douxian6086 2016-07-15 10:55
浏览 15
已采纳

如何获取数组根并添加到数组子

How do I take the root of an array and then add into the value taken his child to become an array of array data

I have an array data like :

array(10) {
  ["2016-07-06"]=>
  array(11) {
    ["nb_uniq_visitors"]=>
    int(795)
    ["nb_users"]=>
    int(0)
    ["nb_visits"]=>
    int(896)
    ["nb_actions"]=>
    int(3134)
  }
  ["2016-07-07"]=>
  array(11) {
    ["nb_uniq_visitors"]=>
    int(878)
    ["nb_users"]=>
    int(0)
    ["nb_visits"]=>
    int(987)
    ["nb_actions"]=>
    int(3975)
  }
}

From the data array above I would like to take in the array of arrays where I want to combine the date data into their array.

Let's just above my input array into a single variable. Call it $allData. Then I loop

foreach ($allData as $data) {
    echo '<pre>';
    var_dump($data);
}

From the results above loop I get a new array data like

array(11) {
    ["nb_uniq_visitors"]=>
    int(795)
    ["nb_users"]=>
    int(0)
    ["nb_visits"]=>
    int(896)
    ["nb_actions"]=>
    int(3134)
}
array(11) {
    ["nb_uniq_visitors"]=>
    int(878)
    ["nb_users"]=>
    int(0)
    ["nb_visits"]=>
    int(987)
    ["nb_actions"]=>
    int(3975)
}

If I want to take the data that previously used the root array array_push(), how to take and put it to use array_push()?

Formed later so that the end result as

array(11) {
    ["date_visitor"]=>
    string(20) "2016-07-06"
    ["nb_uniq_visitors"]=>
    int(795)
    ["nb_users"]=>
    int(0)
    ["nb_visits"]=>
    int(896)
    ["nb_actions"]=>
    int(3134)
}
array(11) {
    ["date_visitor"]=>
    string(20) "2016-07-07"
    ["nb_uniq_visitors"]=>
    int(878)
    ["nb_users"]=>
    int(0)
    ["nb_visits"]=>
    int(987)
    ["nb_actions"]=>
    int(3975)
}

Using what functions to take root array, then I will use array_push to add the data into an array of child?

Thanks

  • 写回答

1条回答 默认 最新

  • dongyang0005 2016-07-15 10:59
    关注

    A simple foreach loop should do the job:

    $new=[];
    foreach($allData as $key=>$value){
        $value['date_visitor']=$key;
        $new[]=$value;
    }
    
    var_dump($new);
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?