variable = conditional

[sabret00the]

how would i go about making sure that a variable shows the results of a conditional?

something like this, i remember being old i couldn't it in such a way before, just don't remember how i was taught to do it.

PHP Code:

        $show_project = if isset($rated) {
             eval("\$rate = \"".fetch_template("projects_rate_show")."\";");
         } else {
             eval("\$rate = \"".fetch_template("projects_rate")."\";");
         } 


[Xenon]

i'd suggest that way:

PHP Code:

$show_project = isset($rated);
if ($show_project) {
            eval("\$rate = \"".fetch_template("projects_rate_show")."\";");
        } else {
            eval("\$rate = \"".fetch_template("projects_rate")."\";");
        } 


[sabret00the]

thanks that code seems to do it, however how would i then call the code?

in my templates it's set as $rate

but i'm not sure what to put in the file for it call to the template

i thought i would be able to just put $show_project below the other template calls for that page but i that don't work

[Xenon]

add $show_project into your template, that should do it.

but i'm not sure, about the output, i think it's TRUE or FALSE, but i may bewrong

[filburt1]

PHP Code:

eval("\$show_project = \"" . fetch_template(isset($rated) ? "projects_rate_show" : "projects_rate") . "\";"); 


( ? : ) is the ternary operator. It is confusing at first but useful for quick and simple inline ifs.

[Xenon]

or use iif as it's the vb pendant

[sabret00the]

well i managed to get the code working bar this last bit i guess that's what i was trying to do in the first place

but i was gonna ask

PHP Code:

        if ($rated >= 0) {
            eval("\$rate = \"".fetch_template("projects_rate")."\";");
        } else {
            eval("\$rate = \"".fetch_template("projects_rate_show")."\";");
        } 


as i tried this a million and one ways and no matter what it will only show the call forth the first version $rate which in this case is the "projects_rate" template and i was getting frustrated, but i'll most definately, give filburts code a test before i start pulling my hair out and throwing myself at your feet for mercy

[sabret00the]

ok i'm at a total loss, it's still not working

PHP Code:

            $rated = $DB_site->query_first("
                SELECT rate
                FROM project_rate
                WHERE projectid = $projectid AND userid = $bbuserinfo[userid]
            ");

            eval("\$rate = \"" . fetch_template(isset($rated) ? "projects_rate_show" : "projects_rate") . "\";"); 


that's the block of code, i'm a total loss as to why it work work, i can only think that maybe i'm selecting the data from the database wrong?

[Velocd]

Try:

PHP Code:

eval("\$rate = \"" . fetch_template(!$rated['rate'] ? "projects_rate" : "projects_rate_show") . "\";"); 


Isset() will always return true because you're always setting $rated to something, whether an empty array or not.

[sabret00the]

:w00t: it worked, i'm so greatful, thank you so so so so so so much