[[SOLVED] Need help optimizing join]

Are you still having trouble with that first query?

[Looking for the most efficient way to do this]

Use DISTINCT or GROUP BY to remove the duplicates.

Definitely group by, DISTINCT is the most useless modifier I've ever come across (except inside a COUNT() ).

 

or GROUP_CONCAT

[Where Are My Error Messages?]

http://www.php.net/errorfunc

[[SOLVED] Selecting duplicate values]

Confirm that Oracle uses the same GROUP_CONCAT function as MySQL does.

 

EDIT: Oracle does not appear have a group_concat function.

[Check if user is registered before inviting]

What is a regged email and what does the array have to do with anything if the query output is fine?

[MYSQL database don't update please help.]

What don't work? - you have to submit the information first.

[varchar(MAX) in MyISAM?]

I doubt that the client used matters but just incase - ALTER queries performed via SQLyog community v5.31

[Looking for the most efficient way to do this]

Use DISTINCT or GROUP BY to remove the duplicates.

[SQL conversion]

Use the php trim function:

 

trim($key)

[[SOLVED] Using CASE to Compare Result and Add Dates]

change the following line:

 

GetSQLValueString($_POST['period'], "text"),

 

to:

 

mysql_real_escape_string($_POST['period']),

 

so that no quoting occurs.

 

[warning mysql_fetch_array]

Change this line:

 

<?php
   $tsel = mysql_query("select * from `users` where `userlevel`='0' and `status`='alive' order by `rank` desc limit 10");
while ($top=mysql_fetch_array($tsel)) { // line 36
      print "<tr><td></center><a href=profile.php?viewuser=$top[username]>$top[username][/url]</td><td align=right>$top[rank]</td></tr>";
   }
   ?>

 

To this:

<?php
   $tsel = mysql_query($query = "select * from `users` where `userlevel`='0' and `status`='alive' order by `rank` desc limit 10") or trigger_error(mysql_error()."<PRE>".$query."</PRE>", E_USER_ERROR);;
while ($top=mysql_fetch_array($tsel)) { // line 36
      print "<tr><td></center><a href=profile.php?viewuser=$top[username]>$top[username][/url]</td><td align=right>$top[rank]</td></tr>";
   }
   ?>

 

... and investigate the mysql error.

[MYSQL database don't update please help.]

Change this line:

 

$sql = "UPDATE school_book SET title='$title', category='$category', type='$type', processed='$processed', image='$image', url='$url', description='$description', username='$username', user='$user' WHERE id=$id";
//replace school_book with your table name above
$result = mysql_query($sql) or die(mysql_error());

 

To this:

 

$sql = "UPDATE school_book SET title='$title', category='$category', type='$type', processed='$processed', image='$image', url='$url', description='$description', username='$username', user='$user' WHERE id=$id";
//replace school_book with your table name above
die($sql);
$result = mysql_query($sql) or die(mysql_error());

 

and post the result contained in $sql;

[[SOLVED] Help with Max dates]

heh...

 

SELECT Max( report_date ) AS LatestDate, loc_id
FROM report_index
GROUP BY loc_id
HAVING max(report_date) < CURRENT_DATE - INTERVAL 1 YEAR
LIMIT 0 , 500

[[SOLVED] Help with Max dates]

unquote LatestDate in the WHERE clause:

 

WHERE LatestDate < CURRENT_DATE - INTERVAL 1 YEAR

[[SOLVED] Using CASE to Compare Result and Add Dates]

I suggest renaming the radio element value attribute to MONTH and DAY and then something like this:

 

adddate(CURRENT_DATE, INTERVAL $_POST['interval'] $_POST['period']);

 

... and perform the appropriate data integrity checks before querying.

[[SOLVED] mySQL database design question]

Basic table schematic:

    repository.id

    repository.date

 

Query:

    SELECT * FROM repository ORDER BY repository.date DESC LIMIT 24

[select statement syntax error]

$Int = addslashes("Pix 'outside' Interface");

[[SOLVED] ID's in one table and output values in other]

Post the three table structures, some test data and a detailed description of what you are trying to achieve.

[[SOLVED] query help]

Something like this:

 

SELECT the_friend.id, the_friend.name
FROM users AS person
    INNER JOIN friends AS friend_of_theres ON person.id = friend_of_theres.friendid 
    INNER JOIN users AS the_friend ON friend_of_theres.userid = the_friend.id
    LEFT JOIN friends AS friend_of_mine ON friend_of_theres.userid = friend_of_mine.friendid AND friend_of_mine.userid = person.id
WHERE person.id = $_SESSION['userID'] AND friend_of_mine.userid IS NULL

[db Design Question]

A table with a user id and product id

 

vote.user_id

vote.product_id

 

To verify if the user has voted on any product:

SELECT * FROM vote WHERE user_id = $user_id

 

To get a total of votes for a product:

SELECT count(*) FROM vote WHERE product_id = $product_id

OR

SELECT product_id, count(*) AS votes FROM vote GROUP BY product_id ORDER BY votes DESC # most votes first

[Which 2 table structure is better for photo order storing]

Using version #2 and the following query to order the records:

 

SELECT FIND_IN_SET(PhotoId, PhotoOrder) AS orderby, Photos.*
FROM Photos
     INNER JOIN `Album Categories` ON Photos,AlbumId = `Album Categories`.AlbumId
ORDER BY orderby;

[drop down menu question]

Predefined variables, depending on how the data is posted... either $_GET['subcat'] or $_POST['subcat'], or regardless of the method used $_REQUEST['subcat'];

 

Have a read: http://au3.php.net/manual/en/language.variables.predefined.php

[Check if user is registered before inviting]

Place: print $query just after the call to mysql_query() and post the results - this is not a mysql issue if there is nothing unusual about the results.

[drop down menu question]

First of all you need to give name to the form and then using the java script get the values of cat and subcat as given below

 

$cat=document.formname.cat.options[document.formname.cat.selectedIndex].value

$sub=document.formname.sub.options[document.formname.sub.selectedIndex].value

A server side variable can not be assigned a client side javascript value.

[drop down menu question]

Did you write that code, imarockstar?