help me :(

kamsmartx

hi
i have form with 2 input type
1- list
2- text

so i wont to when i select the value form "list" for example the list have NO of employee , then we can input the name of the employee on the text befor i press buton submit
how i can do it

Xager

you check if $_POST['list'] is empty and if it is you use the text value

kamsmartx

no
mmmm
i wont when i choose value from list the text must be change into name of the id

benracer

Couls you explain your question in more detail and if possible, post the code you have allready tried 😃

kamsmartx

Welcome
I'm in form and the textfeld containing the name and also add a list of numbers of staff
Be read
Now. I want when selecting a staff member to the officer's name printed in the box adjacent
the code of the list "no of employee"

<select name="FileNo" id="select2" onChange="change_title();">
	<? 
$query="SELECT * FROM employee";
$result=mysql_query($query) or die("error: " . mysql_error());
while($row=mysql_fetch_array($result))
{
	echo "NO";
?>  

<option value="<?php echo $row['FileNo'] ; ?>"><?php echo $row['FileNo'] ; ?></option>   

<?

}  
?>
</select>

the code of text

<input name="Name" type="text" id="title"/>

kamsmartx

any help

kamsmartx

Horizon88

Stop bumping this. No one is helping you because, as they've said, you're not being clear enough about your problem.

kamsmartx

Ok but what can i do

Xager

kamsmartx wrote:
Ok but what can i do

Post more about your problem like the poster above you said...

kamsmartx

the problem is:
i wont to print the name of pationt here

<input name="Name" type="text" id="title"/>

when i select the number of him from here

<select name="FileNo" id="select2" onChange="change_title();"> 
    <? 
$query="SELECT * FROM employee"; 
$result=mysql_query($query) or die("error: " . mysql_error()); 
while($row=mysql_fetch_array($result)) 
{ 
    echo "NO"; 
?>   

<option value="<?php echo $row['FileNo'] ; ?>"><?php echo $row['FileNo'] ; ?></option>    

<? 

}   
?> 
</select>

Horizon88

So where is the "change_title()" JavaScript function you're calling on the select's change?

Xager

Why do you open and close the php tags all the time?

Try using this instead:

while($row=mysql_fetch_array($result))
{
    echo 'NO 
    <option value="'. $row['FileNo'].'">'.$row['FileNo'].'</option>';
}
echo '
</select>
';

kamsmartx

no u r not understant me

i wont when i select value from list print on text the name of this value
the value is number and the print is name

Horizon88

He posted a tip regarding your PHP, kam. I asked about your JavaScript, as you're almost there. You need to use the onChange attribute on your select box, which you do, but you're calling change_title(), a javascript function, that you didn't post - is there code for that you have?

kamsmartx

ok but what do

leatherback

Kams,

If you truely want help, you need to read the questions asked, and answer them. They usually lead up to a solution. We cannot help you unless you let us.

In general: Your onChange in the dropdown should call a javascript function -note that this is a PHP and not a JS forum-. This can then either use a form of AJAX to retrieve the name from the database, or if you already have it somewhere, spit out that value.