[SOLVED] What's Wrong with this code?

[Comptons_Eazy_E]

Note: I'm a newbie at PHP/MYSQL just started 3 days ago

 

Anyway Can someone tell me why this code is not working?:

$sql="UPDATE players SET Linked = '1' WHERE Name = 'PN'
VALUES
('$_POST[fname]')";
mysql_query(sql);

[Alex]

You're missing a $ for the variable $sql on this line:

 

mysql_query(sql);

 

You're also using wrong syntax for updating a database. Try:

 

$sql = "UPDATE players SET Linked='1' WHERE Name='PN'";

[Comptons_Eazy_E]

You're missing a $ for the variable $sql on this line:

 

mysql_query(sql);

 

Thanks, but i have tried that. The row doesn't change at all!

[Alex]

You're also not using the correct syntax for updating (I edited my last post already).

[Comptons_Eazy_E]

You're also not using the correct syntax for updating (I edited my last post already).

 

Thanks again, i inserted it and it looked like this:

$sql = "UPDATE players SET Linked='1' WHERE Name='PN'";
VALUES
('$_POST[fname]')";
mysql_query($sql);

 

I received a error:

Parse error: syntax error, unexpected '"' in /www/110mb.com/s/o/u/t/h/w/e/s/southwestrp/htdocs/link.php on line 31

 

Doing a quick study i commented the two lines that look like this:

	VALUES
('$_POST[fname]')";

Then it worked, but the row still didn't change, Any idea why?

[Alex]

Are you sure the column names in the query are correct (they are case sensitive), and that there is actually a record with the Name 'PN'?

[Comptons_Eazy_E]

Are you sure the column names in the query are correct (they are case sensitive), and that there is actually a record with the Name 'PN'?

 

Sorry i don't know alot about PHP, I know more about the pawn language, can please explain how do i make it a record??

[Alex]

A record is just a row in the database, you do have rows already inserted, right?

[Andy-H]

change:

  mysql_query($sql);
      //to
  mysql_query($sql) or trigger_error(mysql_error() . 'on line: ' . __LINE__, E_USER_ERROR);

[Comptons_Eazy_E]

A record is just a row in the database, you do have rows already inserted, right?

 

Yes i do. This?:

By eazy_efolife at 2009-11-20

[Alex]

Based on that screenshot it doesn't look like you have a record with the Name 'PN', so of course no records will be updated.

[Comptons_Eazy_E]

Based on that screenshot it doesn't look like you have a record with the Name 'PN', so of course no records will be updated.

 

oh, I was a bit confused then, also this is what it looks like now:

$sql = "UPDATE players SET Linked='1' WHERE Name='Linked'";
//VALUES
//('$_POST[fname]')";
mysql_query($sql) or trigger_error(mysql_error() . 'on line: ' . __LINE__, 

E_USER_ERROR);

 

Those are commented because everytime i uncomment it i get errors 

[Alex]

Yea, those two lines aren't supposed to be there. When I posted the query before that's all that was supposed to be in it, you can completely remove those lines.

 

(Again) Based on that screenshot all I see is 1 record with the Name 'Eric_Wright', and do you even have a column named 'Linked'? I can't tell from the screenshot.

[Andy-H]

   $sql = "UPDATE players SET Linked='1' WHERE Name='Linked'";
   mysql_query($sql) or trigger_error(mysql_error() . 'on line: ' . __LINE__, E_USER_ERROR);

 

You only have one record in the table and the "Name" filed does not contain the string "Linked" it contains the string "Eric_Wright".

 

   $sql = "UPDATE players SET Linked='1' WHERE Name='Linked'";
   mysql_query($sql) or trigger_error(mysql_error() . 'on line: ' . __LINE__, E_USER_ERROR);

printf("Query 1: Records updated: %d  ", mysql_affected_rows());

   $sql = "UPDATE players SET Linked='1' WHERE Name='Eric_wright'";
   mysql_query($sql) or trigger_error(mysql_error() . 'on line: ' . __LINE__, E_USER_ERROR);

printf("Query 2: Records updated: %d  ", mysql_affected_rows());

[Gayner]

What akon song are u reading?

[Comptons_Eazy_E]

What akon song are u reading?

 

I wanna love you =)

 

   $sql = "UPDATE players SET Linked='1' WHERE Name='Linked'";
   mysql_query($sql) or trigger_error(mysql_error() . 'on line: ' . __LINE__, E_USER_ERROR);

 

You only have one record in the table and the "Name" filed does not contain the string "Linked" it contains the string "Eric_Wright".

 

   $sql = "UPDATE players SET Linked='1' WHERE Name='Linked'";
   mysql_query($sql) or trigger_error(mysql_error() . 'on line: ' . __LINE__, E_USER_ERROR);

printf("Query 1: Records updated: %d  ", mysql_affected_rows());

   $sql = "UPDATE players SET Linked='1' WHERE Name='Eric_wright'";
   mysql_query($sql) or trigger_error(mysql_error() . 'on line: ' . __LINE__, E_USER_ERROR);

printf("Query 2: Records updated: %d  ", mysql_affected_rows());

 

Sorry for the late reply i was busy, anyway Its hard to explain but everybody has there own account, I was testing the database to see if i can the Linked from 0 to 1. People register there account ingame ( SA-MP is the game http://www.sa-mp.com ) And So I want them to link there account online using my website for my game server. So I just don't want it for my name, for the other people too understand? So when you wrote:

WHERE Name='Eric_wright'

I didnt just want it for me only thats why i put %_post from the input box text!

 

edit:

Yes i do have one coll-um  named "Linked"

[Comptons_Eazy_E]

Thank you all for helping out, I have found the problem and it works now! 

[Alex]

By the way, there's a topic solved feature, you can mark your topic as solved with the button at the bottom left of the page.

[Comptons_Eazy_E]

By the way, there's a topic solved feature, you can mark your topic as solved with the button at the bottom left of the page.

 

Thanks do i have to make a topic for a simple question? ( How do i change php echo function into a color? )

[Alex]

That's HTML/CSS, not PHP. Example:

 

echo "<span style='color: red;'>Red Text!</span>";

[Comptons_Eazy_E]

That's HTML/CSS, not PHP. Example:

 

echo "<span style='color: red;'>Red Text!</span>";

 

=)