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Answer 1

A:

Instead of using hardcoded HTML, print the combo box with a for loop and based on what the user has selected, print "selected" in the relevant "option"

to get the date parts you can do something like:

list($year , $month , $day) = explode("-" , $dob);

Sabeen Malik 2010-09-04 14:03:44

what is $dob? the full birthday? And do i use this with Richard´s answer?

Karem 2010-09-04 14:09:56

oh you added more info the question. so replace $dob with $v["birthday"] and use Richards sample code to do the rest of combo box building.

Sabeen Malik 2010-09-04 14:11:21

`$dob` would be your `1990-01-01` date of birth string. Then yes, use a `for` loop as shown in my answer to output all the elements, with `selected` on the appropriate one.

Richard Fearn 2010-09-04 14:11:46

OK just tried this new method, it works great on $day but as you can see in the HTML i have, at month it has 1-12 at value but then january, february, and so.. with names. How should i do that by making using a for loop?

Karem 2010-09-04 14:40:18

Answer 2

+1 A:

If you use a for loop to generate the <option> elements, such as:

for ($i=1; $i<=31; $i++) {
    echo "<option value=\"$i\">$i</option>\n";
}

Then you can just modify this slightly to insert the selected attribute on the appropriate element:

// Assuming $day is the user's day of birth
for ($i=1; $i<=31; $i++) {
    echo "<option ";
    if ($i == $day) {
        echo "selected=\"selected\" ";
    }
    echo "value=\"$i\">$i</option>\n";
}

You would use similar loops for the month and year. In the case of months, where you don't want to output a number but want the month name, you can store the names in an array and index into that array in the loop:

$month_names = array("Jan", "Feb", "Mar", ...);
for ($i=1; $i<=12; $i++) {
    echo "<option ";
    if ($i == $month) {
        echo "selected=\"selected\" ";
    }
    echo "value=\"$i\">", $month_names[$i-1], "</option>\n";
}

To get the day/month/year from your yyyy-mm-dd date of birth strings, use the explode function, as shown by Sabeen Malik.

Richard Fearn 2010-09-04 14:04:20

but i dont have it stored like that so the day is in one field, $day in one, i have it like this: 1992-01-01, so the day is 01(the last of that)

Karem 2010-09-04 14:06:53

OK just tried this new method, it works great on $day but as you can see in the HTML i have, at month it has 1-12 at value but then january, february, and so.. with names. How should i do that by making using a for loop?

Karem 2010-09-04 14:23:28

Answer updated to describe how to output month names instead of numbers.

Richard Fearn 2010-09-04 14:42:11

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