"Supplied argument is not a valid MySQL result resource"

[jwoods7]

I'm trying to create a function that I can use to pull information from a field in a database.

So far I'm in the testing stage...here's what I have...

function getMemberInfo($login, $field) {
$query="SELECT $field FROM users WHERE login='$login'";
while($row = mysql_fetch_array($query)){
// get information from field:
return $row['$field'];
}
//Didn't find info:
return "empty";
}


The function is called like this:
print(getMemberInfo($_SESSION['login'], 'firstname'));
Where session is a unique login id for each member. This would print the first name of the member that is logged in under the current session ID on the computer.

Why do I get a
Warning: Supplied argument is not a valid MySQL result resource
ERROR AND WHAT DOES THAT MEAN???
Any other ways to do this? (simple please, 4th day learning PHP!!)

Thanks in advance

 

[DRJ478]

Hi!

Why do I get a
Warning: Supplied argument is not a valid MySQL result resource
ERROR AND WHAT DOES THAT MEAN???

1. When you send your SQL query you expect a result. The result is provided as a 'resource'. When there are no rows or the SQL statement fails for some reason there is no result. Trying to retreive rows from such a query produces the error.

What else?
2. You actually never sent the query to the database server, all you did is assign it to a variable called query.
To get a result identifier you need to use mysql_query($query); Also check if the SQL fails or not:

$SQL = "SELECT $field FROM mytable WHERE mycolumn='$myvar'";
$result = mysql_query($sql) OR DIE(mysql_error()); # prints error if fail
# for fun look how many rows were returned
echo(mysql_num_rows($result)." were found");

Good luck in the PHP world!

 

[jwoods7]

Thanks!


Why am I getting "Query was Empty" as a result from this function? The values are set in the database, I can verify that...I'm trying to get a firstname of a member

CALLED FROM:
getMemberInfo('3','firstname')
{verified "firstname" is exactly as in table, login #3 has firstname value....}

function getMemberInfo($login, $field) {
$query="SELECT * FROM users WHERE login = '$login'";
$result = mysql_query($sql) OR DIE(mysql_error());
echo(mysql_num_rows($result)." were found"
while($row = mysql_fetch_array($result)){

return $row['$field'];
}
}

 

[sleipnir214]

That error is probably because you've put the query in a variable called $query, but you pass mysql_query() a variable called $sql.

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TANSTAAFL!

 

[jwoods7]

oh geez I'm retarded. Should be "checking the inline specs on the girders" like Tommy Boy.
(quote from movie Tommy Boy)


Thanks for the help!