How to check key value is exist in an array using PHP

Question

I need to check some key value is exist in an array using PHP. Here is my code:

$comment = json_encode(array(array('day_id' => '1', 'comment' => 'vodka0'),array('day_id' => '', 'comment' => ''), array('day_id' => '3', 'comment' => 'vodka3'),array('day_id'=>'4','comment'=>'hytt')));
$arrComment = json_decode($comment, true);

Here I need to check some day_id key has value or all day_id key has blank value.

jitendrapurohit · Accepted Answer · 2016-10-19 11:04:39Z

1

Use array_column and array_filter to check this:

// extract all day_id columns
$dayId = array_column($arrComment, 'day_id');
// filter the empty values
$filtered = array_filter($dayId);

if (empty($filtered)) {
  echo "All Day Ids are empty.";
}
else {
  echo "Some or all of them have some value.";
}

answered Oct 19, 2016 at 11:04

jitendrapurohit

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Comments

Mahfuzul Alam · Accepted Answer · 2016-10-19 11:03:00Z

0

Do you mean this: var_dump(array_column($arrComment, 'day_id'));

It returns all values of day_id key. Now do what you want.

answered Oct 19, 2016 at 11:03

Mahfuzul Alam

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2 Comments

satya Over a year ago

I need to check this inside if...else statement.

Mahfuzul Alam Over a year ago

Ok above code will provide you array of day_id values. Now filter your array with $someVar = array_filter($values). array_filter() removes empty value. If all are empty $someVar variable will be empty. Hope you got your answer. I can give you idea and you get your hints. Now write your code. You'll get expected output. Happy coding.

smottt · Accepted Answer · 2016-10-20 08:00:45Z

0

for ($i = 0; $i < count($arrComment); $i++) {
    if (isset($arrComment[$i]['day_id'])) {
        //value is set
    } else {
        //value is not set
    }
}

edited Oct 20, 2016 at 8:00

smottt

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answered Oct 19, 2016 at 11:24

kemika

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1 Comment

andreas Over a year ago

While this code snippet may solve the question, including an explanation really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion.

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