[SOLVED] drop down list from an array

[jim.davidson]

I'm trying to teach myself php.

 

Today's mission learn to build and display arrays

 

I want the array to contain a valid year of birth for anyone between the ages of 16 and 80.  I'm pretty sure this code is correct.(though it might be a bit sloppy to all you pros)

 

$dob_dates = array();

$today = date('Y');

$startyear = $today - 80;

$endyear = $today - 16;

$n = 0;

$i = $startyear;

while ($i <= $endyear) {

  $dob_dates[$n] = ($startyear+$n);

  $i++;

  $n++;

}

 

Now that I've built an array containing valid dob years I want to have a drop down list showing the array.  Can someone show me the light on how to do that?

 

Any and all help will be appreciated

[chocopi]

I think you mean something like

 

<select name="select">
<?php
for($i = 0; $i <= $n; $i++)
{
   echo "<option name=\"{$dob_dates[$i]}\">{$dob_dates[$i]}</option>";
}
?>
</select>

 

That should work

 

~ Chocopi

[jim.davidson]

Thanks for the help, now I'll sit here and figure out just how it works.

 

That's one of the ways I learn things, take something that works and figure out how and why.

 

Thanks again

[chocopi]

No problem

 

Does it work how you want it to ???

 

1. Anyways, it is very simple, its a basic for loop which

  i) Sets $i to 0,

  ii) then checks that $i is less than or equal to $n, if not then the loop ends

  iii) lastly after each loop is will increment $i by 1 until $i is greater than $n

 

2. It then gets the $i value of the array and then takes it from the variable and puts it within the <option> tag

 

Hope that helps

 

~ Chocopi

[jim.davidson]

Sorry for the delay in responding...yes it works fine and now I understand it.

 

Thanks again