Why does this not print anything?

[Caffeinenyx]

$query ="SELECT oneID FROM table WHERE table.PersonID = 'game.PlayerA'" ;
$result = mysql_query($query);
$row = mysql_fetch_array($result);
$oneID = $row[0];

[code]

If I then echo "$oneID" why does it not print anything?
$result echos resource7

[Muddy_Funster]

what does

print_r($row);

give you?

[Caffeinenyx]

nothing

[fugix]

try puting this in to see if you receive any errors in your query

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

[salathe]

SELECT oneID FROM table WHERE table.PersonID = 'game.PlayerA'

 

Do you really have a table named table and a row with the PersonID column containing the literal characters game.PlayerA? Should game.PlayerA reference a column from a game database?

[Caffeinenyx]

yes, although the names are not the same as the ones im using, but as this is part of an assignment I remained the tables to post here. Player A is a column name in game.

[Caffeinenyx]

no errors

[PFMaBiSmAd]

Are you getting a completely blank page when you tried the suggestions? If so, what does a 'view source' of the page in your browser show?

[fugix]

i think that it might be because you are referencing the table "game" without selecting the fields from it.

[Caffeinenyx]

SELECT oneID FROM table WHERE table.PersonID = game.PlayerA

 

returns unknown column game.PlayerA

 

but it does exist.

 

Not a blank page, just this value missing, but it may have been the double quotes. Im fairly new to php and have never used table.anything (its a good idea)

[cyberRobot]

To use column names from another table, you'll need to use a MySQL Join.

http://dev.mysql.com/doc/refman/5.0/en/join.html

 

 

You need to join "table" and "game".

[fugix]

yeah, you cant reference a column that is not in the table you specified without joining the tables

[fugix]

SELECT column_name(s)
FROM table_name1
FULL JOIN table_name2
ON table_name1.column_name=table_name2.column_name

[fugix]

so you would have

$query ="SELECT oneID FROM table FULL JOIN games ON table.PersonID = game.PlayerA" ;

[Caffeinenyx]

Unknown column 'table.PersonID' in 'on clause'

 

?? It is correct i can view the table

[fugix]

try

$query ="SELECT oneID FROM table FULL JOIN game ON t.PersonID = g.PlayerA" ;

[cyberRobot]

You need to define the table aliases before using them. So "t." and "g." won't work as the query is currently set up.

[Caffeinenyx]

I have got it now, i did need to define them as well, Thanks everyone!