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i have excuted this script and it does what it should do for the first time but when i click it again it should echo tho because the user does exsit in the db

echo "<b>Sorry you have allready activeted your profile</b>";

 

heres the code

<?php

require_once ('mysqli_connect.php'); 

$userid = 1;
echo $userid;
if(isset($_POST['new'])){

// Query the database:
	$q = "SELECT * FROM construction WHERE user_id = $userid";		
	$r = mysqli_query ($dbc, $q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

	if (@mysqli_num_rows($r) == 1) { // A match was made.

	echo "<b>Sorry you have allready activeted your profile</b>";

	}else{

	$q = "INSERT INTO construction(user_id, complete, hqlevel, progress ) VALUES ($userid,'0000-00-00 00:00:00',1,0)";		
	$r = mysqli_query ($dbc, $q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

	}
	}
?>
<form name="form1" method="post" action="">
  <label>
  <input type="submit" name="new" id="new" value="Activate Space City">
  </label>
</form>

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https://forums.phpfreaks.com/topic/167818-why-is-this-not-working/
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by the looks of your posted code, $userid is always set to 1, so any new user will have the same id... so maybe there are more than 1 rows with that userid, and you are only testing if the record count == 1... maybe try...

 

if (@mysqli_num_rows($r) >= 1) {

I think the problem is that the page must load three times to go through a cycle. The first page load will do nothing except display the form since $_POST['new'] is not set. The second page load, after submitting the form, will add the record. Then the third page load, after again submitting the form, will display the error condition.

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