Re: INSERT Question

Charles Kline <ckline@rightcode.net> · Sat, 29 Mar 2003 17:52:53 -0500

John,

Thanks. Yes that is a big help. The radio button is actually not used 
for that purpose (sorry that was a bit confusing of me to put in 
without an explanation). For each person selected, the radio button Y/N 
relates to whether they are an author of the report or just a 
contributer. I assume, the radio button can be handled the same way and 
I can just extrapolate your example to figure that out - I think ;)

- Charles

On Saturday, March 29, 2003, at 05:46 PM, John W. Holmes wrote:

Since I don't totally understand (newbie) I will show you an example
of
the form.

<form>
<select name="person1">
<option value="1">Joe Smith</option>
etc.
</select>
<input type="radio" name="person1yn" value="Yes"> Y <input
type="radio"
name="person1yn" value="No"> N

So if the 'Y' radio button is chosen, you'll want to take that 'person'
and input them into the table?

First, the easy way would be to lose the radio button and just put a
"Choose Person" option in the <select> dropdown. Then, only process the
names if it's not "Choose Person"...

Even if you can't lose the radio buttons, you'll need to make 
everything

arrays so they are easier to process.

<form>
<select name="person[]">
<option value="0">Choose Person</option>
<option value="1">Joe Smith</option>
...
</select>

Then, to process that (assuming POST) you'd use:

Foreach($_POST['person'] as $person_id)
{
	if($person_id > 0)
	{ $insert[] = "('$person_id')"; }
}
$sql = "INSERT INTO table (person_id) VALUES " . implode(',',$insert);

Then run $sql... The second part loops through each "person" select and
makes sure the value isn't zero before it creates the ('value') part of
the SQL. The implode() at the end of the $sql line takes each ('value')
part and joins them into a string separated by a comma.

Does that help at all?

---John W. Holmes...

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