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";
?>

我是PHP的新手，我收到了这个SQL错误：

查询为空。

我做错了什么？

Answer 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)
{

是正确的。你试图做

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

Answer 2

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";
}
?>

Answer 3

在声明字符串查询变量时，不要只使用$sql = mysql_query(...); $sql = "...";，它应该可以正常工作

SQL错误：查询为空

3 个答案: