Variable to hold IP address

Anon

I have to pass an IP address as a variable in one of my SQL statements.

When i type the IP address in directly 199.222.333.444 it works perfectly.

$result = mysql_query("SELECT * FROM VOTES WHERE IPNUMBER = '199.222.333.444'",$db)

but when i try to use a variable it wont work...

$result = mysql_query("SELECT * FROM VOTES WHERE IPNUMBER = '$domain'",$db);

I've tried every combination of single and double quotes around $domain....i've even compared the variable $domain directly to the 199.222.333.444 and it returned that they were equal.

I've spent 3 days at this and i have no idea what more i can do. I got the IP address using the following call which seems to work fine and when i echo it to the screen it is correct.

$domain = GetHostByName ($REMOTE_ADDR);

i'd be very grateful for any help or ideas on how they might be different...could there be an invisible character somewhere that i'm missing?

Thanks.

kparker

There's two things you can do here to help yourself--yes, you can correctly infer that I don't see anything immediately obvious that might wrong with what you're doing. :-(

(1) Instead of just sending the literal query string, put it into a variable and echo that first. This lets you see exactly what the query is.

(2) If the query fails, be sure to print the value of mysql_error() which is often quite informative as to what went wrong.

Oh, wait! Something did just come to mind: are you doing this inside a function? If so, perhaps you forgot to declare $domain as a global.

cordex

Or try echoing "SELECT * FROM VOTES WHERE IPNUMBER = '$domain'" where you execute this query. See if $domain has a value.

-Ben

Anon

thanks for your suggestions....unfortunately i've already tried printing it to screen and im not using a function.

cordex

have you tried trim() on it? It's possible you have a trailing or leading space.

Anon

yes when i echo "select * from.....$domain"

$domain does have the correct value...so why isnt it accepted as correct sql

Anon

thanks for that suggestion....do you know how to use trim()?

i'll look it up now...thanks!

Anon

I've tried using trim() but to no avail it seems....

I've echoed the select statement to the screen below and its correct...but it isnt valid???

SELECT * FROM VOTES WHERE IPNUMBER = 199.106.172.237

Warning: Supplied argument is not a valid MySQL result resource in /export/elections/login.php on line 101

Anon

I may be for one reason or another you can't use a variable inside a string in the function call. Try separating it out...

$sSQL = "SELECT * FROM VOTES WHERE IPNUMBER = '".$domain."'";

$result = mysql_query($sSQL,$db);

Drew

kparker

I've echoed the select statement to the screen below and its
correct...but it isnt valid???

Did you echo it "manually" or did you do it like this:

  $sql = "SELECT * FROM VOTES WHERE IPNUMBER = '$domain'";
  echo "Sending query: $sql";

  $result = mysql_query( $sql ,$db);

And when it failed (that's the significance of the "not a valid resource" message you get later) did you print the value of mysql_error?

  if (! $result)
    echo "Can't do it: " . mysql_error();

And if so, what did the mysql error message say?

Anon

that worked...thanks so much!