Want to protect your cyber security and still get fast solutions? Ask a secure question today.Go Premium

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 204
  • Last Modified:

PHP Code

The code below is part of a form that fulls data from a mysql database & displays that information in a browser. As you can see in the code, it pulls the employee name. My question is, is there any way I can edit this line of code that will allow me to add a drop down box that shows the other employees but keep the employee it pulls from the mysql database default?
<b>Employee: </b><br><input type="text" name="employee" size="15" maxlength="30" option value="'.$row[4].'" /> <br>

Open in new window

0
wantabe2
Asked:
wantabe2
1 Solution
 
leakim971PluritechnicianCommented:
Check this : http://www.phpro.org/tutorials/Dropdown-Select-With-PHP-and-MySQL.html

build your dropdown first and use :

$dropdown = dropdown(...,....,..

echo '<b>Employee: </b><br><input type="text" name="employee" size="15" maxlength="30" option value="' . $row[4] . '" />' . $dropdown . '<br>';

Open in new window

0
 
Mark BradyPrincipal Data EngineerCommented:
Well yes you can do that very easily. Let's assume a few things as I don't know what your database looks like but I'll make some stuff up. I will assume that you already know the name or ID of the chosen employee so we know who is selected. Try this on your page.

<?php
include("connection-code-or-script-here");
// Get the posted employee
$employee_selected = $_REQUEST['employee'];
########################################## Now we have the chosen employee if any


$employees_select = "<select name='employees' id='employees' />";
$employees_select .= "<option value=''>Please select an employee</option>";

$sql = "SELECT id,first,last FROM employees";
$result = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
$id = "{$row['id']}";
$first = "{$row['first']}";
$last = "{$row['last']}";
if($id == $employee_selected){
$employee_select .= "<option selected value='$id'>$last $first</option>";
}else{
$employee_select .= "<option value='$id'>$last $first</option>";
}
}
$employee_select .= "</select>";
################################# END OF PROCESSING ##########################

now in your form, put all the controls you need to have and add this code.

Employee <?php echo $employee_select; ?><br />

Then continue with your form. If a user changes/selects an employee and submits the form, the selectbox will stay on the selected employee but all the other employees are still in the selectbox. If all you are doing is displaying employee data and only need a selectbox to choose the employee to display the data, then you could add            

onchange='this.form.submit()'

to the <select> tag so it submits the form when it is changed. Then you can do an if statement to see if an employee has been selected. If YES then show the employees data, if NO then just show the selectbox only . If you need help with that form just ask.

If this is not wehat you are looking for, please explain more about the form and perhaps post it here.
0

Featured Post

Free Tool: Subnet Calculator

The subnet calculator helps you design networks by taking an IP address and network mask and returning information such as network, broadcast address, and host range.

One of a set of tools we're offering as a way of saying thank you for being a part of the community.

Tackle projects and never again get stuck behind a technical roadblock.
Join Now