PHP数组：条件和

Question

我有一个如下所示的数组：

$array = array("a1" => 0, "a2" => 2, "a3" => 1, "b1" => 2, "b2" => 3);

我想在仅考虑第一个字符时对每个 "unique" 键的值求和。结果应该是：

$newarray = array("a" => 3, "b" => 5);

我试过在另一个 foreach() 循环中使用一个 foreach() 循环，如下所示：

foreach ($xml->children() as $output) {
  foreach ($array as $key => $value) {
    if (substr($key,0,1) == $output->KEY) {
      $sum += $value; echo $sum;
    }
  }
}

它没有起作用，因为结果显然添加了先前的计算。

Answer 1

一个简单的 isset 应该可以做到：

$array = array("a1" => 0, "a2" => 2, "a3" => 1, "b1" => 2, "b2" => 3);
$result = array();
foreach ($array as $oldkey => $val) {
    $newkey = substr($oldkey, 0, 1);
    if (isset($result[$newkey]) === false)
        $result[$newkey] = $val;
    else
        $result[$newkey] += $val;
}
var_dump($result);

Answer 2

你可以试试这个。它将从您的密钥中获取第一个字母，并使用该第一个字母密钥对所有值求和。

<?php
$sum = [];
foreach ($array as $key => $value) {
    $letter = substr($key,0,1);
    if (!isset($sum[$letter])){$sum[$letter] = 0;}
    $sum[$letter] += $value;
}

var_dump($sum);

Answer 3

试试这个方法

$array = array('a1' => 0, 'a2' => 2, 'a3' => 3, 'b1' => 2, 'b2' => 3);
$result = array();
foreach($array as $key => $value){
    if(isset($result[$key[0]])){
        $result[$key[0]] = $result[$key[0]]+$value;
    } else {
        $result[$key[0]] = $value;
    }
}
print_r($result);

Answer 4

简单快捷，您可以在数组中的字符后添加任意数量的数字。

<?php

$array = ["a1" => 0, "a2" => 2, "a3" => 1, "b1" => 2, "b2" => 3];
$newArray = [];

foreach ($array as $key => $value) {
  $key = preg_replace("/[0-9]*$/", "", $key);
  $newArray[$key] = (isset($newArray[$key])) ? $newArray[$key] + $value : $value;
}

print_r($newArray);

Answer 5

简单的解决方案：

$final_array=[];
foreach($array as $key=>$value){
     $final_array[$key[0]] =  (isset($final_array[$key[0]])) ? $final_array[$key[0]]+$value : $value;
 }
 print_r($final_array);

输出：- https://3v4l.org/tZ4Ei

PHP数组：条件和

PHP array: Conditional sum

php

arrays

calculation