php - Using post in define -

<?php $rootdir = "digiadmin"; //leave if no folder define("config","true"); //    database config    // define('db_host', $_server['server_name']); define("db_user", $_post['db_user']); define("db_pass", $_post['db_password']); define("db_name", $_post['db_name']); 

i have form ask database name, username , password. when click on submit button show me error. have defined config , made function checking config true or false. using post username, password , database name. doing in correct way?

ok, based on comments i'd try answer question.

this should php script db.php (the mysql_* methods deprecated, that's story day keep simple):

<?php define('db_host', $_server['server_name']); define("db_user", $_post['db_user']); define("db_pass", $_post['db_password']); define("db_name", $_post['db_name']);  $link = mysql_connect(db_host, db_user, db_password);  if (!$link) {     die('could not connect: ' . mysql_error()); }  $db_selected = mysql_select_db(db_name, $link);  if (!$db_selected) {     die ("can't use db: " . mysql_error()); } ?> 

and might html page index.html:

<form action="db.php" method="post">  user: <input type="text" name="db_user"><br>  pass: <input type="text" name="db_password"><br>  db: <input type="text" name="db_name"><br>  <input type="submit" value="submit"> </form> 

according question, errors somewhere, db.php die corresponding message. if so, can try this:

<?php $link = mysql_connect("host", "user", "password");  if (!$link) {     die('could not connect: ' . mysql_error()); }  $db_selected = mysql_select_db("db", $link);  if (!$db_selected) {     die ("can't use db: " . mysql_error()); } ?> 

which means not using posted variables strings know work.

if second script fails, data (user, password and/or database) incorrect. if second script succeeds, there may wrong form post.