Question

我试图基于许多键遍历数组和组对象。我尝试过搜索，但我认为我没有使用正确的搜索字词。

数组包含家谱数据，母亲和父亲姓名，姓氏等。 $ births数组中的一些对象显然彼此相关，因为它们具有相同的母亲，父亲，姓氏等，并且我想将它们分组为新的$系列数组。

下面概述了数据的类型以及我想要实现的目标。有关如何做到这一点的任何想法？我对如何接近它感到茫然，

$births = Array (
   [1] => Array  (
            [mother_surname] => Pig
            [mother_firstname] => Mommy
            [father_surname] => Pig
            [father_firstname] => Daddy
            [birth_date] => 2005
            [child_firstname] => Peppa
         )
   [2] => Array  (
            [mother_surname] => Pig
            [mother_firstname] => Mommy
            [father_surname] => Pig
            [father_firstname] => Daddy
            [birth_date] => 2008
            [child_firstname] => George
         )
   [3] => Array  (
            [mother_surname] => Cat
            [mother_firstname] => Mrs
            [father_surname] => Cat
            [father_firstname] => Mr
            [birth_date] => 2005
            [child_firstname] => Candy
         )
)


$families = Array (
   [1] => Array  (
              [1] => Array  (
                [mother_surname] => Pig
                [mother_firstname] => Mommy
                [father_surname] => Pig
                [father_firstname] => Daddy
                [birth_date] => 2005
                [child_firstname] => Peppa
              )
              [2] => Array  (
                [mother_surname] => Pig
                [mother_firstname] => Mommy
                [father_surname] => Pig
                [father_firstname] => Daddy
                [birth_date] => 2008
                [child_firstname] => George
              )
         )
   [2] => Array  (
            [mother_surname] => Cat
            [mother_firstname] => Mrs
            [father_surname] => Cat
            [father_firstname] => Mr
            [birth_date] => 2005
            [child_firstname] => Candy
         )
)

我最接近的是：

$shiftArray = array_shift($births);
foreach($births as $birth){
      print_r(  array_intersect($shiftArray, $birth)  );
}

但那是失败的。

Answer 1

看起来您可能需要第三个数组$ family_pattern数组。这是一些可能有帮助的伪代码

walk through births
  filter $parents_info from birth (father_surname, father_firstname, mother)surname, mother_firstname)
  if($parents_info in $family_pattern get $key)
    $families[$key][] = $birth
  else
    $family_pattern[] = $parents_info
    $key = $family_pattern count - 1 
    $families[$key][] = $birth

Answer 2

泰森，你让我走在正确的轨道上，谢谢你！这是我的解决方案，以防其他人受益。它还考虑了我的原始问题中未提及的家庭住址。

$families = array();
function searchForChildren($MSN,$MFN,$FSN,$FFN,$ADD,$array) {
    global $families;
    $stack = array();
    foreach ($array as $key => $val) {
        if ($val['Mother_Surname'] == $MSN && $val['Mother_Firstname'] == $MFN && $val['Father_Surname'] == $FSN && $val['Father_Firstname'] == $FFN && $val['Family_Address'] == $ADD) {
            $info = array(
                'Mother_Surname' => $MSN,
                'Mother_Firstname' => $MFN,
                'Father_Surname' => $FSN,
                'Father_Firstname' => $FFN,
                'Family_Address' => $ADD,
                'Birth_Year' => $val['Birth_Year'],
                'Child_Surname' => $val['Child_Surname'],
                'Child_Firstname' =>  $val['Child_Firstname']  
            );
            array_push($stack,$info );
        }
    }
   array_push($families,$stack);
}

//The below function strips out any duplicates from the parents array
function arrayUnique ( $rArray ){ 
    $rReturn = array (); 
    while ( list( $key, $val ) = each ( $rArray ) ){ 
        if ( !in_array( $val, $rReturn ) ) 
        array_push( $rReturn, $val ); 
    } 
    return $rReturn; 
} 
$uniqueParents = arrayUnique($parents);

foreach($uniqueParents as $parents){ 
    $MSN = $parents['Mother_Surname'];
    $MFN = $parents['Mother_Firstname'];
    $FSN = $parents['Father_Surname'];
    $FFN = $parents['Father_Firstname'];
    $ADD = $parents['Family_Address'];
    searchForChildren($MSN,$MFN,$FSN,$FFN,$ADD, $births);
}
print_r(json_encode($families));

PHP：基于多个键的多维数组对象

2 个答案: