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如何利用两个数组的key来求交集

WBOY
WBOYOriginal
2016-06-06 20:38:371084browse

<code>A:array(4) {
  ["Api/Index/index"] => string(3) "226"
  ["Main/Index/index"] => string(3) "227"
  ["Main/Index/main"] => string(3) "228"
  ["/Main/Index/test"] => string(3) "229"
}
B:array(22) {
  ["/Main/Index/test"] => int(3)
  ["/Main/Index/liebiao/text/%E5%88%97%E8%A1%A8%E6%B5%8B%E8%AF%95"] => string(7) "0-sec-0"
  ["/Main/Index/test/text/%E7%A8%8E%E6%94%B6%E5%88%86%E6%9E%902"] => string(7) "0-sec-1"
  ["/Main/Index/test/text/%E7%A8%8E%E6%94%B6%E5%88%86%E6%9E%903"] => string(7) "0-sec-2"
  ["/Main/Index/test/text/%E7%A8%8E%E6%94%B6%E6%95%B0%E6%8D%AE1"] => string(7) "1-sec-0"
  ["/Main/Index/test/text/%E7%A8%8E%E6%94%B6%E6%95%B0%E6%8D%AE2"] => string(7) "1-sec-1"
  ["/Main/Index/test/text/%E7%A8%8E%E6%94%B6%E6%95%B0%E6%8D%AE3"] => string(7) "1-sec-2"
  ["/Main/Index/test/text/%E4%BC%81%E4%B8%9A%E7%AE%A1%E7%90%861"] => string(7) "2-sec-0"
  ["/Main/Index/test/text/%E4%BC%81%E4%B8%9A%E7%AE%A1%E7%90%862"] => string(7) "2-sec-1"
  ["/Main/Index/test/text/%E4%BC%81%E4%B8%9A%E7%AE%A1%E7%90%863"] => string(7) "2-sec-2"
  ["/Main/Index/test/text/%E7%B3%BB%E7%BB%9F%E8%AE%BE%E7%BD%AE1"] => string(7) "3-sec-0"
  ["/Main/Tree/index/text/%E6%A0%91%E7%8A%B6%E7%AE%A1%E7%90%86"] => string(7) "3-sec-1"
  ["/Main/Test/index/text/%E5%88%97%E8%A1%A8%E6%B5%8B%E8%AF%95"] => string(7) "3-sec-2"
  ["/Ucenter/User/index/text/%E7%94%A8%E6%88%B7%E7%AE%A1%E7%90%86"] => string(7) "3-sec-3"
  ["/System/Config/index/text/%E9%85%8D%E7%BD%AE%E7%AE%A1%E7%90%86"] => string(7) "3-sec-4"
  ["/System/Index/index"] => int(4)
  ["/System/Subject/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-0"
  ["/System/Industry/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-1"
  ["/System/Area/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-2"
  ["/System/FlatRate/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-3"
  ["/System/Settlement/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-4"
  ["/System/Split/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-5"
}
</code>

有以上两个数组，现需要知道其中重复的有哪些
如何算是重复

第一种情况 比如a数组权限中的key为/Main/Index/test b数组菜单的key为/Main/Index/test 可算是重复的

第二种情况 比如a数组权限中的key为Main/Index/test b数组菜单的key为/Main/Index/test/text/fdsfsdfljl 也算作是重复数据

以上两种情况均算作将b数组的改key放入到c数组中作为交集使用
请问如何使用较少的时间来求这两个数组的交集。
我能想到的是foreach 进行循环 单条单条的匹配，但是当a中的规则有几千上百的时候，b中的菜单项也几十上百时，就会有较明显的卡顿现象，不知道谁有更好的方法能够减少循环次数节省运行时间的方法

回复内容：

<code>A:array(4) {
  ["Api/Index/index"] => string(3) "226"
  ["Main/Index/index"] => string(3) "227"
  ["Main/Index/main"] => string(3) "228"
  ["/Main/Index/test"] => string(3) "229"
}
B:array(22) {
  ["/Main/Index/test"] => int(3)
  ["/Main/Index/liebiao/text/%E5%88%97%E8%A1%A8%E6%B5%8B%E8%AF%95"] => string(7) "0-sec-0"
  ["/Main/Index/test/text/%E7%A8%8E%E6%94%B6%E5%88%86%E6%9E%902"] => string(7) "0-sec-1"
  ["/Main/Index/test/text/%E7%A8%8E%E6%94%B6%E5%88%86%E6%9E%903"] => string(7) "0-sec-2"
  ["/Main/Index/test/text/%E7%A8%8E%E6%94%B6%E6%95%B0%E6%8D%AE1"] => string(7) "1-sec-0"
  ["/Main/Index/test/text/%E7%A8%8E%E6%94%B6%E6%95%B0%E6%8D%AE2"] => string(7) "1-sec-1"
  ["/Main/Index/test/text/%E7%A8%8E%E6%94%B6%E6%95%B0%E6%8D%AE3"] => string(7) "1-sec-2"
  ["/Main/Index/test/text/%E4%BC%81%E4%B8%9A%E7%AE%A1%E7%90%861"] => string(7) "2-sec-0"
  ["/Main/Index/test/text/%E4%BC%81%E4%B8%9A%E7%AE%A1%E7%90%862"] => string(7) "2-sec-1"
  ["/Main/Index/test/text/%E4%BC%81%E4%B8%9A%E7%AE%A1%E7%90%863"] => string(7) "2-sec-2"
  ["/Main/Index/test/text/%E7%B3%BB%E7%BB%9F%E8%AE%BE%E7%BD%AE1"] => string(7) "3-sec-0"
  ["/Main/Tree/index/text/%E6%A0%91%E7%8A%B6%E7%AE%A1%E7%90%86"] => string(7) "3-sec-1"
  ["/Main/Test/index/text/%E5%88%97%E8%A1%A8%E6%B5%8B%E8%AF%95"] => string(7) "3-sec-2"
  ["/Ucenter/User/index/text/%E7%94%A8%E6%88%B7%E7%AE%A1%E7%90%86"] => string(7) "3-sec-3"
  ["/System/Config/index/text/%E9%85%8D%E7%BD%AE%E7%AE%A1%E7%90%86"] => string(7) "3-sec-4"
  ["/System/Index/index"] => int(4)
  ["/System/Subject/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-0"
  ["/System/Industry/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-1"
  ["/System/Area/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-2"
  ["/System/FlatRate/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-3"
  ["/System/Settlement/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-4"
  ["/System/Split/index/text/%E5%9F%BA%E7%A1%80%E8%AE%BE%E7%BD%AE"] => string(7) "4-sec-5"
}
</code>

有以上两个数组，现需要知道其中重复的有哪些
如何算是重复

第一种情况 比如a数组权限中的key为/Main/Index/test b数组菜单的key为/Main/Index/test 可算是重复的

第二种情况 比如a数组权限中的key为Main/Index/test b数组菜单的key为/Main/Index/test/text/fdsfsdfljl 也算作是重复数据

以上两种情况均算作将b数组的改key放入到c数组中作为交集使用
请问如何使用较少的时间来求这两个数组的交集。
我能想到的是foreach 进行循环 单条单条的匹配，但是当a中的规则有几千上百的时候，b中的菜单项也几十上百时，就会有较明显的卡顿现象，不知道谁有更好的方法能够减少循环次数节省运行时间的方法

有官方函数可以直接干这个事情啊... 随手写大概是这个样子的：

<code>array_intersect_uassoc($a, $b, function($ka, $kb) {
    if( preg_match("%$ka%i", $kb) ) return 0;
    return $ka - $kb;
});
</code>
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