CODEWITHSUNDEEP
php
ProEvilz
ProEvilz

Reputation: 5445

How to exclude a parameter within an if statement?

I'm trying to create somewhat of a fuzzy search with PHP. Consider the following pseudo code/scenario below.

Take these $_POSTed values:

name
price
location
age

And I have an array of items (lets assume cars), of which each item contains the following properties:

name
price
location
age

So in my function, I use an if() within a foreach() to check if any of the posted data matches with an item and if it does, echo the item.

function search() {

 $items = array(...);

 $name = $_POST['name'];
 $price = $_POST['price'];
 $location = $_POST['location'];
 $age = $_POST['age'];

 foreach($items as $item) {
   if(
    $item['name']  ==  $name &&
    $item['price'] == $price &&
    $item['location'] == $location &&
    $item['age'] == $age
    ){
      echo json_encode($item);
    } 

 }

}

The issue is that not every posted value is filled in, so for example $name could be empty i.e. $name = ''.

Question: How can I exclude any of the items in the if() statement if the initial $_posted key is empty without creating an if() inception type scenario? I considered using || but I'm pretty sure that wouldn't be the same as excluding a comparison all together. A

Upvotes: 1

Views: 430

Answers (1)

Alexis
Alexis

Reputation: 463

You can achieve that by taking advantage of PHP's lazy interpreter:

if(
    (!$name || ($name && $item['name']  ==  $name)) &&
    (!$price || ($price && $item['price'] == $price)) &&
    (!$location || ($location && $item['location'] == $location)) &&
    (!$age || ($age && $item['age'] == $age))
    ){
      echo json_encode($item);
    }

This will first check that each variable is not null, nor empty. (How exactly does if($variable) work?) Also look at Does PHP have short-circuit evaluation?

Upvotes: 2

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