php foreach 二维数组的问题

Question

数据库中查询出的结果是二维数组，然后foreach后出现以下结果
foreach：

foreach ($users as $k=>$v){
        var_dump($v);
}

结果：

  array(12) {["id"]=> string(1) "7" ["name"]=> string(9) "小红" ["province"]=> string(6) "四川"   ["activation"]=> int(5) ["money"]=> string(5) "25.00" ["count"]=> int(1000) ["total"]=> string(1) "4" ["coltd"]=> string(1) "2" ["station"]=> string(1) "0" ["totalMoney"]=> string(8) "13001.01" } 
  array(12) { ["id"]=> string(1) "9" ["name"]=> string(9) "蒋磊"  ["province"]=> string(6) "浙江"   ["activation"]=> int(10) ["money"]=> string(5) "50.00" ["count"]=> int(1000)["total"]=> string(1) "3" ["coltd"]=> string(1) "2" ["station"]=> string(1) "0" ["totalMoney"]=> string(4) "0.51" } 
  array(12) { ["id"]=> string(1) "3" ["name"]=> string(9) "魏天" ["province"]=> string(6) "天津"   ["activation"]=> int(0) ["money"]=> string(4) "0.00" ["count"]=> int(0) ["total"]=> string(1) "4" ["coltd"]=> string(1) "2" ["station"]=> string(1) "2" ["totalMoney"]=> string(4) "0.44" } 
  array(12) { ["id"]=> string(1) "1"  ["name"]=> string(9) "张小明"  ["province"]=> string(6) "广东"   ["activation"]=> int(1) ["money"]=> string(4) "5.00" ["count"]=> int(1000) ["total"]=> string(1) "1" ["coltd"]=> string(1) "2" ["station"]=> string(1) "0" ["totalMoney"]=> string(4) "0.04" } 
  array(12) { ["id"]=> string(2) "12" ["name"]=> string(9) "张小明"  ["province"]=> string(6) "湖北"   ["activation"]=> int(1) ["money"]=> string(4) "5.00" ["count"]=> int(1000) ["total"]=> string(1) "1" ["coltd"]=> string(1) "0" ["station"]=> string(1) "0" ["totalMoney"]=> string(4) "0.04" } 
  array(12) { ["id"]=> string(2) "26" ["name"]=> string(9) "张小明" ["province"]=> string(6) "北京"   ["activation"]=> int(1) ["money"]=> string(4) "5.00" ["count"]=> int(1000) ["total"]=> string(1) "1" ["coltd"]=> string(1) "0" ["station"]=> string(1) "7" ["totalMoney"]=> string(4) "0.04" } }

想得到的结果是：将名字【name】相同的人的对应的值相加，但是总的人数不变【数组的元素个数不变】。[例：张小明的数据， 将三个元素中的["activation"]，["money"]，["count"]，["total"]，["totalMoney"]的值相加]
想要得到以下的结果：

  array(12) { ["id"]=> string(1) "7" ["name"]=> string(9) "小红" ["province"]=> string(6) "四川"   ["activation"]=> int(5) ["money"]=> string(5) "25.00" ["count"]=> int(1000) ["total"]=> string(1) "4" ["coltd"]=> string(1) "2" ["station"]=> string(1) "0" ["totalMoney"]=> string(8) "13001.01" } 
  array(12) { ["id"]=> string(1) "9" ["name"]=> string(9) "蒋磊"  ["province"]=> string(6) "浙江"   ["activation"]=> int(10) ["money"]=> string(5) "50.00" ["count"]=> int(1000)["total"]=> string(1) "3" ["coltd"]=> string(1) "2" ["station"]=> string(1) "0" ["totalMoney"]=> string(4) "0.51" } 
  array(12) { ["id"]=> string(1) "3" ["name"]=> string(9) "魏天" ["province"]=> string(6) "天津"   ["activation"]=> int(0) ["money"]=> string(4) "0.00" ["count"]=> int(0) ["total"]=> string(1) "4" ["coltd"]=> string(1) "2" ["station"]=> string(1) "2" ["totalMoney"]=> string(4) "0.44" } 
  array(12) { ["id"]=> string(1) "1"  ["name"]=> string(9) "张小明"  ["province"]=> string(6) "广东"   ["activation"]=> int(3) ["money"]=> string(4) "15.00" ["count"]=> int(3000) ["total"]=> string(1) "3" ["coltd"]=> string(1) "2" ["station"]=> string(1) "0" ["totalMoney"]=> string(4) "0.12" } 
  array(12) { ["id"]=> string(2) "12" ["name"]=> string(9) "张小明"  ["province"]=> string(6) "湖北"   ["activation"]=> int(3) ["money"]=> string(4) "15.00" ["count"]=> int(3000) ["total"]=> string(1) "3" ["coltd"]=> string(1) "0" ["station"]=> string(1) "0" ["totalMoney"]=> string(4) "0.12" } 
  array(12) { ["id"]=> string(2) "26" ["name"]=> string(9) "张小明" ["province"]=> string(6) "北京"   ["activation"]=> int(3) ["money"]=> string(4) "15.00" ["count"]=> int(3000) ["total"]=> string(1) "3"  ["coltd"]=> string(1) "0" ["station"]=> string(1) "7" ["totalMoney"]=> string(4) "0.12" } }

Answer 1

//数组自己补吧，不想写那么多了
$arr=[['id'=>7,'name'=>'php','activation'=>1,'money'=>10,'province'=>'北京'],
       ['id'=>1,'name'=>'js','activation'=>12,'money'=>80,'province'=>'广东'],
       ['id'=>10,'name'=>'js','activation'=>2,'money'=>20,'province'=>'天津']
      ];

$res=[];
foreach($arr as $k=>$v){
    if(isset($res[$v['name']])){
        // 自己补剩下字段
        $res[$v['name']]['activation']+=$v['activation'];
        $res[$v['name']]['money']+=$v['money'];
    } else {
       $res[$v['name']] = $v;
    }
}
foreach($res as $k=>$v){
     foreach($arr as $key=>&$value){
        if($v['name']==$value['name']){
            // 自己补剩下字段
            $value['activation']=$v['activation'];
            $value['money']=$v['money'];
        }
     }
}
echo '<pre>';print_r($arr);
Array
(
    [0] => Array
        (
            [id] => 7
            [name] => php
            [activation] => 1
            [money] => 10
            [province] => 北京
        )

    [1] => Array
        (
            [id] => 1
            [name] => js
            [activation] => 14
            [money] => 100
            [province] => 广东
        )

    [2] => Array
        (
            [id] => 10
            [name] => js
            [activation] => 14
            [money] => 100
            [province] => 天津
        )

)

Answer 2

sum(activation),sum(money),sum(count),sum(total),sum(totalMoney).....group by name

Answer 3

能做，有点麻烦。

但更好的办法是：
通过数据库查询的时候直接计算出所需要的值。
像楼上说的：用group by name
出来后，当然，相同名字的只会出现一行，如果你真的需要多行的话，你可以另外再查询一次:
select id,name from table;
然后循环把两次的结果结合起来处理。

然后：
我还是不明白，什么时候需要把相同名字的求和，又要显示多条数据？