使用 PHP 基于键连接两个关联数组
Join two associative array based on key using PHP
我这里有两个数组:
<?php
$array1 = array(
array('id' => 1, 'name' => 'John Doe', 'age' => 20),
array('id' => 2, 'name' => 'Mae Doe', 'age' => 17),
array('id' => 3, 'name' => 'Mark Smith', 'age' => 35),
array('id' => 4, 'name' => 'Selena Smith', 'age' => 15),
array('id' => 5, 'name' => 'Shane Doe', 'age' => 26),
array('id' => 6, 'name' => 'Will Smith', 'age' => 45)
)
$array2 = array(
array('id' => 1, 'address' => 'Singapore'),
array('id' => 4, 'address' => 'Japan'),
array('id' => 5, 'address' => 'Korea')
)
?>
我想加入他们,有点左加入mysql所以结果会是这样
$result = array(
array('id' => 1,'name' => 'John Doe','age' => 20, 'address' => 'Singapore'),
array('id' => 2,'name' => 'Mae Doe','age' => 17),
array('id' => 3,'name' => 'Mark Smith','age' => 35),
array('id' => 4,'name' => 'Selena Smith','age' => 15, 'address' => 'Japan'),
array('id' => 5,'name' => 'Shane Doe','age' => 26, 'address' => 'Korea'),
array('id' => 6,'name' => 'Will Smith','age' => 45)
)
地址会根据id合并到$array1
这可以使用 PHP 吗?请帮忙。谢谢
你可以通过id
值使用array_column
到re-index$array2
来简化这个问题;那么这只是一个检查 $array1
中的 id
值是否存在数据的问题:
$array1 = array(
array('id' => 1, 'name' => 'John Doe', 'age' => 20),
array('id' => 2, 'name' => 'Mae Doe', 'age' => 17),
array('id' => 3, 'name' => 'Mark Smith', 'age' => 35),
array('id' => 4, 'name' => 'Selena Smith', 'age' => 15),
array('id' => 5, 'name' => 'Shane Doe', 'age' => 26),
array('id' => 6, 'name' => 'Will Smith', 'age' => 45)
);
$array2 = array(
array('id' => 1, 'address' => 'Singapore', 'phone_number' => '123', 'store_name' => 'ABC'),
array('id' => 4, 'address' => 'Japan', 'phone_number' => '456', 'store_name' => 'DEF'),
array('id' => 5, 'address' => 'Korea', 'phone_number' => '789', 'store_name' => 'GHI')
);
$addresses = array_column($array2, null, 'id');
foreach ($array1 as &$arr) {
if (isset($addresses[$arr['id']])) $arr = array_merge($arr, $addresses[$arr['id']]);
}
print_r($array1);
输出:
Array
(
[0] => Array
(
[id] => 1
[name] => John Doe
[age] => 20
[address] => Singapore
[phone_number] => 123
[store_name] => ABC
)
[1] => Array
(
[id] => 2
[name] => Mae Doe
[age] => 17
)
[2] => Array
(
[id] => 3
[name] => Mark Smith
[age] => 35
)
[3] => Array
(
[id] => 4
[name] => Selena Smith
[age] => 15
[address] => Japan
[phone_number] => 456
[store_name] => DEF
)
[4] => Array
(
[id] => 5
[name] => Shane Doe
[age] => 26
[address] => Korea
[phone_number] => 789
[store_name] => GHI
)
[5] => Array
(
[id] => 6
[name] => Will Smith
[age] => 45
)
)
是的,可以。这只是另一个如何完成的例子:
<?php
$array1 = array(
array('id' => 1, 'name' => 'John Doe', 'age' => 20),
array('id' => 2, 'name' => 'Mae Doe', 'age' => 17),
array('id' => 3, 'name' => 'Mark Smith', 'age' => 35),
array('id' => 4, 'name' => 'Selena Smith', 'age' => 15),
array('id' => 5, 'name' => 'Shane Doe', 'age' => 26),
array('id' => 6, 'name' => 'Will Smith', 'age' => 45)
);
$array2 = array(
array('id' => 1, 'address' => 'Singapore'),
array('id' => 4, 'address' => 'Japan'),
array('id' => 5, 'address' => 'Korea')
);
// The address will be merged to $array1 based on the id.
array_walk($array1, function(&$item) use ($array2) {
array_walk($array2, function($arr2) use (&$item) {
if($item['id'] == $arr2['id']) {
$item['address'] = $arr2['address'];
}
});
});
print_r($array1);
?>
我这里有两个数组:
<?php
$array1 = array(
array('id' => 1, 'name' => 'John Doe', 'age' => 20),
array('id' => 2, 'name' => 'Mae Doe', 'age' => 17),
array('id' => 3, 'name' => 'Mark Smith', 'age' => 35),
array('id' => 4, 'name' => 'Selena Smith', 'age' => 15),
array('id' => 5, 'name' => 'Shane Doe', 'age' => 26),
array('id' => 6, 'name' => 'Will Smith', 'age' => 45)
)
$array2 = array(
array('id' => 1, 'address' => 'Singapore'),
array('id' => 4, 'address' => 'Japan'),
array('id' => 5, 'address' => 'Korea')
)
?>
我想加入他们,有点左加入mysql所以结果会是这样
$result = array(
array('id' => 1,'name' => 'John Doe','age' => 20, 'address' => 'Singapore'),
array('id' => 2,'name' => 'Mae Doe','age' => 17),
array('id' => 3,'name' => 'Mark Smith','age' => 35),
array('id' => 4,'name' => 'Selena Smith','age' => 15, 'address' => 'Japan'),
array('id' => 5,'name' => 'Shane Doe','age' => 26, 'address' => 'Korea'),
array('id' => 6,'name' => 'Will Smith','age' => 45)
)
地址会根据id合并到$array1
这可以使用 PHP 吗?请帮忙。谢谢
你可以通过id
值使用array_column
到re-index$array2
来简化这个问题;那么这只是一个检查 $array1
中的 id
值是否存在数据的问题:
$array1 = array(
array('id' => 1, 'name' => 'John Doe', 'age' => 20),
array('id' => 2, 'name' => 'Mae Doe', 'age' => 17),
array('id' => 3, 'name' => 'Mark Smith', 'age' => 35),
array('id' => 4, 'name' => 'Selena Smith', 'age' => 15),
array('id' => 5, 'name' => 'Shane Doe', 'age' => 26),
array('id' => 6, 'name' => 'Will Smith', 'age' => 45)
);
$array2 = array(
array('id' => 1, 'address' => 'Singapore', 'phone_number' => '123', 'store_name' => 'ABC'),
array('id' => 4, 'address' => 'Japan', 'phone_number' => '456', 'store_name' => 'DEF'),
array('id' => 5, 'address' => 'Korea', 'phone_number' => '789', 'store_name' => 'GHI')
);
$addresses = array_column($array2, null, 'id');
foreach ($array1 as &$arr) {
if (isset($addresses[$arr['id']])) $arr = array_merge($arr, $addresses[$arr['id']]);
}
print_r($array1);
输出:
Array
(
[0] => Array
(
[id] => 1
[name] => John Doe
[age] => 20
[address] => Singapore
[phone_number] => 123
[store_name] => ABC
)
[1] => Array
(
[id] => 2
[name] => Mae Doe
[age] => 17
)
[2] => Array
(
[id] => 3
[name] => Mark Smith
[age] => 35
)
[3] => Array
(
[id] => 4
[name] => Selena Smith
[age] => 15
[address] => Japan
[phone_number] => 456
[store_name] => DEF
)
[4] => Array
(
[id] => 5
[name] => Shane Doe
[age] => 26
[address] => Korea
[phone_number] => 789
[store_name] => GHI
)
[5] => Array
(
[id] => 6
[name] => Will Smith
[age] => 45
)
)
是的,可以。这只是另一个如何完成的例子:
<?php
$array1 = array(
array('id' => 1, 'name' => 'John Doe', 'age' => 20),
array('id' => 2, 'name' => 'Mae Doe', 'age' => 17),
array('id' => 3, 'name' => 'Mark Smith', 'age' => 35),
array('id' => 4, 'name' => 'Selena Smith', 'age' => 15),
array('id' => 5, 'name' => 'Shane Doe', 'age' => 26),
array('id' => 6, 'name' => 'Will Smith', 'age' => 45)
);
$array2 = array(
array('id' => 1, 'address' => 'Singapore'),
array('id' => 4, 'address' => 'Japan'),
array('id' => 5, 'address' => 'Korea')
);
// The address will be merged to $array1 based on the id.
array_walk($array1, function(&$item) use ($array2) {
array_walk($array2, function($arr2) use (&$item) {
if($item['id'] == $arr2['id']) {
$item['address'] = $arr2['address'];
}
});
});
print_r($array1);
?>