CODEWITHSUNDEEP
php
Alfred
Alfred

Reputation: 1063

PHP: array_push in loop results only the last record repeat

I thought I have some basic PHP skills, at least enough to do the development for simple applications, but I found myself to be stuck with very basic array operation problem.

I have results of curl operation set on $results variable. For easier code, (blame me it is too primitive.), I have used the basic loop to loop through $results variable to produce the favored results for my next operation.

$data = array();
for ($i = 0; $i < count($results); $i++) {
    $row = $results[$i];
    $atom->objectId = $row['objectId'];
    $atom->username = $row['username'];
    array_push($data, $atom);
}
var_dump(json_encode($data));

Sample data:

{ 
  [0] => { ["objectId"]: "12345", ["username"]: "user1" },
  [1] => { ["objectId"]: "12567", ["username"]: "user2" },
  [2] => { ["objectId"]: "12789", ["username"]: "user3" },
  [3] => { ["objectId"]: "13579", ["username"]: "user4" }
}

You can simply figure out the expected output, but the real output turned out to be:

[
  { "objectId": "13579", "username": "user4" }, 
  { "objectId": "13579", "username": "user4" },
  { "objectId": "13579", "username": "user4" },
  { "objectId": "13579", "username": "user4" }
]

Any idea why this is giving me ridiculous result? Any help would be appreciated. Thanks in advance.

Upvotes: 2

Views: 1895

Answers (2)

Akshay Kalose
Akshay Kalose

Reputation: 777

What your code is doing right now:

Every loop, it is changing the properties of one object, and appending the object's reference (pass by value, but the value is the reference.) to the $data array, so you have 4 spots in the $data array pointing to only one object. Each time you loop, you change the value, the last time it is assigned is on the last value. This is why all your values are the same.

What you want to do:

Create a new object every iteration with the data and point to the new object each time.

Other solution:

If you just want an array containing the values and not a reference to an object(like your output), you could try this:

$data = array();
for ($i = 0; $i < count($results); $i++) {
    $row = $results[$i];
    $data[$i]['objectId'] = $row['objectId'];
    $data[$i]['username'] = $row['username'];
}
var_dump(json_encode($data));

Upvotes: 5

Jeroen
Jeroen

Reputation: 16865

Akshay2598's answer might work out, but the explanation isn't quite right. This has nothing to do with pass-by-value or pass-by-reference: no variables are being passed.

The reason you are getting the same results is indeed related to references. When you initially create an object, the $atom variable isn't set to the object itself, but to a reference to that object, a memory address.

What you basically are doing, is changing the properties of the object $atom is pointing to, and later add that reference to that object to your array.

Because you never assign a new object to $atom, it'll keep pointing to the same object. Because it'll keep pointing to the same object, you will keep adding more references to the same object to your array.

To fix this you can simply do: 

$data = array();
for ($i = 0; $i < count($results); $i++) {
    $atom = new Atom(); // Add this bit to assign a new object for $atom to point to
    $row = $results[$i];
    $atom->objectId = $row['objectId'];
    $atom->username = $row['username'];
    array_push($data, $atom);
}
var_dump(json_encode($data));

Upvotes: 1

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