php - count elements in array

Question

I am trying to count elements in an array, but it doens't work as intended:

I have a while loop, which loops through my user table:

while($refsData=$refs->fetch()){
    $new_array = array($refsData['id']);
    print_r($new_array);

        $outcome = $rentedrefs->_paying($new_array);
    }

The print_r($new_array); gives me:

Array
(
    [0] => 90427
)
Array
(
    [0] => 90428
)
Array
(
    [0] => 90429
)
Array
(
    [0] => 90430
)
Array
(
    [0] => 90431
)
Array
(
    [0] => 90432
)
Array
(
    [0] => 90433
)
Array
(
    [0] => 90434
)
Array
(
    [0] => 90435
)
Array
(
    [0] => 90436
)

Inside the _paying function, I count the number of values from the array:

function _paying($referrals_array){

echo count($referrals_array);


}

The problem is, that the above count($referrals_array); just gives me: 1, when it should be 10

What am I doing wrong?

On each iteration of your while loop you're overwriting $new_array. — user1864610
– user1864610, Commented Nov 2, 2014 at 12:24
Your array is only returning one element. Look at your output. Each array is Array[0]. — user3542456
– user3542456, Commented Nov 2, 2014 at 12:24

raina77ow · Accepted Answer · 2014-11-02 12:23:48Z

2

You create a new array at each step of the loop. Instead it should be written like this:

$new_array = array();
while($refsData=$refs->fetch()){
    $new_array[] = $refsData['id'];
    // print_r($new_array);
}

$outcome = $rentedrefs->_paying($new_array);

Note that I moved the _paying call outside the loop, as it seems to be the aggregating function. If not, you'd most probably make it process $refsData['id'] instead - not the whole array.

As a sidenote, I'd strongly recommend using fetchAll() method (instead of fetch when you need to fill a collection with results of a query. It'll be trivial to count the number of the resulting array.

answered Nov 2, 2014 at 12:23

raina77ow

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2 Comments

oliverbj Over a year ago

The while loop I showed in the question, is actually a nested loop. Meaning there is another while loop outside that loop. Will that be a problem, if I am placing the $outcome outside of that nested loop, but in the first loop?

raina77ow Over a year ago

You didn't show the outer loop, but the way I see it, the code I've shown is a direct replacement for what you have at the moment.

andy · Accepted Answer · 2014-11-02 12:24:14Z

1

You are creating $new_array as a new array with the single element $refsData['id']. The count of 1 is therefore correct.

To get the number of results, either use a COUNT(*) select to ask your sql server, or add a counter to your loop, like this:

$entries = 0;
while($refsData=$refs->fetch()){
    $new_array = array($refsData['id']);
    print_r($new_array);
    $entries++;

    $outcome = $rentedrefs->_paying($new_array);
}

echo $entries;

answered Nov 2, 2014 at 12:24

andy

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4 Comments

oliverbj Over a year ago

This gives me "100" when I echo $entries

andy Over a year ago

Then you have 100 results, I guess.

oliverbj Over a year ago

there is only 10 rows in the table.

andy Over a year ago

Since i don't know your query, I can only guess. You could be generating a much larger number of results by joining the table with itself.

Sean Doe · Accepted Answer · 2014-11-02 12:23:52Z

0

It works properly, in each circulation of loop you have one array, so in first you have:

Array
(
    [0] => 90427
)

in 2nd:

 Array
(
    [0] => 90428
)

and so on.

It should work properly:

var $count = 0;
while($refsData=$refs->fetch()){
    $new_array = array($refsData['id']);
        $count += count($new_array);    
        $outcome = $rentedrefs->_paying($new_array);
    }

answered Nov 2, 2014 at 12:23

Sean Doe

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Comments

Deimoks · Accepted Answer · 2014-11-02 12:24:58Z

0

You are not adding elements to an array, but creating a new array each iteration. To add elements, just do:

$new_array[] = $refsData['id'];

answered Nov 2, 2014 at 12:24

Deimoks

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