Marcel1993

November 23, 2011

Hey guys,

I've got a table with the following contents:

CID VALUE TIME

1 A 500

1 B 700

1 C 400

2 D 100

2 E 300

3 F 100

Now I want to output this table without having entries with the same CID. If a entry got a unique CID it is supposed to get displayed. If several entries got the same CID, only the entry with the highest time is supposed to get displayed:

Example:

1 B 700

2 E 300

3 F 100

Is there a way to solve it in SQL?

Thanks :-)

July 30, 2011

Hey ;-)

thank you for creating the query!

I tried it but there is a malfunction in the script. It shows as many results as available profiles, in reference to the USED_ID. If the user got 3 profiles, the result contains 3 rows, independent whether the WORLD_ID matches the column "WORLDS" in "comp". The rows that are not matching with "comp" are empty, the right ones are filled with the correct content.

Thanks ;-)

July 29, 2011

Hey,

thank you for answering. I read a lot of examples on the internet for join but I don't know how to start in my case :-(

I will be very happy if you can help me

July 29, 2011

Hey guys,

I want to creat a SQL-Query and I don't know how to construct the part of "where".

There are four tables

1: users - important column ID (USER_ID)

2: profiles (each user is able to create more than one profile) - important column WORLD_ID, USER_ID

(3: comp_participants (each user is able to create more than one entry in the table, each comp_participant is connected to "comp"))

4: comp - important column WORLDS (e.g. "2, 4, 20")

$user_id is set (e.g. 44)

Now I want to show all entries of "comp" if the user got a profile on a world which is listed in WORLDS in the table "comp". I thing it is a very difficult kind of SQL-query and I look forward to a answer ;-)

Best regards ;-)

July 18, 2011

Hey guys,

I've got a if-statement, which is supposed to be true by some kinds of combinations:

$true = "1,2,3";

$value = 1 (TRUE)

$value = 2 (TRUE)

$value = 3 (TRUE)

$value = 4 (FALSE)

How can I put it in an if-statement?

Best regards ;-)

January 6, 2011

Hey guys,

I want to create stats of visitors and some other values in the database depending on time.

The stats are supposed to be created one time per hour. What can I use for it? Cronjob? What will be if it doesn't work??

Thanks for helping!

December 27, 2010

Sorry, there were little mistakes in. I corrected these and now I think it is working :-)

Thanks for helping! I try this script seriously and then I post again :-*

December 27, 2010

Okay, I tried, but that doesn't work:

<?php

include 'includes/settings/mysql.php';
mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbbase);

$time = time();

$update_weed_world_data_sql = "SELECT id, weed_factor, weed_basis FROM worlds";
$update_weed_world_data_res = mysql_query($update_weed_world_data_sql) OR DIE (mysql_error());
while($update_weed_world_data_while = mysql_fetch_assoc($update_weed_world_data_res)){
$update_weed_world_id = $update_weed_world_data_while['id'];
$update_weed_world_basis[$update_weed_world_id] = $update_weed_world_data_while['weed_basis'];
$update_weed_world_factor[$update_weed_world_id] = $update_weed_world_data_while['weed_factor'];
}


function do_update_weed($profile_id,$world,$update,$level,$time,$ba,$fa){
$update_weed_period = $time - $update;
$update_weed_add = ($update_weed_period / 3600 * $ba[$world] * pow($fa[$world],$level)) +1;
echo $update_weed_world_basis[$world]*5;
echo "<br>";
mysql_query("UPDATE profiles SET weed = weed +$update_weed_add, weed_update = $time WHERE id = $profile_id LIMIT 1");
}

function todo_update_weed($time,$ba,$fa){
$todo_update_weed_sql = "SELECT * FROM profiles WHERE weed_update < $time";
$todo_update_weed_res = mysql_query($todo_update_weed_sql) or die (mysql_error());
while($todo_update_weed_while = mysql_fetch_assoc($todo_update_weed_res)){
do_update_weed(			$todo_update_weed_while['id'],
							$todo_update_weed_while['world_id'],
							$todo_update_weed_while['weed_update'],
							$todo_update_weed_while['level_farm'],
							$time,
							$ba,
							$fa
				);
}
}

todo_update_weed($time, $update_weed_world_basis[$update_weed_world_id], $update_weed_world_factor[$update_weed_world_id]);

December 27, 2010

Thanks,

but $update_weed_add has to be set inside the function. :shrug:

December 27, 2010

Hey,

I've been developing a browsergame. In this you are able to produce something ("weed"). Now I've got some kind of trouble with the function that is responsible for updating the "weed" on the user accounts.

Im going to explain, how my script is working

First step:

Loading the datas from "worlds". In this case id, weed_factor, weed_basis. What this is required for you will see in the next steps :-*

<?php

include 'includes/settings/mysql.php';
mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbbase);

$time = time();

$update_weed_world_data_sql = "SELECT id, weed_factor, weed_basis FROM worlds";
$update_weed_world_data_res = mysql_query($update_weed_world_data_sql) OR DIE (mysql_error());
while($update_weed_world_data_while = mysql_fetch_assoc($update_weed_world_data_res)){
$update_weed_world_id = $update_weed_world_data_while['id'];
$update_weed_world_basis[$update_weed_world_id] = $update_weed_world_data_while['weed_basis'];
$update_weed_world_factor[$update_weed_world_id] = $update_weed_world_data_while['weed_factor'];
global $update_weed_world_basis;
}

Second step:

This is the function which is editing each account by the data given as parameters.

function do_update_weed($profile_id,$world,$update,$level,$time){
$update_weed_period = $time - $update;
$update_weed_add = ($update_weed_period / 3600 * $update_weed_world_basis[$world] * pow($update_weed_world_factor[$world],$level)) +1;
echo $update_weed_world_basis[$world];
echo "<br>";
mysql_query("UPDATE profiles SET weed = weed +$update_weed_add, weed_update = $time WHERE id = $profile_id LIMIT 1");
}

Note:

echo $update_weed_world_basis[$world];

doesnt result an output.

In step three, the accounts are loaded:

function todo_update_weed($time){
$todo_update_weed_sql = "SELECT * FROM profiles WHERE weed_update < $time";
$todo_update_weed_res = mysql_query($todo_update_weed_sql) or die (mysql_error());
while($todo_update_weed_while = mysql_fetch_assoc($todo_update_weed_res)){
do_update_weed(			$todo_update_weed_while['id'],
							$todo_update_weed_while['world_id'],
							$todo_update_weed_while['weed_update'],
							$todo_update_weed_while['level_farm'],
							$time
				);
}
}

And the last step is calling function "todo_update_weed" with the time.

todo_update_weed($time);

I script calculate the time between the last update and the current time. This make a difference in time. With this value I calculate something else. And then the update time has to be saved.

The script has just working fine, when I set the variable "$update_weed_add" with a fix value. But by now the script doesn't work, because I don't know how to make the array available inside functions.

I considered whether I put the database-function for the world_data inside the function that updates the profiles. But if I do this, the database will be called thousands of times...

Hope anybody has a nice idea to fix the problem

December 27, 2010

Hey,

thanks for helping :-). I've just made an embarrassing mistake :-*

I didn't select "id" in the SQL-Query...

Don't be cross with me

December 27, 2010

Thank you but that didn't lead to any difference:

$update_weed_world_data_sql = "SELECT weed_factor, weed_basis FROM worlds";
$update_weed_world_data_res = mysql_query($update_weed_world_data_sql) OR DIE (mysql_error());
while($update_weed_world_data_while = mysql_fetch_assoc($update_weed_world_data_res)){
$update_weed_world_id = $update_weed_world_data_while['id'];
$update_weed_world_basis[$update_weed_world_id] = $update_weed_world_data_while['weed_basis'];
$update_weed_world_factor[$update_weed_world_id] = $update_weed_world_data_while['weed_factor'];
}

echo $update_weed_world_factor[1];

December 27, 2010

Hey,

my code doesnt work and I dont know why:

$update_weed_world_data_sql = "SELECT weed_factor, weed_basis FROM worlds";
$update_weed_world_data_res = mysql_query($update_weed_world_data_sql) OR DIE (mysql_error());
while($update_weed_world_data_while = mysql_fetch_assoc($update_weed_world_data_res)){
$update_weed_world_basis[''.$update_weed_world_data_while['id'].''] = $update_weed_world_data_while['weed_basis'];
$update_weed_world_factor[''.$update_weed_world_data_while['id'].''] = $update_weed_world_data_while['weed_factor'];
}

I want to save some rows in variables identifiable by the value in the array.

As example. When I want to echo $update_weed_world_factor['1'] - I want to get the row with ID 1 and the column "world_factor".

I've used this code long time ago - but know something was going wrong.

Thanks for helping :-)

December 24, 2010

Hey guys,

I don't know what to do. I've got a script used for gamble:

$gamble_rand = rand($number_of_tickets,$gamble_chance);
if($gamble_chance == $gamble_rand){

If $number_of_tickets is set by 99 and $gamble_chance is set by 100 the chance is 1:100 that the If-Clause is not true. But it seems to me that PHP is making its on "random". Independent of the values PHP is interpreting that 20 times of 100 times the correct logic is false...

If $number_of_tickets is set by 1 there's no big different. Its just a little bit less, but not the chance that it has to be.

Thanks for helping and Merry Christmas :-)

December 23, 2010

Hey,

I've got a table with thousands of entries. No I want to select some of them. At first I want to get some of them with a more exact condition. The other ones I want to pick with using ORDER BY RAND()

Its there a way to put this in one query?

December 23, 2010

Hey, thats was it :-)

Thank you very much! :-)

December 23, 2010

Hey,

$upgrade_next_sql = "SELECT * FROM todo_upgrades WHERE profile_id = ".$profile_id." AND level = ".$profile_data['level_'.$show.'']+1." AND type = '$show'";

doesn't work because there is "+1". How can I add this value to the variable without forming a extra variable before?

December 22, 2010

Hey,

I've got a query in which a variable is interpreted as a column and I don't why this is caused.

$upgrade_time_sql = "SELECT * FROM todo_upgrades WHERE profile_id = ".$profile_id." AND level = ".$profile_data['level2_'.$show.'']." AND type = ".$show."";
$upgrade_time_res = mysql_query($upgrade_time_sql) or die (mysql_error());

$show is filled with the content "storage"

The mysql_error is

"Unknown column 'storage' in 'where clause'"

Thanks for helping.

December 22, 2010

Hey,

thanks. As you said above the variable was without content. It was because the query was surrounded by a function - so I made a condition of the var filled with the time-content.

Thank you very much!

December 22, 2010

Hey,

$time has been defined before as

$time = time();

December 22, 2010

Hey,

$todo_upgrades_sql = "SELECT * FROM todo_upgrades WHERE time_completion >= $time";
$todo_upgrades_res = mysql_query($todo_upgrades_sql) or die (mysql_error());

results:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

December 22, 2010

Hey,

there is no different between including PHP in HTML or JavaScript. I recommend to use this Format

<?php echo $var; ?>

With the version above you won't get trouble. In some cases you will get trouble with " or ' if you use mixed code. echo "<javasc..."

December 21, 2010

Hey guys,

I dont know whether this is the right place for my question. I've just installed XAMPP on a computer. Now I want to reach XAMPP on another one in my network. So I've just used the IP-adress from the computer where XAMPP is installed. This has been working very fine.

Now I want to open "htdocs" using FTP. When I used the same IP-adress as stated above my computer didn't show htdocs to me as supposed. There is a directory "outgoing" and the file "onefile.html".

Thank you for answering :shy:

December 11, 2010

Thanks, that's a very important aspect - I think I'll write it in files

December 11, 2010

Hey guys,

I am looking for a possibily to create a multilingual application. The languages I want to provide are English, German, and Spainish. At first I thougth it is going to be very easy. I just create a extra table for the language-contents.

ENGLISH GERMAN SPANISH

TABLE TISCH MESA

So far so good - but what will be if there is a phrase with included variables? I think in English it is quiet easy to form correct sentences. In German and Spanish there are some specifications what lead to garbled sentences.

So LANGUAGE_PART_1 VARIABLE_1 LANGUAGE_PART_2 LANGUAGE_PART_2 wouldn't work as desired.

Can I make PHP just set in a place any variable? So when I've got the sentence "At the moment there are #NUMBER_OF_TABLES# tables available", "Momentan sind #NUMBER_OF_TABLES# Tische verfügbar" or "Actualmente hay #NUMBER_OF_TABLES# mesas disponibles".

In this case PHP is supposed to set in #NUMBER_OF_TABLES# the variable $number_of_tables.

Perhaps there are some other ways for developing a multilingual app.

Thanks a lot for helping ::)

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Marcel1993

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