address FROM deals LEFT JOIN property ON deals.property_id = property.id results: id select_type table type possible_keys key key_len ref rows Extra 1 SIMPLE deals ALL NULL NULL NULL NULL 7 1 SIMPLE property eq_ref PRIMARY PRIMARY 4 test.deals.property_id 1 Using index With only the index property_id in the SELECT values resulted in the type `index`: EXPLAIN SELECT deals.property_id # , deals.price # , deals.sales_pitch # , property.address FROM deals LEFT JOIN property ON deals.property_id = property.id; resulted in: id select_type table type possible_keys key key_len ref rows Extra 1 SIMPLE deals index NULL PRIMARY 8 NULL 7 Using index 1 SIMPLE property eq_ref PRIMARY PRIMARY 4 test.deals.property_id 1 Using index

First, a question: Do you understand the following sentence? I cant figure out how to go about this as I don't be able to put in ore than one WHERE statement? And now the answer: SELECT user.id, user.bandname, user.bio, users.history, users.discography, images.path FROM users, images WHERE users.id = images.id AND users.bandid = $bandid or SELECT user.id, user.bandname, user.bio, users.history, users.discography, images.path FROM users INNER JOIN images ON users.id = images.id WHERE users.bandid = $bandid

Aliases will be required to distinguish identical column names, something like this: SELECT person.pid , person.name_first , baclabor.encounter_nr AS baclabor_encounter_nr , patho.encounter_nr AS patho_encounter_nr , chemlab.encounter_nr AS chemlab_encounter_nr , .... etc FROM care_encounter AS encounter... And then in PHP: <?php echo $rows['chemlab_encounter_nr']; ?> Consider trimming the selected values to only those that are required.

Having a read of the following might be helpful in regards to old_password. http://dev.mysql.com/doc/refman/5.0/en/old-client.html

The following query: SELECT person.pid , person.name_first , baclabor.* , patho.* , chemlab.* FROM care_encounter AS encounter INNER JOIN care_person AS person ON encounter.pid = person.pid LEFT JOIN care_test_findings_baclabor AS baclabor ON encounter.encounter_nr = baclabor.encounter_nr LEFT JOIN care_test_findings_patho AS patho ON encounter.encounter_nr = patho.encounter_nr LEFT JOIN care_test_findings_chemlab AS chemlab ON encounter.encounter_nr = chemlab.encounter_nr WHERE baclabor.encounter_nr IS NOT NULL OR patho.encounter_nr IS NOT NULL OR chemlab.encounter_nr IS NOT NULL Produces the following result: pid name_first batch_nr encounter_n room_nr dept_nr notes findings_in findings_cu findings_fi type entry_nr rec_date doctor_id findings_da findings_ti status modify_id modify_time create_id create_time batch_nr encounter_n room_nr dept_nr type doctor_id findings_da findings_ti status modify_id modify_time create_id create_time batch_nr encounter_n job_id test_date test_time group_id type validator validate_dt status modify_id modify_time create_id create_time 10000000 pig 1 1 10000001 2007-03-13 10:26:50 priority Chemical Test 0000-00-00 00:00:00 admin 2007-05-05 10:44:52 admin 2007-03-13 10:26:50 10000000 pig 3 9 10000002 2007-04-04 09:57:49 priority Chemical Test 0000-00-00 00:00:00 hidden 2007-05-05 11:02:22 admin 2007-04-04 09:57:49 10000000 pig 5 9 10000002 2007-04-04 14:38:21 priority Chemical Test 0000-00-00 00:00:00 admin 2007-04-13 14:38:24 admin 2007-04-13 14:38:21 10000000 pig 6 12 10000001 2007-05-06 00:00:00 priority Chemical Test 0000-00-00 00:00:00 2007-05-03 14:35:59 0000-00-00 00:00:00 10000005 kib 4 13 0 0 0 0 Bacteriological Test 0000-00-00 0000-00-00 00:00:00 2007-05-05 11:00:01 0000-00-00 00:00:00 10000006 test 4 14 10000003 2007-04-11 16:59:25 priority Chemical Test 0000-00-00 00:00:00 2007-05-05 12:07:20 admin 2007-04-11 16:59:25 10000005 kib 1 15 0 Pathological Test 2007-05-18 00:00:00 2007-05-07 16:10:01 0000-00-00 00:00:00 10000007 benson 5 16 25 dftf 0 0 0 Bacteriological Test 30000001 2007-04-16 2007-04-16 09:57:00 done admin 2007-05-11 08:54:12 admin 2007-04-16 09:57:04 If the above results are not what is required then what is required???????

maliary... I'd suggest that Barand would appreciate more than one tables worth of information.. more like all the tables that are described in the query and some test data for each table.

This is pretty much the same as your original query Azu: SELECT count(*) FROM Z LEFT JOIN a ON Z.1 = a.1 LEFT JOIN b ON Z.1 = b.1 LEFT JOIN c ON Z.1 = c.1 WHERE a.1 IS NULL AND b.1 IS NULL AND c.1 IS NULL

Use the JOIN syntax

Possibly?: SELECT count(*) FROM Z WHERE X NOT IN ( SELECT 1 FROM a UNION SELECT 2 FROM b UNION SELECT 3 FROM c ) EDIT: the above query is not faster.

I do not understand your question, galone.

There is probably an issue with the parenthesis used around the select columns.

Use a direct link (foreign key): player.id player.name player.team_id :: identifies which team the player record is associated to player.foot team.id team.name team.coach A player is linked to a team via player.team_id = team.id

By adding an ORDER BY clause...

looks ok'ish

Make the following code changes and post the results here: $query = "INSERT INTO mailinglist (email) VALUES ('$email')"; mysql_query($query) or die('Error, insert query failed: '.$query.' error: '.mysql_error());