Question

我有一组由于mysql表的查询而生成的数组，在“while”循环中生成：

while ($row = mysql_fetch_array($result)) {
    $k_id =  $row['kw'];
    $path[$level] = $k_id;

    if(array_search('7', $path)) {
        $it = new CachingIterator(new ArrayIterator($path));
        $t1 = array();
        foreach ( $it as $v ) {
            $var = array("no" => $v,"to" => 0);
            if ($it->hasNext()) {
                $var["to"] = $it->getInnerIterator()->current();
            }
            $t1[] = $var;
        }

        extract($t1, EXTR_PREFIX_ALL, "t1");

        for ($i = 0; $i < sizeof($t1); $i++) {
            $dmp = print_r($t1[$i], result);
            echo $dmp."<br />";
        }
    }     
}

echo $ dmp的结果是一组数组：

Array ( [no] => 2 [to] => 1 )
Array ( [no] => 1 [to] => 4 )
Array ( [no] => 4 [to] => 7 )
Array ( [no] => 7 [to] => 0 )
Array ( [no] => 2 [to] => 1 )
Array ( [no] => 1 [to] => 5 )
Array ( [no] => 5 [to] => 7 )
Array ( [no] => 7 [to] => 0 )
Array ( [no] => 2 [to] => 1 )
Array ( [no] => 1 [to] => 6 )
Array ( [no] => 6 [to] => 7 )
Array ( [no] => 7 [to] => 0 )
Array ( [no] => 2 [to] => 1 )
Array ( [no] => 1 [to] => 3 )
Array ( [no] => 3 [to] => 7 )
Array ( [no] => 7 [to] => 0 )

我的问题是如何使用相同的“no”键值合并数组

mysql_fetch之前的var_dump：

array(0) { } array(1) { [0]=> string(1) "2" } array(2) { [0]=> string(1) "2" [1]=> string(1) "1" } array(3) { [0]=> string(1) "2" [1]=> string(1) "1" [2]=> string(1) "4" }

因此，我想获得数组，我可以轻松地从以下代码生成这样的代码：

 "nodeTo": "4", 
    "nodeFrom": "1"

"nodeTo": "5", 
    "nodeFrom": "1"

"nodeTo": "6", 
    "nodeFrom": "1"

"nodeTo": "3", 
    "nodeFrom": "1"

    "id": "1"

Answer 1

您可以替换

 $t1[] = $var;

用

$t1[$v][] = $var;

它将根据ID进行分组...有关如何输出结果的示例：

foreach ( $t1 as $no => $group ) {
    echo "Merge $no <br />" ;
    foreach ( $group as $list ) {
        var_dump($list);
    }
}

Answer 2

我真的建议你在从sql中获取结果时直接创建正确的数组结构。这将为您节省额外的foreach循环。

为了完整，我为你准备了一个例子，你怎么能像你想象的那样做:(为了好玩;））

$a = array (
    array ('no' => 2, 'to' => 1), //  |___ will be merged
    array ('no' => 2, 'to' => 1), //  |
    array ('no' => 2, 'to' => 2),
    array ('no' => 3, 'to' => 2)
);

$result = array();
foreach($a as $item) {
    if(!array_key_exists($item['no'], $result)) {
        $result[$item['no']] = array($item['to']);
    } else if(!in_array($item['to'], $result[$item['no']], TRUE)) {
        $result[$item['no']][] = $item['to'];
    }
}

// get all to values with no=2 :
$elements = $result[2];

转换后的结果数组的var_dump可能有助于理解发生了什么：

//
var_dump($result);

/* Output:

array(2) {
  [2] =>  <---- 'no' value as index
  array(2) {
    [0] =>    -
    int(1)     |__ different 'to' values
    [1] =>     |
    int(2)    -
  }
  [3] => <---- 'no' value as index
  array(1) {
    [0] =>
    int(2)   <------ 'to' value for 'no'==3
  }
}

*/

正如我所说，你应该在从sql中获取结果时直接创建这个数组结构。不幸的是，如果没有$path的内容，我就无法说明如何实现这一目标。

Answer 3

foreach ($t1 as $dmp) {
    $combined[$dmp["no"]][] = $dmp["to"];
}

结果：

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

    [1] => Array
        (
            [0] => 4
            [1] => 5
            [2] => 6
            [3] => 3
        )

    [4] => Array
        (
            [0] => 7
        )

    [7] => Array
        (
            [0] => 0
            [1] => 0
            [2] => 0
            [3] => 0
        )

    [5] => Array
        (
            [0] => 7
        )

    [6] => Array
        (
            [0] => 7
        )

    [3] => Array
        (
            [0] => 7
        )

)

合并循环中的数组

3 个答案: