Question

我有两个数组，我想匹配两个数组。如果array 1中没有array 2的值，那么我想为该元素添加0。

阵列1

$weeksArr = array("p1","p2","p3","p4");

数组2

$dailyArr = array(
    "0"=>array("p1","123"),
    "1"=>array("p2","125"),
    "2"=>array("p4","126")
);

加入后我希望最终的数组是

$finalArr = array(
    "0"=>array("p1","123"),
    "1"=>array("p2","125"),
    "2"=>array("p3","0"),
    "3"=>array("p4","126")
);

所以我想在最终数组中插入p3。

非常欢迎任何提示/建议。提前谢谢。

Answer 1

我不知道到达那里的最短途径，但你可以先得到所有的p，然后比较周。

获得差异后，循环分配，然后再次对它们进行排序。

$weeksArr = array("p1","p2","p3","p4");
$dailyArr = array("0"=>array("p1","123"),"1"=>array("p2","125"),"2"=>array("p4","126"));
$finalArr = $dailyArr;

$temp = array_map(function($piece){ return $piece[0]; }, $finalArr); // get all p's
$diff = array_diff($weeksArr, $temp); // get the missing p
foreach($diff as $w) { 
    $finalArr[] = array($w, 0); // assign missing p
}
usort($finalArr, function($a, $b){ // you do not need this is you do not care about the order
    return strcmp($a[0], $b[0]); // if you need to order them, p1, p2, p3, p4, then i guess you need to sort
});

echo '<pre>';
print_r($finalArr);

Answer 2

比我想的有点棘手，但这个简单的嵌套foreach循环将使它成为：

$weeksArr = array("p1","p2","p3","p4");
$dailyArr = array("0"=>array("p1","123"),"1"=>array("p2","125"),"2"=>array("p4","126"));

foreach ($weeksArr as $i => $value) {
  foreach ($dailyArr as $a) {
    if ($a[0] == $value) $finalArray[$i] = array($a[0],$a[1]);
  }
  if (!isset($finalArray[$i])) $finalArray[$i] = array($value,0);
}

var_dump($finalArray);

Answer 3

排序$ weeksArr
按$ dailyArr排序$ dailyArr [i] [0]＆lt; = $ dailyArr [i + 1] [0]
初始化两个变量i，j为0，$ finalArray为空数组

DO

4.a如果$ weeksArr [i]等于$ dailyArr [j] [0]，将$ dailyArr [j]添加到$ finalArr并增加i和j

4.b if $ weeksArr [i]＆lt; $ dailyArr [j] [0]，为$ weekArr [i]添加$ finalArr的新条目并增加i

取消循环，如果i或j超出范围

LOOP

如果我小于$ weekArr的长度，则将剩余的条目添加到$ finalArr

使用php组合单维和多维数组

3 个答案: