mysql error

Anon

hello,
i've been stuck trying to debug an SQL error i keep getting
ERROR 1064 you have an error in your SQL syntax near 'WHERE custid = 'scotty" at line 1

i've looked in my code but cant get rid of it.
can anyone please help as i've been stuck on this for 2 hours

the link is http://netlab.csd.anglia.ac.uk/~csd99rwp/project/userman_pr.php

you can view the records but when you VIEW them and try to change the details the error appears
many thanks

Anon

Iv'e spotted your error straight away...
ERROR 1064 you have an error in your SQL syntax near 'WHERE custid = 'scotty" at line 1

custid = 'scotty" <-- there shouldn't be a double quote at the end there. It should be a single quote.

Regards, Stezz.

Anon

this is my line of sql
$query = "UPDATE $cust_tablename SET $field_str WHERE custid = '$custid'";

as far as i can see i have got equal quotes

kevinsequeira

why dont you print out the query to see what input goes into the database.
that usually helps

Anon

OK kevin, i'm a little lost as to what you mean, could you please break that down a little.

thanks

Anon

Okay.. now Iv'e seen the full query, I know what the problem is.

The basic syntax for an UPDATE query is as follows...

UPDATE (tablename) SET (fieldname) = (value) WHERE (conditions)

In the query you showed in one of the previous messages, you're doing this...

UPDATE (value) SET (value) WHERE (conditions)

Your probelm is that you've put a $ before your table name and done the same with the field name.

Try this...

$query = "UPDATE cust_tablename SET field_str = $a_value WHERE custid = '$custid'";

Hope it helps.

Regards, Stezz.

Anon

i'm still stuck, anymore ideas??

kevinsequeira

sorry if i was a bit cryptic, i just meant print($query) to see what the query looks like with all the $ variables filled in

Anon

Are you trying to dynamicly specify the table name and field name here??
If so your query should look like

$query = "UPDATE $cust_tablename SET $field_str = $value WHERE custid = '$custid'";

Otherwise the update query should look like

$query = "UPDATE tablename SET fieldname = $value WHERE custid = '$custid';

either way you are missing either the fieldname part or the value part and the equal sign.

Do an echo or print of $query after you have assigned it to see what the query is evaluating to after variable substitution.

Greg