PHP array returns different value than expected

Question

With the piece of PHP code I have been working on, I wanted to return only:

value1
value2

of an array.

However when I run this code, it returns the following:

value1
value2
value2
value2
value2

This is my code:

<?php
    $inputname = array("value1", "value2", "value3", "value4", "value5");
    foreach($inputname AS $i){
        if($i == "value1" || $i = "value2")
        {
            echo($i."<br />");
        }
    }
?>

Am I overlooking something?

n-dru · Accepted Answer · 2015-06-30 09:56:08Z

4

Add extra = before value2:

if($i == "value1" || $i == "value2")
                        ^

otherwise you assign value2 to every variable $i that is not value1.

answered Jun 30, 2015 at 9:56

n-dru

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Comments

Yada · Accepted Answer · 2015-06-30 09:56:21Z

1

if($i == "value1" || $i = "value2")

Suppose to be

if($i == "value1" || $i == "value2")

answered Jun 30, 2015 at 9:56

Yada

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Comments

Devsome · Accepted Answer · 2015-06-30 09:56:36Z

1

you need to set the $i = to $i ==

if($i == "value1" || $i == "value2")

//edit: to slow :c

answered Jun 30, 2015 at 9:56

Devsome

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1 Comment

Barrosy Over a year ago

Ah thanks, didn't expect that, I didn't even see that.

Vinod Kirte · Accepted Answer · 2015-06-30 12:20:52Z

0

You need to check if($i == "value1" || $i == "value2") instead of the if($i == "value1" || $i = "value2")

answered Jun 30, 2015 at 12:20

Vinod Kirte

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