Return of a if using empty() function in PHP

Question

I have following IF in my code to check if array are empty,

if (!empty($data['id']) && (empty($data['student_no']) || empty($data['batch']) ) ) {
   print_r("inside if ");
}

$data['id'],$data['student_no'],$data['batch'] are arrays.

array values are like below,

1.$data['id'] -> Array ( [0] => 1 [1] => [2] => )

2.$data['student_no'] -> Array ( [0] => [1] => [2] => )

3.$data['batch'] -> Array ( [0] => )

But this does not got to inside the if and print the "inside if" string.

Please tell me what is wrong ?

Empty array is array with NO values. Array with one empty value is not empty. — u_mulder
– u_mulder, Commented Mar 28, 2017 at 10:20
change your condition to if (true). That's how you would get inside the if with these arrays. — zerkms
– zerkms, Commented Mar 28, 2017 at 10:21

Nishant Nair · Accepted Answer · 2017-03-28 10:20:40Z

2

You forgot ! in if

!empty($data['student_no']

answered Mar 28, 2017 at 10:20

Nishant Nair

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2 Comments

zerkms Over a year ago

This is that rare case, when the answerer knows the task better than the OP*. (* actually not) @Nishant Nair how do you know if they have not forgotten something else? Like a condition to check whether it's a Tuesday today?

Thamilhan Over a year ago

OP says: check if array are empty

S. Schenkel · Accepted Answer · 2017-03-28 10:21:33Z

0

$data['student_no'] and $data['batch'] are not empty but your condition wants one of them empty, so it does not work

answered Mar 28, 2017 at 10:21

S. Schenkel

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Comments

codenathan · Accepted Answer · 2017-03-28 10:23:33Z

0

Try wrapping another set of brackets in the !empty($data['id'])

if ((!empty($data['id'])) && (empty($data['student_no']) || empty($data['batch']) ) ) {
   print_r("inside if ");
}

edited Mar 28, 2017 at 10:23

answered Mar 28, 2017 at 10:21

codenathan

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1 Comment

zerkms Over a year ago

More parentheses for the god of parentheses!

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