Php 如果信息存在于两个数组之一中，如何不显示信息

情况是，我有两个数组通过API收集JSON数据：

$players  = getAPI("http://xx.xxx.xxx.xx:xxxxx/players.json?apiKey=xxxxxxxxxxxxxxxxxxxxxxxx");
$recents  = getAPI("xx.xxx.xxx.xx:xxxxx/recent.json?apiKey=xxxxxxxxxxxxxxxxxxxxxxxx");

方法是获取内容并将JSON解码为数组。 对于玩家数组，我们在数组中有以下数据：

美元玩家

[
  {
    "id": "76561198033377272",
    "name": "PitMonk",
    "position": {
      "x": -339,
      "y": 26,
      "z": 191
    },
    "rotation": 128,
    "time": 418310,
    "ip": "",
    "inventory": {
      "main": [],
      "belt": [
        {
          "name": "rock",
          "amount": 1,
          "blueprint": false,
          "condition": 100
        },
        {
          "name": "torch",
          "amount": 1,
          "blueprint": false,
          "condition": 100
        }
      ],
      "wear": []
    }
  },
  {
    "id": "76561198088638439",
    "name": "Pippa",
    "position": {
      "x": -337,
      "y": 25,
      "z": 177
    },
    "rotation": 73,
    "time": 419136,
    "ip": "",
    "inventory": {
      "main": [
        {
          "name": "arrow.wooden",
          "amount": 12,
          "blueprint": false
        },
        {
          "name": "bow.hunting",
          "amount": 1,
          "blueprint": false,
          "condition": 93
        },
        {
          "name": "blueprint_fragment",
          "amount": 25,
          "blueprint": false
        },
        {
          "name": "metal.fragments",
          "amount": 1366,
          "blueprint": false
        },
        {
          "name": "metal.refined",
          "amount": 48,
          "blueprint": false
        },
        {
          "name": "charcoal",
          "amount": 1120,
          "blueprint": false
        },
        {
          "name": "lowgradefuel",
          "amount": 738,
          "blueprint": false
        }
      ],
      "belt": [
        {
          "name": "rock",
          "amount": 1,
          "blueprint": false,
          "condition": 100
        },
        {
          "name": "torch",
          "amount": 1,
          "blueprint": false,
          "condition": 100
        },
        {
          "name": "pickaxe",
          "amount": 1,
          "blueprint": false,
          "condition": 76
        },
        {
          "name": "pickaxe",
          "amount": 1,
          "blueprint": false,
          "condition": 17
        },
        {
          "name": "pickaxe",
          "amount": 1,
          "blueprint": false,
          "condition": 100
        },
        {
          "name": "pickaxe",
          "amount": 1,
          "blueprint": false,
          "condition": 100
        }
      ],
      "wear": [
        {
          "name": "burlap.shirt",
          "amount": 1,
          "blueprint": false
        },
        {
          "name": "attire.hide.skirt",
          "amount": 1,
          "blueprint": false
        }
      ]
    }
  }
]

$recents

[
  {
    "id": "76561198039206786",
    "name": "JakeGroves"
  },
  {
    "id": "76561198088638439",
    "name": "Pippa"
  },
  {
    "id": "76561198033377272",
    "name": "PitMonk"
  },
  {
    "id": "76561198146864439",
    "name": "YepWellDone"
  },
  {
    "id": "76561198164836207",
    "name": "Baz"
  },
  {
    "id": "76561198076406281",
    "name": "xwalnutx"
  },
  {
    "id": "76561197985716090",
    "name": "Darkflame134"
  },
  {
    "id": "76561198263423842",
    "name": "XitaikiznerX"
  },
  {
    "id": "76561198129952244",
    "name": "NatanGamer"
  },
  {
    "id": "76561198071842055",
    "name": "Baha Bey"
  }
]

正如你所看到的，玩家是有联系的人，而recents是最近有联系的人的总列表

我尝试过这样做：

 foreach ($players as $player) {
       echo $players->name;
    }
    echo "</br></br>";
    foreach ($recent as $rec) {
      if ($rec->name != $player->name) {
       echo $rec->name . "</br>";
     }
    }

---

因此，它只是忽略了“pippa”，我不确定是否可以与两个数组进行交互以获得唯一值？

您感兴趣的是列出

$users

中不存在于

$players

或

$recents

中的所有名称，对吗

假设您只对名称感兴趣：

// First, let's get a new array with all names from both arrays (can contain dups)
$pcNames = array_map($players + $recents, function($playerObject) {
    return $playerObject->name;
});

// Next, let's remove all dups
$pcNames = array_unique($pcNames);

// === At this point you have an array with all names from `$players` and `$recents` ===
// === You may do something else with those, but I'll now create another array with  ===
// === all users not in the players/recents lists.                                   ===

// Now let's also get a list of names of users in the `$users` variable
$userNames = array_map($users, function($playerObject) {
    return $playerObject->name;
});

// And finally let's get all names which are not in players or recents
$diffNames = array_diff($userNames, $pcNames);

// Let's output those to see whether it worked
var_dump($diffNames);

---

当然，根据您的用例，还有其他方法。例如，我们可以提取所有三个数组的名称，然后只使用带有3个参数的

array_diff

（但是您没有

$pcArray

副产品），或者如果您确实想要比较ID，但打印名称，我们必须更改所有内联函数以提取ID而不是名称，并进一步向下引用用户数组以获取实际名称，等等。

您的算法似乎还可以，到底是什么问题？输出看起来很奇怪，我不明白它怎么能在同一行显示两个玩家的名字，因为你在每个名字后面都回显一个br。“Pitmank Pippa”如何显示在同一行上？也许你应该在ID上匹配用户，这比比较名字更干净。@OlivierH yaa我正在使用名字，这样我就可以知道它现在是谁，如果你在第一行看到Pitmank和Pippa都在$players数组中，它们都在$recents数组中，但是我要做的是遍历$recents数组，忽略$players数组中存在的任何条目什么是

$users

数组？你的代码中根本没有使用

$players

。我写错了$users=$players。。。。我已经做了这个代码，我认为它是工作

foreach（$rec）{foreach（$users as$user）{if（$user->name==$rec->name）{break；}}}if（$user->name！=$rec->name）{echo$rec->name；}}

为什么使用中间数组

$userNames

？如果他的用户数据量很大，这可能会消耗大量内存。如果在

$pcNames

中找不到

$user->name

，则只在

$users

数组上循环，并为另一个数组提供数据，会更有效；你说得对，它可能更有效。请随意在回答中发表您的想法。我的答案是尝试使用数组函数将其分解为多个步骤，这些函数描述了它们在做什么，这样数据的“流”就更清晰了。

// First, let's get a new array with all names from both arrays (can contain dups)
$pcNames = array_map($players + $recents, function($playerObject) {
    return $playerObject->name;
});

// Next, let's remove all dups
$pcNames = array_unique($pcNames);

// === At this point you have an array with all names from `$players` and `$recents` ===
// === You may do something else with those, but I'll now create another array with  ===
// === all users not in the players/recents lists.                                   ===

// Now let's also get a list of names of users in the `$users` variable
$userNames = array_map($users, function($playerObject) {
    return $playerObject->name;
});

// And finally let's get all names which are not in players or recents
$diffNames = array_diff($userNames, $pcNames);

// Let's output those to see whether it worked
var_dump($diffNames);