Question

我有以下数组

    foreach ($reviews as $review){

$sqlreviews[] = array (

'score' => $review['value'] + $review['location'] + $review['service'] + $review['rooms'] + $review['cleanliness'],

);

以下回声输出11 25 25 25

foreach ($sqlreviews as $review1) { 
echo"
".$review1['score']."
";
}

如何将上述输出求和并将总数86传递给变量？

Answer 1

$sum = 0;
foreach ($sqlreviews as $review1) { 
    $sum += $review1['score'];
}
echo $sum;

Answer 2

使用PHP 5.5＆＃39; array_column()功能，您可以执行以下操作：

$total = array_sum(
    array_column(
        $reviews,
        'score'
    )
);

对于早期版本的PHP，您可以执行以下操作：

$total = array_sum(
    array_map(
        function($value) {
            return $value['score'];
        },
        $reviews
    )
);

或

$total = array_reduce(
    $reviews,
    function($runningTotal, $item) {
        return $runningTotal + $item['score'];
    }
);

Answer 3

我没有真正了解您的阵列设置方式，但您可以使用array_sum函数输出数组总和：

$a = array_sum($array);
echo $a;

Answer 4

我猜你有一个像这样的数组=＆gt; [＆＃39;评论＆＃39;] =＆gt; （＆＃39;值＆＃39; =＆gt; 0，＆＃39; location＆＃39; =＆gt; 1，＆＃39; service＆＃39; =＆gt; 1，＆＃39; rooms＆＃39; =＆gt; ; 2，＆＃39; cleanlines＆＃39; =＆gt; 2）;

$sqlreviews['score'] = $arrayname['reviews']['value'] + $arrayname['reviews']['location'] + $arrayname['reviews']['service'] + $arrayname['reviews']['rooms'] + $arrayname['reviews']['cleanliness'];

如果您只有一组数据，则不需要迭代....

计算数组中的和

4 个答案: