CODEWITHSUNDEEP
phparrays
John
John

Reputation: 3

Variable In Array

I am trying to add variable in an array but failed to do so please help me out.

Below is Original code which works fine

$array1 = array("a" => "green", "red", "blue");
$array2 = array("b" => "green", "yellow", "red");
$result = array_intersect($array1, $array2);
print_r($result);

and it output this 

Array
(
    [a] => green
    [0] => red
)

But I want to add variable in array so array gets value from the variable but it's not working

What I am trying

$a = '"green", "red", "blue"';
$b = '"green", "yellow", "red"';
$array1 = array("a" => $a);
$array2 = array("b" => $b );  
$result = array_intersect($array1, $array2);
print_r($result);

I want it to output like this

Array
(
    [a] => green
    [0] => red
)

What I am getting

Array ( )

Any help will be appreciated. Thanks

Upvotes: 0

Views: 70

Answers (2)

Matheus Santos
Matheus Santos

Reputation: 680

You are adding a string '"green", "red", "blue"' not an array. In your snippet

$a = '"green", "red", "blue"'; $b = '"green", "yellow", "red"'; $array1 = array("a" => $a); $array2 = array("b" => $w );
$result = array_intersect($array1, $array2); print_r($result);

PHP will understand $a and $b as strings. If you meant to pass an array to $a and $b you will need to change it to 

$a = array("green", "red", "blue");
$b = array("green", "yellow", "red");

Then make your intersection. If you use an var_dump($a) you will see that $a is storing an String variable.

Upvotes: 1

Joe
Joe

Reputation: 28316

The code you have tried is assigning the string '"green", "red", "blue"' to the array element "a" when what you appear to want is to split the string like so that "green red blue" becomes array("green","red","blue")

$a = "green red blue";
$array1 = split(" ",$a);

see http://php.net/manual/en/function.split.php

Upvotes: 1

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