SQL error: query empty

Question

mysql_select_db("musicDB", $con);

$sql = mysql_query("INSERT INTO Users (username, fname, lname, email, dob, password, occupation, genre )
VALUES ('$_POST[username]', '$_POST[fname]', '$_POST[lname]', '$_POST[email]', '$_POST[dob]', sha1('$_POST[password]'), '$_POST[occupation]', '$_POST[genre]')");

if (!mysql_query($sql,$con))
{
    die('Error: ' . mysql_error());
}
echo "1 record added";
?>

I am new to PHP and I am getting this SQL error:

The query is empty.

What am I doing wrong?

Welcome SQLI... read about SQL Injection

genesis
– genesis

2011-09-03 09:47:34 +00:00
Commented Sep 3, 2011 at 9:47 — genesis
– genesis, Commented Sep 3, 2011 at 9:47

genesis · Accepted Answer · 2011-09-03 09:48:18Z

1

$sql = mysql_query("INSERT INTO Users (username, fname, lname, email, dob, password, occupation, genre )
VALUES ('$_POST[username]', '$_POST[fname]', '$_POST[lname]', '$_POST[email]', '$_POST[dob]', sha1('$_POST[password]'), '$_POST[occupation]', '$_POST[genre]')");

if (!sql)
{

is correct one. You were trying to do

if(!mysql_query(mysql_query("....")))
{

answered Sep 3, 2011 at 9:48

genesis

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Comments

Your Common Sense · Accepted Answer · 2011-09-03 10:07:20Z

1

mysql_select_db("musicDB", $con);

// that's dirty but at least something to protect that silly code
$_POST['password'] = sha1($_POST['password'].$_POST['username']);
foreach($_POST as $key => $value) $_POST[$key] = mysql_real_escape_string($value);

$sql = "INSERT INTO Users (username, fname, lname, email, dob, password, occupation, genre )
        VALUES ('$_POST[username]', '$_POST[fname]', '$_POST[lname]', '$_POST[email]',
                '$_POST[dob]', '$_POST[password]', '$_POST[occupation]', '$_POST[genre]')");

if (!mysql_query($sql,$con))
{
    trigger_error(mysql_error()." ".$sql);
} else {
    echo "1 record added";
}
?>

edited Sep 3, 2011 at 10:07

answered Sep 3, 2011 at 9:52

Your Common Sense

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Comments

genesis · Accepted Answer · 2011-09-03 09:49:53Z

0

when declaring the string query variable, dont use $sql = mysql_query(...); just simply $sql = "..."; and it should work fine

edited Sep 3, 2011 at 9:49

genesis

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answered Sep 3, 2011 at 9:44

mgabor

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