[SOLVED] How to use "select from" to store a result

[chinwan]

I have two tables:

 

One is 'families', the other is 'buildings'.

 

In the families, there is a field called 'buildings' that relates to entries in the 'buildingid' table. As well as a field called 'phonenumber'.

 

What code would I use to pull up the the phone number of the familes when I am viewing a particular building.

 

I have the first part:

"SELECT phonenumber FROM families WHERE buildingid='"$building";

But how do I tell it to store the results.

 

Thanks

 

[bruce080]

I think you might need to provide more information.  If I am reading your post correctly, you want to display all the phone numbers that correspond to a particular building from the 'families' table.  Try the following sql statement if that is correct.

 

SELECT phonenumber FROM families WHERE buildings='<building name>';

[chinwan]

The result will only be one phone number.

How do I store that result, so I can pull it it up to display it?

 

Thanks

[bruce080]

Again, you are still being slightly vague, but if there will only be one result, this will work.

 

$result = mysql_query("SELECT phonenumber FROM families WHERE buildingid='"$building";");
while($row = mysql_fetch_array($result))
{
$number=$row['phonenumber'];
}
echo "$number";

[bruce080]

There was a slight error with the quotes I used in the query statement of my code.  It is fixed below.

 

$result = mysql_query("SELECT phonenumber FROM families WHERE buildingid='$building';");
while($row = mysql_fetch_array($result))
{
   $number=$row['phonenumber'];
}
echo "$number";

[chinwan]

I tired that but now I am getting:

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

 

 

Thanks

[bruce080]

Okay, this error means that your SQL query is failing.  Try the following code.  The "or die(mysql_error())" will tell us what the problem is.

 

$sql = "SELECT phonenumber FROM families WHERE buildingid='$building';";
$result = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
     $number=$row['phonenumber'];
}
echo "$number";

[chinwan]

Okay, I got that fixed.

 

How would I have it set up, if I would want it that instead of just pulling up 'phonenumber', it should pull up 'phonenumber' 'contactname' and  'emailaddress'.

 

How would I structure the select command?

 

$sql = "SELECT ???????? FROM families WHERE buildingid='$building';";

 

 

Thanks

[bruce080]

$sql = "SELECT phonenumber, contactname, emailaddress FROM families WHERE buildingid='$building';"

[chinwan]

Got it. Thanks a million.