Help: Require HTML form to submit to same page to insert data into database

flash99

Hi, I am trying to create a form in order to allow users to enter information directly into the database. I have the HTML form and the PHP code to send the SQL queries to the server all on one page. The problem here is that I am not sure how to triger the PHP code, I have tried many if-else lines but keep getting an error saying that I have an undefined variable in that line, some lines don't show any errors but still doesn't work. I hope someone might be able to help me.

*THE LINES ARE IN BOLD

THANKS FOR ANY HELP GIVEN!

<html>
<head>
<title>Employee List</title>
</head>
<body>
<?php
// Connection variables
$host = "localhost";
$username = "root";
//$password = "pass";
$db = "mydb";
$table = "employees";

// Connection to host
$connect = @mysql_connect($host, $user) or die("could not connect to host!");
// Select database
$db_select = @mysql_select_db($db) or die("could not connect to database!");

// Process form
//if ($submit)
//if (@$submit)
//if ($submit=="Submit")
//if ($REQUEST_METHOD=="POST")
//if ($HTTP_POST_VARS['sent'])

if (isset($sent)) {

$sqlreceive = "INSERT INTO employees SET first='$POST[first]',last='$POST[last]',address='$POST[address]',position='$POST[address]'";
echo "New Link Added\n";
} else {

// SQL
$sqldisplay = "SELECT * FROM $table";
// Query the database
$resultdisplay = @($sqldisplay) or die("could not complete query!");
// Display results - loop required as results are in an array
if ($row = @mysql_fetch_array($resultdisplay)) {
echo "<table border=1>\n";
echo "<tr><td>Id</td><td>First name</td><td>Last name</td><td>Address</td><td>Position</td></tr>\n";

do	{
	printf("<tr><td>%s</td><td>%s</td><td>%s</td><td>%s</td><td>%s</td></tr>",$row["id"],$row["first"],$row["last"],$row["address"],$row["position"]);
	} while($row = @mysql_fetch_array($resultdisplay));
	echo "</table><br>";

} else {
echo "Sorry, no records were found!\n";
}
?>
<form name="employee" action="employee.php" method="post" enctype="text/plain">
<table>
<tr><td><b>First name:</b></td><td><input type="text" name="first" size="40" maxlength="40"><br></td></tr>
<tr><td><b>Last name:</b></td><td><input type="text" name="last" size="40" maxlength="40"><br></td></tr>
<tr><td><b>Address:</b></td><td><input type="text" name="address" size="40" maxlength="40"><br></td></tr>
<tr><td><b>Position:</b></td><td><input type="text" name="position" size="40" maxlength="40"><br></td></tr>
</table><br>
<input type="submit" value="Add Record">
<input type="reset" value="Reset form">
</form>
<?php }
?>
</body>
</html>

nemonoman

$_POST contains the values of the variables you "name" in your form.

as in:

input type="text" name="first"

When you submit this form, PHP can access the value of $first:

echo $first:

It can also access $_POST['first']

DON'T FORGET THE QUOTES around the array key your you'll get warnings and/or failure.

Check to see if any of your NAMED variables have been submitted.

if($first){}

You might add a "name" to your submit button

input type=submit name=itsasubmit

then test

if($itsasubmit)

This way you can have 2 submit buttons calling different PHP functions from the same page

The reason you haven't triggered any PHP in the code you've shown is that the variable names you've checked as triggering events have no relation to the 'name' of any form variables.

once you set the 'name' of a var in the form, and check the same name as a $var in your PHP code (as shown above) the rest of your code will execute.

Don't for get to change add "" to the $_POST[] keys in your INSERT statement or that will fail as well.