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mr msg

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  1. Hi, When you save the visual basic project as an exe file is there a way where when the exe file opened it just executes (opens) the file (software/program created) instead of installing the software/program onto the computer. Can you let me know as soon as possible as I need to save and do it as soon as please. Mr MSG
  2. thanks all, I completed the code in the end.
  3. mmmmm I did try it but nothing getting done right the code is not repeating and sometimes causes errors. i have a seperate program that updates the value of the field in a table from "1" to "0" all the time when someone logs in and out so its allways getting updated i need php code to get this ALL the time, can someone write the code giving indication what to do, been on this for 3 days got 2 days to finish this if you could would be much appriciated...
  4. oh im not sure on any of these - mysql_fetch_array() mysql_num_rows() mysql_affected_rows() i tired to add it to the code but nothing happens or just one image would appear and they are not responding to the actual databse bieng updating. i am a bit of a NOOB unfortunantly.
  5. oh okay thanks i shal read upon it.
  6. Hi PFMaBiSmAd, oh rite so what I am trying to do is, I have made a visual basic program were when the member log s in it sends a value "1" to the mysql n wen the member logs out it sends a value "0" to the mysql. So im using this php and mysql to connect to databse so when user log s in, the php will pik up on the databse and the value wil be"1" it will show a red image in the php, then when user logs out it that value in mysql change to "0" then the colours goes green in the php, this is what im trting to do but not working this is the code I got so far, please let me know. i dont think the "mysql_num_rows()" will work for what I am trying to do. <?php $con = mysql_connect("*****","*****","*****"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("table"); $query = mysql_query("SELECT * FROM 'people' WHERE 'Access'='1'"); if ($query) { echo'<img src="red.jpg">'; } else { echo'<img src="green.jpg">'; } ?>
  7. I also forgot to add that When I run this query it only displays one output either "Access" or "No Access". When I change the value to "0" it still stays the same output as before. so the query does work it just doesn't show either option when the value is changed to either "1" or "0", if you get me, thanks.
  8. abdbuet , I also tried the quotes but It didnt work
  9. Hi ix.selken, this is the code: (I am using Phpmyadmin for mysql database and dreamweaver to make the php in) <?php $con = mysql_connect("********","********","********"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("names", $con); $result = mysql_query("SELECT * FROM People WHERE Access='1'"); if ($result) { echo"Access"; } else { echo"No Access"; } ?> Hi abdbuet oh rite thats cool but I need an IF statement in there as I need 2 optional outcomes, thats the thing. I just need it to say if value in the "Access" colum equlas "1" then it will "Access" the user if not "No Access", Thanks.
  10. Hi I have done this IF statment using PHP and MYSQL so far and its not showing the results I want its 1:18am and I need to get it done now! can anybody help, please.. $result = mysql_query("SELECT * FROM People WHERE Access='1'"); if ($result) { echo"Access"; } else { echo"No Access"; } ?> :shrug:
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