PHP比较两个日期数组并检查是否匹配

时间:2018-09-13 09:06:34

标签: php arrays

我想要两个带有日期对象的arrays要比较。第一个数组包含今天的日期和接下来的五天:

$today = date('Y-m-d');
$fiveDays = [];
for($i=0; $i <= 5; $i++){
    $today = date('Y-m-d', strtotime('+1 day', strtotime($today)));
    $fiveDays[] = date('Y-m-d', strtotime($today));
}

给出结果:

0: "2018-09-14"
1: "2018-09-15"
2: "2018-09-16"
3: "2018-09-17"
4: "2018-09-18"
5: "2018-09-19"

另一个数组可以包含多个对象:

[{
   days: (4) ["2018-09-13", "2018-09-14", "2018-09-15", "2018-09-16"]
   duration: 4
   end: "2018-09-16"
   name: "vacation blabla"
   start: "2018-09-13"
},
{
   days: (5) ["2018-09-20", "2018-09-21", "2018-09-22", "2018-09-23", "2018-09-24"]
   duration: 5
   end: "2018-09-24"
   name: "vacation blabla"
   start: "2018-09-20"
}]

现在,我要检查第一个数组的日期/日期中的某一个是否将在Vacations_array中。我该如何实现?

编辑

找到匹配项后,必须将匹配项为true的第二个数组(带有休假期)分配给$vacation数组

3 个答案:

答案 0 :(得分:2)

使用array_intersect()

基本上,您可以在数组2的循环内执行此操作:

$array_of_same_elements = array_intersect($array_1, $array_2[$i]['days']);

$ array_of_same_elements现在将包含您要查找的日期。

对此有一个很好的描述: https://www.w3schools.com/php/func_array_intersect.asp

答案 1 :(得分:1)

您也可以这样操作:

$result_array = array();
$found = false;
for($i=0; $i<count($vacation_array); $i++) {
  $found = false;
  for($j=0; $j<count($vacation_array[$i]['days']); $j++) {
    if(in_array($vacation_array[$i]['days'][$j], $array_1) && !$found) {
      $result_array[] = $vacation_array[$i];
      $found = true;
    }
  }
}
print_r($result_array);

答案 2 :(得分:0)

这有点简单,但它可能满足您的需求:

$event1 = ['days'=>["2018-09-13", "2018-09-14", "2018-09-15", "2018-09-16"]];
$event2 = ['days'=>["2018-09-20", "2018-09-21", "2018-09-22", "2018-09-23", "2018-09-24"]];

function checkEvents($ev1, $ev2){
    foreach($ev1 AS $day){
       if(in_array($day, $ev2)){
          return $day." is found in both events\n";
       }
    }
    return "no match found\n";
}

echo checkEvents($event1['days'], $event2['days']);