benanamen

Simplest thing to do is install VirtualBox and load Linux in a Virtual Machine

function MaFonction($x){ for ($i=0 ; $i<=360 ; $i+=$x){ echo sprintf("%02d", $i) . " ==> " . deg2rad($i) . "<br />"; } } MaFonction(15 );

The proper syntax is not else, else, else.... https://www.php.net/manual/en/control-structures.elseif.php

Just use var_dump if you just need to see what the array contains.

The the left join would have worked if you switched the tables around.

You would do a JOIN on ProspectID

It never hurts to learn new things. There are plenty of Laravel tutorials out there. Make sure the tutorial matches the version of Laravel you will be using or you will get stuck, as substantial changes have been made between particular versions.

It depends. Do you want to learn how to code or do you want to learn how to use Laravel? Read the manual and watch the Laracast videos they provide. You should really learn the basics of coding before you attempt a Framework.

It' simple really. You are providing an array, it matches the 1 and returns back to the caller so the rest of the if's are not processed. Try this... <?php declare(strict_types=1); function property_structure_type ($data) { if (in_array("1", $data)) { return "House"; } if (in_array("2", $data)) { return "Cabin"; } if (in_array("3", $data)) { return "Mobile Home"; } if (in_array("4", $data)) { return "Apartment"; } if (in_array("5", $data)) { return "Condominium"; } if (in_array("6", $data)) { return "Townhome"; } if (in_array("7", $data)) { return "None"; } } echo property_structure_type ([2]);

A minor comment. This line if(count($result_contest) > 0) Can simply be if($result_contest) The return value for fetchALL is either an array with results or an empty array, or false on failure. There is no need to count anything. The default truthy check of the if is all you need.

Well..., if that isn't a "how about that" LOL. I get the explanation by Andrea, but still.......

Just a reminder, in PHP 7 with strict_types, the second parameter of print_r MUST be a boolean and not an integer or it will fail with a fatal error. Correct <?php declare(strict_types=1); echo '<pre>', print_r($array, true), '</pre>'; Fails Fatal Error <?php declare(strict_types=1); echo '<pre>', print_r($array, 1), '</pre>'; Fatal error: Uncaught TypeError: print_r() expects parameter 2 to be boolean, integer given TypeError: print_r() expects parameter 2 to be boolean, integer given It does "work" but it will give you an Undefined variable notice. It doesn't matter if it is an array or not.

The \ is escaping the /. Why is your data in the DB like that in the first place?

No. https://dev.mysql.com/doc/refman/8.0/en/logical-operators.html

You pretty much covered it in the first sentence so I will just tell you what my standard is. snake_case = DB Columns, Variable Names & Function Names (same as Php. i.e: set_exemption_handler) camelCase = OOP methods, method calls, & OOP Properties PascalCase = Class Names. (i.e: class UserAccount) dash-seperator = Filenames (For SEO i.e: my-file.php)