compare submitted value to a table

[socializer]

hope someone can help.

i have a mysql table that includes blocked numbers -

$blockednum

i would like to compare a value submitted in a form and do execute a command only if the number submitted does not appear in the blocked numbers list.

hope i made myself clear.

thanks

[aalicki]

Well, assuming blocked numbers means a set of numbers that you do not want (or, for instance, blocked from a phone?)

$get_all_nums = mysql_query("SELECT * FROM `BLOCKEDNUMBERS` WHERE `$blockednum` = '$row[name]' ASC");
while ($allnums = mysql_fetch_array($get_all_nums)){
 
 echo "$allnums <br />";
 
}

[socializer]

thanks,


i was able to print out all blockednum with the following code

$result = mysql_query("SELECT * FROM sample")
or die(mysql_error());

$row = mysql_fetch_array($result) or die(mysql_error());

while($row = mysql_fetch_array($result)){


echo $row['blockednum'];
echo "<br />";
}


what i am trying to achieve is to compare a variable called $number that is submitted in a form to the blocked numbers.

if the $number variable appears in the list of blocked numbers i would like the code to execute something like echo "the number is blocked" and at that point to discontinue executing the rest of the code.

if the $number does not match the list of blockednum's i would like the code to continue.

thanks

[r937]

would you kindly run this query and show us the results

SHOW CREATE TABLE sample

[socializer]

hi

i was able to solve the problem with the following

$result = mysql_query("SELECT * FROM sample where blocked =
'$number'");
if(! $result) //if error in SQL display the error
die(mysql_error()); //do not proceed beyond this point

//see if we got any results
if(mysql_num_rows($result) > 0) {

//get phone number info to show below
$row = mysql_fetch_array($result);

echo "The number - " .$row['blocked']. " - has been blocked";


thanks!!!