Question

我今天对阵列太累了 - 把我扔到了这个地方。

所以，这是数组的输出：

Array
(
[2010091907] => Array
    (
        [home] => Array
            (
                [score] => Array
                    (
                        [1] => 7
                        [2] => 17
                        [3] => 10
                        [4] => 7
                        [5] => 0
                        [T] => 41
                    )

                [abbr] => ATL
                [to] => 2
            )

Array
(
[2010091909] => Array
    (
        [home] => Array
            (
                [score] => Array
                    (
                        [1] => 7
                        [2] => 17
                        [3] => 10
                        [4] => 7
                        [5] => 0
                        [T] => 41
                    )

                [abbr] => ATL
                [to] => 2
            )

Array
(
[2010091901] => Array
    (
        [home] => Array
            (
                [score] => Array
                    (
                        [1] => 7
                        [2] => 17
                        [3] => 10
                        [4] => 7
                        [5] => 0
                        [T] => 41
                    )

                [abbr] => ATL
                [to] => 2
            )

我将编写一个preg_match来迭代每个[2010091907]，但在我可以之前，我不明白如何获取这些信息，或者如何调用它。我会做的事情如下：

$json=json_decode($data,true);
foreach ($json['dont-know-what-to-call-it'] as $key => $value) {
            echo "Key: ".$key."; Value: ".$value."<br />";
}

我只是不知道如何调用那些[2010091901]块中的每一个，就像我想要称之为的名称一样。我知道如何将这些东西称为得分，因为它被命名为“得分”，数据就在所有这些之下。我不知道如何获取数组的初始“部分”的键/值。最后，我想要抓住每个[2010091901]，操纵/使用[2010091901]中每一个之间的数据，然后转到下一个“记录”。

Answer 1

$date_keys = array_keys($json)会给(0 => 2010091907, 1 => 2010091909, ...)。那么你可以做到

foreach ($date_keys as $d) {
   foreach ($json[$d] as $key => $value) {
...

另外，如果你实际上不需要外部数组的索引（日期值 - 2010091907等），那么

foreach ($json as $j) {
   foreach ($j as $key => $value) {
...

忽略$json

的键

Answer 2

你不能只嵌套foreach()吗？

foreach($jsondata as $somedate => $value){
    //do you actually need $somedate?
    foreach($value['home']['score'] as $score){
        echo $score.PHP_EOL;
    }
}

Answer 3

你可以做到

$json = json_decode($data, true);
foreach($json as $ymd => $data)
{
    // $ymd = [2010091907, 2010091909,… ]
    // $data is the array starting with the key home. so $data['home']['score'][1] = 7 for the first iteration
}

这回答你的问题？你所要求的并不是100％清楚

PHP数组处理第一个孩子

3 个答案: