Php 通过键连接2个数组

假设我有两个数组：

$arr1 = array (
        array (
                "animal" => "cat",
                "color" => "grey" 
        ),
        array (
                "animal" => "dog",
                "color" => "black" 
        ),
        array (
                "animal" => "mouse",
                "color" => "white" 
        ) 
);

$arr2 = array (
        array (
                "title" => "mouse",
                "name" => "Abby" 
        ),
        array (
                "title" => "dog",
                "name" => "Rex" 
        ),
        array (
                "title" => "dog",
                "name" => "Barney" 
        ),
        array (
                "title" => "mouse",
                "name" => "Georgie" 
        ) 
);

我现在想通过键（

animal

from

$arr1

和

title

from

$arr2

）加入它们，这样我就有了：

$arr1 = array (
        array (
                "animal" => "cat",
                "color" => "grey" 
        ),
        array (
                "animal" => "dog",
                "color" => "black",
                "name" => "Rex",
                "name" => "Barney" 
        ),
        array (
                "animal" => "mouse",
                "color" => "white",
                "name" => "Abby",
                "name" => "Georgie" 
        ) 
);

因此，只需从

$arr2

添加

狗

和

鼠标

的名称即可

我的解决办法是：

foreach ( $arr1 as $key1 => $entry1 ) {
    foreach ( $arr2 as $key2 => $entry2 ) {
        if ($entry1 ["animal"] == $entry2 ["title"]) {
            $arr1 [$key1] ["name"] = $entry2 ["name"];
        }
    }
}

但它会覆盖第一个条目，因此我只能得到姓氏：

$arr1 = array (
        array (
                "animal" => "cat",
                "color" => "grey" 
        ),
        array (
                "animal" => "dog",
                "color" => "black",
                "name" => "Barney"
        ),
        array (
                "animal" => "mouse",
                "color" => "white",
                "name" => "Georgie"
        ) 
);

---

我查看了，但它只是合并了数组，而没有比较它们的键。

所以名称索引必须是数组：）

尝试如下

foreach ( $arr1 as $key1 => $entry1 ) {
    foreach ( $arr2 as $key2 => $entry2 ) {
        if ($entry1 ["animal"] == $entry2 ["title"]) {
            if (!isset($arr1[$key1]["name"])) {
                $arr1[$key1]["name"] = array();
            }
            $arr1 [$key1]["name"][] = $entry2 ["name"];
        }
    }
}

输出

$arr1 = array (
            array (
                    "animal" => "cat",
                    "color" => "grey" 
            ),
            array (
                    "animal" => "dog",
                    "color" => "black",
                    "name" => ["Rex", "Barney" ] 
            ),
            array (
                    "animal" => "mouse",
                    "color" => "white",
                    "name" => ["Abby", "Georgie"]
            ) 
);

---

在地图中不能多次使用同一关键点。Halayem Anis的解决方案建议在单个“name”键后面使用一个数组，这是一个很好的解决方案因为不能有相同的密钥name@Uchiha：哇，太容易了！不客气。很高兴它对您有所帮助。这是一个很好的解决方案，既适用于使用数组，也适用于在尚未设置密钥时使用正确的初始化。