SQL syntax error: unknown column - mysql

Question

I'm trying to run this query:
$result = db_query("INSERT INTO `timesheets` (clientid, candid, weekending, department, orderno, basicpay, basiccharge, otpay, otcharge, ot2pay, ot2charge, status, hue, huc) VALUES (`$client`, `$cand`, `$week_ending`, `$department`, `$order_no`, `$basic_pay`, `$basic_charge`, `$ot_pay`, `$ot_charge`, `$ot2_pay`, `$ot2_charge`, `$status`, `$hue`, `$huc`)");
if($result){
print 'Success! ID of last inserted record is';
}
else {
die('Error : ' . db_error());
}
These are the values from the form
$client = db_quote($_POST['client']);
$cand = db_quote($_POST['cand']);
$order_no = db_quote($_POST['order_no']);
$department = db_quote($_POST['department']);
$week_ending = db_quote($_POST['week_ending']);
$basic_pay = db_quote($_POST['basic_pay']);
$hue = db_quote($_POST['hue']);
$basic_charge = db_quote($_POST['basic_charge']);
$huc = db_quote($_POST['huc']);
$ot_pay = db_quote($_POST['ot1_pay']);
$ot_charge = db_quote($_POST['ot1_charge']);
$ot2_pay = db_quote($_POST['ot2_pay']);
$ot2_charge = db_quote($_POST['ot2_charge']);
$status = 'cand';
The value of $client in this instance is "725". The error I am getting is Error : Unknown column ''725'' in 'field list'.
How can this be - the column name in the timesheets table is clientid. I'm trying to put the value 725 into the clientid column. Is there a syntax error somewhere? The error message doesn't seem to make sense?

Answer 1

When setting the values in MySQL you can't use the backtick character to denote a field, you'll need to use either ' or " to enclose your variables.
Use of a back-tick is for naming a field. It's a quirk of MySQL.
Your query line should be:
$result = db_query("INSERT INTO `timesheets` (clientid, candid, weekending, department, orderno, basicpay, basiccharge, otpay, otcharge, ot2pay, ot2charge, status, hue, huc) VALUES ('$client', '$cand', '$week_ending', '$department', '$order_no', '$basic_pay', '$basic_charge', '$ot_pay', '$ot_charge', '$ot2_pay', '$ot2_charge', '$status', '$hue', '$huc')");

Answer 2

The correct query would be like below one:
"INSERT INTO 'timesheets' (clientid, candid, weekending, department, orderno, basicpay, basiccharge, otpay, otcharge, ot2pay, ot2charge, status, hue, huc)
VALUES ('$client', '$cand', '$week_ending', '$department', '$order_no', '$basic_pay', '$basic_charge', '$ot_pay', '$ot_charge', '$ot2_pay', '$ot2_charge', '$status', '$hue', '$huc')

Answer 3

Please pay attention to the quotation marks during the creation of the query.
The INSERT statement should something like that:
INSERT INTO table_name (col_name_1, col_name_2, col_name_3) VALUES('$param_1', '$param_2', '$param_3')
Note that the table_name is written without quotation marks, while each value is delimited by a single quotes.
You can find the official documentation here and a full example here.
By the way, the query for your code should be something like that:
$sqlQuery = "INSERT INTO timesheets (clientid, candid, weekending, department, orderno,basicpay, basiccharge, otpay, otcharge, ot2pay, ot2charge, status, hue, huc) VALUES ('$client', '$cand', '$week_ending', '$department', '$order_no', '$basic_pay', '$basic_charge', '$ot_pay', '$ot_charge', '$ot2_pay', '$ot2_charge', '$status', '$hue', '$huc')";

Answer 4

Try this way
$result = db_query("INSERT INTO `timesheets` (clientid, candid, weekending, department, orderno, basicpay, basiccharge, otpay, otcharge, ot2pay, ot2charge, status, hue, huc) VALUES ('".$client."', '".$cand."', '".$week_ending."', '".$department."', '".$order_no."', '".$basic_pay."', '".$basic_charge."', '".$ot_pay."', '".$ot_charge."', '".$ot2_pay."', '".$ot2_charge."', '".$status."', '".$hue."', '".$huc."')");