output a row?

[Guest]

How do I got about outputing one row that can only hold one value? Any reason why the function below doesn't work?

 

<?php
include "database.php";
session_start();
$name = $_session['outname'];

function output_wins(){
$sql = 'SELECT * FROM account_info WHERE username ="' . $name .'"';
$result = mysql_query($sql);
$wins = mysql_fetch_array($result);
echo $wins['win'];
}



output_wins();

?>

 

Is there an easier way I can output this one row?

[litebearer]

Variables do NOT automatically pass to functions.

 

Try...

$name = $_session['outname'];
function output_wins($name){
  $sql = "SELECT * FROM account_info WHERE username = '$name'";
  $result = mysql_query($sql);
  $wins = mysql_fetch_array($result);
  return $wins['win'];
}
echo output_wins($name);

[Guest]

Ehh, it seems to work but it's not outputting anything. It's blank.

[Guest]

Ah, here's the error:

mysql_fetch_array(): supplied argument is not a valid MySQL result resource

[BlueSkyIS]

change the mysql_query line to this:

 

$result = mysql_query($sql) or die(mysql_error());

[Guest]

change the mysql_query line to this:

 

$result = mysql_query($sql) or die(mysql_error());

I did and this is the error:

mysql_fetch_array(): supplied argument is not a valid MySQL result resource

 

[BlueSkyIS]

can you post your latest code?

[Guest]

<?php
include "database.php";
session_start();

$name = $_session['outname'];
function output_wins($name){
  $sql = "SELECT * FROM account_info WHERE username = '$name'";
  $result = mysql_query($sql) or die(mysql_error());
  $wins = mysql_fetch_array($result);
  return $wins['win'];
}



?>

[BlueSkyIS]

mysql_fetch_array(): supplied argument is not a valid MySQL result resource

 

on line 8, correct?

 

I guess we need to look at the sql. try this or something similar:

 

function output_wins($name){
  $sql = "SELECT * FROM account_info WHERE username = '$name'";

echo "sql: $sql <br />";
  $result = mysql_query($sql) or die(mysql_error());
echo "result: ".$result."<br />";
  $wins = mysql_fetch_array($result);
echo "after mysql_fetch_array <br />";
  return $wins['win'];
}

[Guest]

Edit:

 

got this:

 

sql: SELECT * FROM account_info WHERE username = "tony"

result: Resource id #3

after mysql_fetch_array

0

[Guest]

Everything is working great now! I figured out what happen!

 

Thanks!