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[SOLVED] Getting data to form


june_c21

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<?php

#check to see if the submit button has been pressed.
if (isset($_POST['submit'])) {
#if so then output staff number is ... then the value of the textbox sent via post variable
echo "Your stff number is..." . $_POST['staff_no'];

} else {
#if the submit button hasn't been pressed then display the form for the user
echo "<form name=\"Form1\" method=\"POST\" action=$_SERVER['PHP_SELF']>
Staff Number: <input type=text name=staff_no><br>
<input type=submit name=submit value=submit>
</form>";

}

?>

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use mysql to query your database where stuff_no = employees_name

 

then create a hidden field in your form and just echo the variable out for your employees name.

 

something like:

 

<?php

// connect to db
// select db
$mystaff = $_POST['staff_no'];
$results = mysql_query("select employee_name from tblName where staff_no='$mystaff'");

while ($mychoice=mysql_fetch_array($results)) {

$employee=$mychoice["employee_name"];

echo "<form name=\"Form1\" action=\"". $_SERVER['PHP_SELF'] ."\">
Staff Number: <input type=text name=staff_no><br>
<input type=\"hidden\" value=\"$employee\" name=\"currentemployee\">
<input type=submit name=submit value=submit>
</form>";

}

?>

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well you have to have the employee enter the number some how - if your not using a form for them to enter their employee number in; then how are they entering it?

 

what i want is when user enter their staff no

 

if your using a link with each employees name in them and a predefine employee number in the link; the just change $_POST to $_GET.

 

$mystaff = $_POST['staff_no'];

 

Link To Employee Form

<a href="employeeform.php?staff_no=123">John Doe</a>

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