Question

我一直在为这个php类工作。我无法相信我会得到左右错误。现在我收到了这个错误：

未选择数据库表连接错误

但我检查一下，一切都在我的桌子上，这是我的代码：

if ($_POST['submit']!=""){
if ($_POST['username']==""||$_POST['password1']==""||$_POST['password2']==""||$_POST['firstname']==""||$_POST['lastname']==""||$_POST['address']==""||$_POST['email']==""||$_POST['city']==""||$_POST['state']==""||$_POST['zip']==""||$_POST['phone']=="");
$error=1;
  }
        else if ($_POST['password1']!=$_POST['password2']){
    $error=2;
   }
else{
$hostname="localhost";
$database="Contacts";
$mysql_login="Web_User";
$mysql_password="my1230";

if (!($db = mysql_connect($hostname, $mysql_login , $mysql_password))){
    echo "error on connect";
}
   else{
  if (!(mysql_select_db($databse,$db))){
    echo mysql_error();
    echo "<br>error on table connection";
}
   else{
    $SQL="Insert into tblUsers(username,password,firstname,lastname,email,address,city,state,zip, phone,signupDate)values)'".$_POST['username']."',PASSWORD('".$_POST['password1']."'),'".$_POST['firstname']."','".$_POST['lastname']."','".$_POST['address']."','".$_POST['city']."','".$_POST['state']."','".$_POST['zip']."','".$_POST['phone']."',NOW())";
    mysql_query($SQL);
    if (is_numeric(mysql_insert_id())){
        header("Location:member-content.php?name=".$_POST['username']);
    }
    else{
        echo "Sorry, there was an errot.Please try again ot contact the administrator";
    }
    mysql_close($db);//closeing out connection,done for now
  }
}

}

＆GT;

我一直坐在这里，当我开始工作时，接下来的事情就不起作用了......

Answer 1

您在此行中拼错了数据库：mysql_select_db($databse,$db)

Answer 2

改变：

if (!(mysql_select_db($databse,$db))){

by：

if (!(mysql_select_db($database,$db))){

php的形式错误

2 个答案: