Form issue

[petenyce102]

Hello
Iam new to this forum and hope i can get this awnsered. I have a form that lets users submit data online. Now i check my sql database to see if they arent submitting a duplicate username.

How can i get the error message to appear on the form? You have this done when you register for this forum. I need to connect to the database and if there is a error display it on the form??? Below is a simple form can i have an example with this form where and how would i add the php to A check the database field(userid) and 😎 Submit the form into the database.
C) to display a error message on the form if duplicate userid. I think i have to use php echo self

thanks
pete

Form:

<meta http-equiv="Content-Language" content="en-us">
<form method="POST" action="--WEBBOT-SELF--">
<p>Name: <input type="text" name="T1" size="20"></p>
<p>Username: <input type="text" name="T3" size="20"></p>
<p><input type="submit" value="Submit" name="B1"><input type="reset" value="Reset" name="B2"></p>
</form>

[travelbuff]

Hi Pete:

What you are wanting to do is fairly common and easy but judging by your post a bit more complicated than you think.

Here are the basic steps involved:

1. check and see if the form has been submitted or if the form needs to be displayed;
2. make sure there is valid data in the form fields;
3. check the DB to make sure the submitted data is not duplicate data;
4. if so, display an error message and show the form again;
5. if not, insert the data into the database and display a confirmation message

It is really more than can be covered in a forum post. There are some great tutorials available on the subject - take a look here: http://www.hotscripts.com/PHP/Tips_and_Tutorials/Form_Processing/index.html

[petenyce102]

I need help with steps
3. check the DB to make sure the submitted data is not duplicate data;
4. if so, display an error message and show the form again;

I dont know how to display the error code on the Form. Could you provide a basic example with my code below tied into the error message showing on the form.

Below is the cod e im using now to check the database

$query1 = "SELECT * from test WHERE name = '$user'";
$result1 = mysql_query($query1) or exit(mysql_error());
if (mysql_num_rows($result1))
{

echo ' <script language = "javascript">
alert("You have entered a taken username. Please try another username")
</script>';
print '<meta http-equiv="refresh" content="0;URL=register2.html">';

}

This works but wipes out all the forms data, I would like to get the message to display on the form

[travelbuff]

A couple more issues:

1. I assume you are assigning the value of $_POST[T3] to $user before running that query?
2. I think the if statement you want is:
   if (mysql_num_rows($result1)>0) {
   3.Do you have an if statement to determine whether or not to display the form?

There is a faster way:

$count_users = mysql_result(mysql_query("SELECT COUNT(user_id) FROM users WHERE user_name = '$_POST[T3]'"),0);
  if ($count_users > 0) {
     //the name is taken, display the error and show the form
  } else {
    // the name is ok
  } 

To make the form "sticky" add the values to the form fields as follows:
<p>Name: <input type="text" name="T1" size="20" value = "<? echo $_POST['T1']; ?>></p>

I still suggest reading one of the tutorials on the topic.

[petenyce102]

You will see a zip file containing my files. New.php and registerform.html and maybee this will help you undrstand. This is a hard topic for me and soemthing i would like to accomplish

The errors i would like to display on the form are as folowws

if userid is a duplicate and if ehp id is already taken.

Hope this helps

[travelbuff]

Originally posted by petenyce102
This is a hard topic for me and soemthing i would like to accomplish

That's why I suggested reading a tutorial on the topic. If I rewrite your code for you, you won't learn anything.