好的我有这个函数(我得到了这个question的答案),它合并了一个数组,如下所示:
function readArray( $arr, $k, $default = 0 ) {
return isset( $arr[$k] ) ? $arr[$k] : $default ;
}
function merge( $arr1, $arr2 ) {
$result = array() ;
foreach( $arr1 as $k => $v ) {
if( is_numeric( $v ) ) {
$result[$k] = (int)$v + (int) readArray( $arr2, $k ) ;
} else {
$result[$k] = merge( $v, readArray($arr2, $k, array()) ) ;
}
}
return $result ;
}
$basketA = array( "fruit" => array(), "drink" => array() ) ;
$basketA['fruit']['apple'] = 1;
$basketA['fruit']['orange'] = 2;
$basketA['fruit']['banana'] = 3;
$basketA['drink']['soda'] = 4;
$basketA['drink']['milk'] = 5;
$basketB = array( "fruit" => array(), "drink" => array() ) ;
$basketB['fruit']['apple'] = 2;
$basketB['fruit']['orange'] = 2;
$basketB['fruit']['banana'] = 2;
$basketB['drink']['soda'] = 2;
$basketB['drink']['milk'] = 2;
$basketC = merge( $basketA, $basketB ) ;
print_r( $basketC ) ;
Array
(
[fruit] => Array
(
[apple] => 3
[orange] => 4
[banana] => 5
)
[drink] => Array
(
[soda] => 6
[milk] => 7
)
)
确定这适用于1个漏洞我无法弄清楚如何修复: 如果$ arr1缺少$ arr2所具有的东西,它应该只使用$ arr2中的值,而是将它们全部省略:
$basketA = array( "fruit" => array(), "drink" => array() ) ;
$basketA['fruit']['apple'] = 1;
$basketA['fruit']['orange'] = 2;
$basketA['fruit']['banana'] = 3;
$basketA['drink']['milk'] = 5;
$basketB = array( "fruit" => array(), "drink" => array() ) ;
$basketB['fruit']['apple'] = 2;
$basketB['fruit']['orange'] = 2;
$basketB['fruit']['banana'] = 2;
$basketB['drink']['soda'] = 2;
$basketB['drink']['milk'] = 2;
$basketC = merge( $basketA, $basketB ) ;
print_r( $basketC ) ;
Array
(
[fruit] => Array
(
[apple] => 3
[orange] => 4
[banana] => 5
)
[drink] => Array
(
[milk] => 7
)
)
注意新的数组中没有[soda],因为第一个数组没有它。
我该如何解决这个问题?
感谢!!!
答案 0 :(得分:2)
快速修复,将merge()
功能更改为如下所示:
function merge( $arr1, $arr2 ) {
$result = array() ;
foreach( $arr1 as $k => $v ) {
if( is_numeric( $v ) ) {
$result[$k] = (int)$v + (int) readArray( $arr2, $k ) ;
} else {
$result[$k] = merge( $v, readArray($arr2, $k, array()) ) ;
}
}
foreach( $arr2 as $k => $v ) {
if( is_numeric( $v ) ) {
$result[$k] = (int)$v + (int) readArray( $arr1, $k ) ;
} else {
$result[$k] = merge( $v, readArray($arr1, $k, array()) ) ;
}
}
return $result ;
}
输出:
Array
(
[fruit] => Array
(
[apple] => 3
[orange] => 4
[banana] => 5
)
[drink] => Array
(
[soda] => 2
[milk] => 7
)
)
值得注意的是array_merge_recursive()
单独表现几乎相同:
$basketC = array_merge_recursive($basketA, $basketB);
输出:
Array
(
[fruit] => Array
(
[apple] => Array
(
[0] => 1
[1] => 2
)
[orange] => Array
(
[0] => 2
[1] => 2
)
[banana] => Array
(
[0] => 3
[1] => 2
)
)
[drink] => Array
(
[milk] => Array
(
[0] => 5
[1] => 2
)
[soda] => 2
)
)
因此,如果您想知道$basketC
中有多少橙子,您只需要这样做:
array_sum($basketC['fruit']['orange']); // 4
这样您就不需要使用任何hackish,slow和未经证实的自定义函数。