将二数组转换成字符，并将其变成字符 列出所有，有序唯一的组合

Array
(
[time] => Array
(
[0] => 1
[1] => 3
[2] => 5
)

[people] => Array
    (
        [0] => 2
        [1] => 4
        [2] => 3
    )

[start] => Array
    (
        [0] => 1
        [1] => 3
    )

[other] => Array
    (
        [0] => 2
        [1] => 3
    )
……

)
把这样的数组变成，规则1：有序-由上到下的顺序，2：唯一 最后的结果是唯一不重复的

time:1;people:2;start:1;other:2;

time:1;people:4;start:1;other:2;

time:1;people:3;start:1;other:2;

time:1;people:2;start:3;other:2;

time:1;people:4;start:3;other:2;

time:1;people:4;start:3;other:2;

time:1;people:2;start:1;other:3;

time:1;people:4;start:1;other:3;

time:1;people:3;start:1;other:3;

time:1;people:2;start:3;other:3;

time:1;people:4;start:3;other:3;

time:1;people:4;start:3;other:3;
……

http://www.cnblogs.com/majiang/p/5990632.html

这个应该有36中变化,最笨的方法,无线枚举就可以了,4层for循环
for(int i=0;i<time.length;i++)
for(int j=0;j<people.length;j++)
for(int m=0;m<start.length;m++)
for(int n=0;n<other.length;n++)
System.out.println("time:"+i+";people:"+j+";start:"+m+";other:"+n+";");

你好，这里直接上代码了

public  function combination($arr) {
    $data = array();
    foreach ($arr as $key => $val) {
        $data = $this->_combination($data, $key, $val);
    }
    return $data;
}

public function _combination($data, $key, $arr)
{
    $_data = array();
    if (!$data) {
        //第一次组合
        foreach ($arr as $val) {
            $_data[] = "{$key}:{$val};";
        }
    } else {
        foreach ($data as $val) {
            foreach ($arr as $v) {
                $_data[] = $val . "{$key}:{$v};";
            }
        }
    }
    return $_data;
}

public function actionTestData() {
    $arr = array(
        'time'   => array(1,3,5),
        'people' => array(2,4,3),
        'start'  => array(1,3),
        'other'  => array(2,3),
    );

    $data = $this->combination($arr);
    echo json_encode($data);
}


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