Reputation: 1177
Ok, so I have some MySQL tables as follows:
Buildings
Building-ID Building-Name
===========----=============
1 Building-1
2 Building-2
3 Building-3
4 Building-4
Building-1
Mroom State
=====----======
1 Booked
2 Empty
3 Empty
4 Empty
Building-2
Mroom State
=====----======
1 Booked
2 Empty
3 Empty
4 Empty
And a query in PHP as follows (Ignore the hard coded while, I've simplified the code a bit):
$sql = "select * from Buildings";
$result = mysql_query ($sql) or die(mysql_error());
while ($row = mysql_fetch_array($result))
{
$building[] = $row['ward_name'];
}
$v1 = 0;
while ($v1 < 4)
{
$sql = "SELECT COUNT(*) FROM `$building[$v1]` WHERE state = 'Empty'";
$result = mysql_query($sql) or die(mysql_error());
$count = mysql_result($result, 00);
var_dump($count[$v1]);
$v1 = $v1 + 1;
}
To my way of thinking this should create an array of the buildings contained in the "Buildings" table, start a loop, load the building name from the array and provide a row count for the table of how many rows contain "Empty" in the state column. What it actually does is provide a count for the first table and then provides "NULL" for the rest.
I'd appreciate any help you can give me.
Cheers!
Upvotes: 0
Views: 1354
Reputation: 73
What about changing your data model?
Table buldings can be kept as is:
Buildings
Building-ID Building-Name
===========----=============
1 Building-1
2 Building-2
3 Building-3
4 Building-4
New table:
Rooms
Building-ID Mroom State
===========-=====-=====
1 1 1
1 2 0
2 1 0
State 0 = Empty, State 1 = Booked
Then use a join with group by:
select count(*) from buildings b inner join rooms r on r.bid = b.id where r.state = 0 group by b.id;
Then you will get a row for each building with the count of empty rooms. You won't need a table for each building.
Upvotes: 1
Reputation: 2332
mysql_result() returns a string, not an array. Modify the code and check that now it works as expected.
var_dump($count);
Upvotes: 0
Reputation: 5868
This does noit make sense:
$count = mysql_result($result, 00);
var_dump($count[$v1]);
you mean to write:
$count[$v1] = mysql_result($result, 00);
var_dump($count[$v1]);
Also do not use several tables with names matching columns of other tables. You can use one table with a primary key that spans two columns instead, for example create primary key on($buildingid,$roomid) so that the table has columns $buildingid,$roomid, and $state.
Upvotes: 0