select statement query

susi

hi..

was wondering if anyone can help me..

i have a login system using id which works... but when the person is logged in.. i want to display the information from the database which is relevant to that id..

i have

$query = "SELECT forename, surname, addL1, addL2, postalCode, county FROM student where id =[post_$id]" ;

while($row = @ mysql_fetch_array($result))
{
echo "\n<tr>\n<tg bgcolor=\"silver\">" .
"" .
"Forename:" . " " . $row["forename"] . " " . " " .
"Surname:" . $row["surname"] . " " . " " .
"Address Line1:" . $row
["addL1"] . " " . " " .
"Address Line2:" . $row
["addL2"] . " " . " " .
"Postal Code:" . $row
["postalCode"] . " " . " " .
"County:" . $row["county"] . " " .
"</td>\n<tr>";

echo "\n<tr>\n\t<td></td>\n</tr>";

}

Any suggestions??
Thanks,
Susi

mmilano

are you getting any errors? what is happening when you try to execute that code?

take the '@' sign out and then tell us what errors you are getting.

susi

i get no errors...
just blank page

i tried calling the information by using

$query = "SELECT forename, surname, addL1, addL2, postalCode, county FROM student where id = '50326595'";

and it displayed the information... so i think da problem is in the where id bit..

ultraslacker

You are suppressing the error output with @ - remove it or just check the logs.

susi

i tried that... still get no errors..
??

help???

dartcol

$query = "SELECT forename, surname, addL1, addL2, postalCode, county FROM student where id = " . $_POST['id'];

$result = mysql_query($query) or die(mysql_error());
while($row =  mysql_fetch_array($result))
{

Few problems. Use $_POST['']. Always run a ... or die(mysql_error()) .. after doing a mysql_query() when testing - remove when putting the page up. And don't forget to do the mysql_query! 🙂

susi

hi..

very silly of me to forget to do the mysql query.. thanks for that..
unfortunatly it is still not working...

i am totally baffled as to what im doing wrong... going crazy! 🙂

what i hav now is :-

$query = "SELECT forename, surname, addL1, addL2, postalCode, county FROM student where id =" . $_POST['id'];

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

while($row = mysql_fetch_array($result))
{
echo "\n<tr>\n<tg bgcolor=\"silver\">" .
"" .
"Forename:" . " " . $row["forename"] . " " . " " .
"Surname:" . $row["surname"] . " " . " " .
"Address Line1:" . $row["addL1"] . " " . " " .
"Address Line2:" . $row["addL2"] . " " . " " .
"Postal Code:" . $row["postalCode"] . " " . " " .
"County:" . $row["county"] . " " .
"</td>\n<tr>";

echo "\n<tr>\n\t<td></td>\n</tr>";

}

any ideas at all???!!
thanks,
susi

dartcol

Do you get an error message? It works fine here. What do you see when the page tries to load?

susi

Thats strange.... i get the following error.

Father's Information is as follows:-

You have an error in your SQL syntax near '' at line 1

Its fine when i use an actual id in the select statement.. prints it out fine..??

tekky

sounds like you arent passing the proper $_POST data...

after your $query = line....

do echo "query: $query\n";

and show us what that comes out with

susi

I finally got it working....
it was a silly mistake again...

ill have to try and be more careful!

thank you so much for all your help! 🙂