Jay2391

So in my local server i have a folder call Assets and Public In the assets folder I have a PHP file call "123.php" I am trying to get my "index.php" file that is located in the public folder to include that file. the issue is that all do they are on the same server folder there not in the same folder Host -> asset -> 123.php -> public -> index.php how do I get that done. Thanks...J PS: First time I use PHP in 4 years can believe how quickly i can for get things

Can someone help me understand what technology and connections you need to create a Ladder site for third party games? My Plan… I am working on a site that you can create tourneys and register results, but what will it take to get results from the actual game servers. Is apache sufficient? Is it a simple feed? Do I need to know Java? Thx

i included my user name and this is what i got know Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\web\xampp\htdocs\mdo_main\t.php on line 20

i got a database error??? Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\web\xampp\htdocs\mdo_main\t.php on line 7 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\web\xampp\htdocs\mdo_main\t.php on line 7 Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\web\xampp\htdocs\mdo_main\t.php on line 8 Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\web\xampp\htdocs\mdo_main\t.php on line 12 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\web\xampp\htdocs\mdo_main\t.php on line 12 Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\web\xampp\htdocs\mdo_main\t.php on line 13 Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\web\xampp\htdocs\mdo_main\t.php on line 19 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\web\xampp\htdocs\mdo_main\t.php on line 19 Warning: mysql_fetch_assoc(): supplied argument is not

I have a date 11-07-2007 22:06 and i want to know how to calculate 4 hours ahead and create a count down system $date_now $date_4hours $countdown = $date_now - $date_4hours;

So i have this code that i go to this page and i display teh data from a table so i click a link that take sme to this page .. home.php?i=$1&a=$a7t=$t now i display the info and i have a submit button that have a post to say yes or no and update a table now when i click the submit button it take me to ,,,, home.php so what i did is i made it that when you post you go to home.php?i=$1&a=$a&t=$t but the code dosen't get the info back THE BOTTON LINE IS I WANT TO CLICK POST AND KEEP THE DATA FROM CLERING OUT LIKE A STICKY FORM BUT APERANTLY MY GET STATEMENT DO NOT GET THE DATA THE SECOND TIME(WHEN I CLICK POST) AND I TRY PHP_SELF that did not work :-\

sorry what is rss a web site ????

??? is there any script that pulls score from a central source that you can put in you site??? example nba scores I do not want to insert the scores i want to pick them up from a location and display them on my site Thanks Jay

Hope this clarifies you question on what I am looking for.... /////the code is inside md_challenge.php <?php $i =100; ////the browser shows... /////md_challenge.php?i=&a= /////the second time I click submit i get /////md_challenge.php?i=100&a=TRUE $i= $_GET['i']; $b = $_GET['b']; echo "$i and $b"; print "<form action=md_challenge.php?i=$i&a=TRUE method='post'>"; <select class="newsmall" name="i"> <option value="1">opt1</option></p> <option value="2">opt2</option></p> <option value="3">3</option></p> </select <input type="submit" name="submit" size"20" value="submit">"; } ?>

all my code is inside php FYI

that won't work i still have to click submit twice to get the result my issu is not the diplay is why the first time i click submit I do not get the result post on the browser???

is a ticky form so i use the same page ....md_challenge.php in this case the issue is that the first time click submit .. the browser shows... $i =100; md_challenge.php?i=&a= the secint time i click submit i get md_challenge.php?i=100&a=TRUE

Okay here is my issue , I am weak at the posting and HTML deal... So i have a simple drop down and I want that when i select from the drop down and click submit that my browser will change to website.php?get_something=$i&another=$b the issue is that the first time i click submit it sytays the same ...the second time it changes to the prior selection... I do not want to click twice to get my result any Ideas...here is the code i clean it up a lot but that is the main idea $i= $_GET['i']; $b = $_GET['b']; echo "$i and $b"; print "<form action=md_challenge.php?i=$i&a=TRUE method='post'>"; <select class="newsmall" name="i"> <option value="1">opt1</option></p> <option value="2">opt2</option></p> <option value="3">3</option></p> </select <input type="submit" name="submit" size"20" value="submit">"; }

never mind i got it my mistake wrong testing file :-\ Im stupid

I have a list that comes in during a while statement.. Jane john pete ..ect and i have a hyperlink so i can view their stats... and i want to past their name so i can get it in the next page...but the variable is not comming up ??? <a href=info.php?name=$name>".$name."</a> this is what I get http://localhost/mdo_main/info.php?name=