Another Parse error (unexpected T_VARIABLE) [SOLVED]

[James142]

After about 30 minutes (no exaggeration) of trying to fix this error, I am still coming up empty handed. Hopefully someone here can help me...

When I use this code:

<?php $getid = $_GET['user']
$rs = mysql_query("select 'name','age','country' from users where id='$getid' LIMIT 1"); <--- THIS IS LINE 9
list($name, $age, country) = mysql_fetch_row($rs);    ?>

I get the following message: Parse error: syntax error, unexpected T_VARIABLE in "bla bla bla" on line 9. (line 9 is the second line in the code above)

I do know that the error means that I am missing one of these characters: ' ; ' " ) ( [ ] ...But I dont know which one(s) I am missing and where they belong..

Perhaps someone could point out what I have done wrong?

Kindest Regards,

James

[mark1million]

I am no expert but have you tried `1`

so, LIMIT `1`"

Also shouldn't the select be SELECT and also where be WHERE

[James142]

I am no expert but have you tried `1`

so, LIMIT `1`"

Also shouldn't the select be SELECT and also where be WHERE

Thanks Mark but I'm still getting the same error

[mark1million]

Have you tried in phpmyadmin? then creating the php code?

[James142]

Have you tried in phpmyadmin? then creating the php code?

Do you mean create the code using phpmyadmin?

[Strider]

If I am not mistaken in the last line, shouldn't country have the dollar sign in front too? Also the first line is not ended you forgot to put the ';' at the end of $_GET['user'].

[James142]

Would you look at that, it works!

Thanks guys!

<?php $getid = $_GET['user'];
$rs = mysql_query("SELECT 'name','age','country' FROM users WHERE id='$getid' LIMIT 1"); 
list($name, $age, $country) = mysql_fetch_row($rs);    ?>

Kindest Regards,

James

[Nabeel]

BTW you should do

$getid = mysql_real_escape_string($_GET['user']);
$rs = mysql_query("SELECT 'name','age','country' FROM users WHERE id='$getid' LIMIT 1"); 
list($name, $age, $country) = mysql_fetch_row($rs);

Otherwise, you're ripe for SQL injection attack.

[James142]

BTW you should do

$getid = mysql_real_escape_string($_GET['user']);
$rs = mysql_query("SELECT 'name','age','country' FROM users WHERE id='$getid' LIMIT 1"); 
list($name, $age, $country) = mysql_fetch_row($rs);

Otherwise, you're ripe for SQL injection attack.

Thanks Nabeel!