How do you include a variable as part of the name of a class

[N. David]

In bash, you can include a variable as part of a filename. For
example:

date=$(date)
cat log.txt > $date"_log.txt"

Which, when setting the right options in date, creates a file named
20040705_log.tx t

I am now trying to call a class based on a previously defined variable
set by an array:

1 $field[1] = "first_name ";
7 $field[2] = "mi";
8 $field[3] = "last_name" ;
12 foreach ($field as $field_name) {
13 $var_validate_f ield = &New "cls_validate_" $field_name;
14 $var_strip_fiel d = &New "cls_strip_"$fi eld_name;
15 $record[$field_name] = $_POST[$field_name];
16 $record_check[$field_name] =
$var_validate_f ield->fn_validate($r ecord[$field_name]);
17 $record[$field_name] =
$var_strip_fiel d->fn_strip($reco rd[$field_name]);
18 print "<p>$field_ name is $record[$field_name] and its check is
$record_check[$field_name]";
19 }

As you can see, I am trying to make $field_name part of the name of
the class, but it is not working. Everything else works fine; I've
tested the classes individually; but the array method will not work
without this. Suggestions are welcome. Thank you.

[Janwillem Borleffs]

Re: How do you include a variable as part of the name of a class

N. David wrote:[color=blue]
> As you can see, I am trying to make $field_name part of the name of
> the class, but it is not working. Everything else works fine; I've
> tested the classes individually; but the array method will not work
> without this. Suggestions are welcome. Thank you.[/color]

foreach ($field as $field_name) {
$cls_validate = "cls_valida te_" . $field_name;
$var_validate_f ield = &New $cls_validate;
...
}

Please read:

http://www.php.net/manual/en/language.operators.string.php

JW

[Tony Marston]

Re: How do you include a variable as part of the name of a class

Your code is terribly inefficient. Instead of iterating through the array
and calling a different method to validate each field individually why don't
you pass the whole array to a single validation method which can iterate
through the array internally. As the $_POST array is an associative array of
name=value pairs then the validation method can easily identify which fields
have been passed to it for validation.

This is explained in more detail in my articles at
http://www.tonymarston.co.uk/php-mys...seobjects.html and
http://www.tonymarston.co.uk/php-mys...eobjects2.html.

--
Tony Marston

http://www.tonymarston.net

"N. David" <ndavidg@yahoo. com> wrote in message
news:6fa7cbb4.0 408051056.11aa6 091@posting.goo gle.com...[color=blue]
> In bash, you can include a variable as part of a filename. For
> example:
>
> date=$(date)
> cat log.txt > $date"_log.txt"
>
> Which, when setting the right options in date, creates a file named
> 20040705_log.tx t
>
> I am now trying to call a class based on a previously defined variable
> set by an array:
>
> 1 $field[1] = "first_name ";
> 7 $field[2] = "mi";
> 8 $field[3] = "last_name" ;
> 12 foreach ($field as $field_name) {
> 13 $var_validate_f ield = &New "cls_validate_" $field_name;
> 14 $var_strip_fiel d = &New "cls_strip_"$fi eld_name;
> 15 $record[$field_name] = $_POST[$field_name];
> 16 $record_check[$field_name] =
> $var_validate_f ield->fn_validate($r ecord[$field_name]);
> 17 $record[$field_name] =
> $var_strip_fiel d->fn_strip($reco rd[$field_name]);
> 18 print "<p>$field_ name is $record[$field_name] and its check is
> $record_check[$field_name]";
> 19 }
>
> As you can see, I am trying to make $field_name part of the name of
> the class, but it is not working. Everything else works fine; I've
> tested the classes individually; but the array method will not work
> without this. Suggestions are welcome. Thank you.[/color]

[N. David]

Re: How do you include a variable as part of the name of a class

"Tony Marston" <tony@NOSPAM.de mon.co.uk> wrote in message news:<ceud2c$kp q$1$830fa79f@ne ws.demon.co.uk> ...[color=blue]
> Your code is terribly inefficient. Instead of iterating through the array
> and calling a different method to validate each field individually why don't
> you pass the whole array to a single validation method which can iterate
> through the array internally. As the $_POST array is an associative array of
> name=value pairs then the validation method can easily identify which fields
> have been passed to it for validation.
>
> This is explained in more detail in my articles at
> http://www.tonymarston.co.uk/php-mys...seobjects.html and
> http://www.tonymarston.co.uk/php-mys...eobjects2.html.
>[/color]

So what you are telling me is that the "for each" is unecessary and I
can just pass the whole array one time to the class file? I will try
this, thanks.

[Chung Leong]

Re: How do you include a variable as part of the name of a class


"N. David" <ndavidg@yahoo. com> wrote in message
news:6fa7cbb4.0 408051056.11aa6 091@posting.goo gle.com...[color=blue]
> In bash, you can include a variable as part of a filename. For
> example:
>
> date=$(date)
> cat log.txt > $date"_log.txt"
>
> Which, when setting the right options in date, creates a file named
> 20040705_log.tx t
>
> I am now trying to call a class based on a previously defined variable
> set by an array:
>
> 1 $field[1] = "first_name ";
> 7 $field[2] = "mi";
> 8 $field[3] = "last_name" ;
> 12 foreach ($field as $field_name) {
> 13 $var_validate_f ield = &New "cls_validate_" $field_name;
> 14 $var_strip_fiel d = &New "cls_strip_"$fi eld_name;
> 15 $record[$field_name] = $_POST[$field_name];
> 16 $record_check[$field_name] =
> $var_validate_f ield->fn_validate($r ecord[$field_name]);
> 17 $record[$field_name] =
> $var_strip_fiel d->fn_strip($reco rd[$field_name]);
> 18 print "<p>$field_ name is $record[$field_name] and its check is
> $record_check[$field_name]";
> 19 }
>
> As you can see, I am trying to make $field_name part of the name of
> the class, but it is not working. Everything else works fine; I've
> tested the classes individually; but the array method will not work
> without this. Suggestions are welcome. Thank you.[/color]

An inconsistency in PHP. Variable class name works with variables but not
literal strings or evaluated values. Do this:

$class = "cls_validate_$ field_name";
$obj = new $class;