PHP in_array() 无法正常工作

Question

I have an if statment below that checks if a value is present in an array and adds it in there if it's not. Apparently even if the value is in the array, it triggers it again.

As far as I understand, it should only display 1 of each value, because it will only trigger 3 times, like this:

Digital Photography -> 0
Step by Step Macintosh Training -> 0
How to become a Powerful Speaker -> 0

The code:

if (!in_array($unit['course_name'], $courseList)) {

  $courseList[$unit['course_name']]['name'] = $unit['course_name'];
  $courseList[$unit['course_name']]['seconds'] = 0;
  echo $courseList[$unit['course_name']]['name'] . ' -> ' . $courseList[$unit['course_name']]['seconds'];
  echo "<BR>";

}

But it outputs:

Digital Photography -> 0
Step by Step Macintosh Training -> 0
Step by Step Macintosh Training -> 0
Step by Step Macintosh Training -> 0
How to become a Powerful Speaker -> 0
How to become a Powerful Speaker -> 0

Here is the var_dump($unit) :

array(8) { ["author_name"]=> string(10) "tuiuiu_dev" [0]=> string(10) "tuiuiu_dev" ["course_name"]=> string(19) "Digital Photography" [1]=> string(19) "Digital Photography" ["unit_id"]=> string(3) "181" [2]=> string(3) "181" ["unit_quantity"]=> string(1) "1" [3]=> string(1) "1" } 
array(8) { ["author_name"]=> string(15) "William Merussi" [0]=> string(15) "William Merussi" ["course_name"]=> string(31) "Step by Step Macintosh Training" [1]=> string(31) "Step by Step Macintosh Training" ["unit_id"]=> string(3) "227" [2]=> string(3) "227" ["unit_quantity"]=> string(1) "1" [3]=> string(1) "1" } 
array(8) { ["author_name"]=> string(15) "William Merussi" [0]=> string(15) "William Merussi" ["course_name"]=> string(31) "Step by Step Macintosh Training" [1]=> string(31) "Step by Step Macintosh Training" ["unit_id"]=> string(3) "231" [2]=> string(3) "231" ["unit_quantity"]=> string(1) "1" [3]=> string(1) "1" } 
array(8) { ["author_name"]=> string(15) "William Merussi" [0]=> string(15) "William Merussi" ["course_name"]=> string(31) "Step by Step Macintosh Training" [1]=> string(31) "Step by Step Macintosh Training" ["unit_id"]=> string(3) "233" [2]=> string(3) "233" ["unit_quantity"]=> string(1) "1" [3]=> string(1) "1" } 
array(8) { ["author_name"]=> string(10) "tuiuiu_dev" [0]=> string(10) "tuiuiu_dev" ["course_name"]=> string(32) "How to become a Powerful Speaker" [1]=> string(32) "How to become a Powerful Speaker" ["unit_id"]=> string(4) "1080" [2]=> string(4) "1080" ["unit_quantity"]=> string(1) "1" [3]=> string(1) "1" } 
array(8) { ["author_name"]=> string(10) "tuiuiu_dev" [0]=> string(10) "tuiuiu_dev" ["course_name"]=> string(32) "How to become a Powerful Speaker" [1]=> string(32) "How to become a Powerful Speaker" ["unit_id"]=> string(4) "1084" [2]=> string(4) "1084" ["unit_quantity"]=> string(1) "1" [3]=> string(1) "1" } 

Thank you for any help!

Answer 1

in_array() method check array values, not keys. From your example I can see that as value you have an array with name and seconds fields. I think you want to check if that id exists in your array.

So this if :

if (!in_array($unit['course_name'], $courseList)) {

Should be change like this:

if (!isset($courseList[$unit['course_name']])) {

This way you check if the $unit['course_name'] is in your array as key.

Answer 2

If you have an array with only values you should use in_array function as this example:

$array = array("A", "B", "C");
in_array($string, $array) ? "found" : "not found";

Else if you have an array with keys you should use isset function as this example:

$array = array("A" => "Result A", "B" => "Result B", "C" => "Result C");
isset($array[$string]) ? "found" : "not found";