How to return value of an array inside a loop and use it outside in PHP

Question

I have a code like this: Lets assume that this arrays has this values:

$arr1 = array();
$arr2 = array();
$result = array();

$arr1[] = array( 'grade' => [1,2,3,4] );
$arr2[] = array( 'grade' => [1,2,3,4] );

foreach($arr1 as $a1){
$set1 = $a1['grade'];
    foreach($arr2 as $a2){
    $set2 = $a2['grade'];
    }
$result[] = array('show_result' => $set1+$set2);
}

foreach{$result as $res){
    echo $res['show_result'];
}

The output of the array $res['show_result'] must be:

2, 4, 6, 8

But I get the wrong addition of this arrays. Help will be much appreciated.

Your first error is on line 3: ' should be ;. Clearly the code you actually have has different errors than what you've posted here. We can help you only if the code you post here has the same errors as the code you're asking about. — Joni
– Joni, Commented Mar 15, 2019 at 1:04
That is my actual code sir. All I wanted to do is how to display an summed array — Mark C
– Mark C, Commented Mar 19, 2019 at 23:36

PHPnoob · Accepted Answer · 2019-03-15 01:23:52Z

1

As Joni said, your first error is on line 3: ' should be ;
Then, you're not filling arrays like you wanted : array( 'grade' => 1,2,3,4 ); creates an array with first key is 'grade' with value '1', then second key is '0' with value '2' etc...
Your last foreach loop has a syntax error similar to your first error.

See a working correction here

$arr1 = array();
$arr2 = array();
$result = array();

array_push($arr1, 1, 2, 3, 4); //fill array with 4 values (integers)
array_push($arr2, 1, 2, 3, 4); //fill array with 4 values (integers)

//so $arr1 & $arr2 are now a 4 elements arrays

$length = count($arr1); //size of array, here 4 

for ($i = 0; $i < $length; $i++) { //loop over arrays
    array_push($result, ($arr1[$i] + $arr2[$i])); //fill the results array with sum of the values from the same position 
}

var_dump($result);

answered Mar 15, 2019 at 1:23

PHPnoob

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2 Comments

Mark C Over a year ago

Got it sir. But I have another problem. How do I use the $result can be shown in table <td> ? Because the var_dump shows result not applicable in table <td>

PHPnoob Over a year ago

$result is an array, so simply loop over it with foreach ($result as $sum) { echo $sum; }

waterloomatt · Accepted Answer · 2019-03-15 01:19:31Z

1

You have quite a few syntax errors in your code.

Although this solution works, the idea behind using the same counter, $i, to extract a value from both arrays is brittle. For example, you'll get an Undefined offset if the first array has 5 grades instead of 4. If you take a step back and explain your problem in the larger context, perhaps we can provide a better solution. I get the sneaking suspicion you're asking an XY Problem.

http://sandbox.onlinephpfunctions.com/code/bb4f492c183fcde1cf4edd50de7ceebf19fe343a

<?php

$gradeList1 = ['grade' => [1,2,3,4]];
$gradeList2 = ['grade' => [1,2,3,4]];
$result = [];

for ($i = 0; $i < count($gradeList1['grade']); $i++) {

    $first = $gradeList1['grade'][$i];
    $second = $gradeList2['grade'][$i];

    $result['show_result'][] = (int)$first + (int)$second;
}

var_dump($result);

answered Mar 15, 2019 at 1:19

waterloomatt

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2 Comments

Mark C Over a year ago

Got it sir. But I have another problem. How do I use the $result can be shown in table <td> ? Because the var_dump shows result not applicable in table <td>

Mark C Over a year ago

I got it sir. Thank you so much

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