Question

下面我有2个PHP数组。 $all_tags和$list_tags

我需要迭代$all_tags数组，并为'class' => 'added'中的每个数组项添加一个键/值$all_tags，该数组项具有匹配的数组项$list_tags $all_tags 1}}数组。

在下面的示例数据中，id数组中class值为1,2和5的项目将具有名为added的值，其值为{ {1}}。

$all_tags = array();

$all_tags[] = array(
    'id' => 1,
    'title' => 'title 1',
    'description' => 'description 1'
);
$all_tags[] = array(
    'id' => 2,
    'title' => 'title 2',
    'description' => 'description 2'
);
$all_tags[] = array(
    'id' => 3,
    'title' => 'title 3',
    'description' => 'description 3'
);
$all_tags[] = array(
    'id' => 4,
    'title' => 'title 4',
    'description' => 'description 4'
);
$all_tags[] = array(
    'id' => 5,
    'title' => 'title 5',
    'description' => 'description 5'
);



$list_tags = array();
 $list_tags[] = array(
    'id' => 1,
    'title' => 'title 1',
    'description' => 'description 1'
);
$list_tags[] = array(
    'id' => 2,
    'title' => 'title 2',
    'description' => 'description 2'
);
$list_tags[] = array(
    'id' => 5,
    'title' => 'title 5',
    'description' => 'description 5'
);



foreach($all_tags as $key => $val){
    echo $val;
}

Answer 1

在您的简单情况下，使用带foreach函数的常规in_array循环就足够了：

...
foreach ($all_tags as &$v) {
    if (in_array($v, $list_tags)) {
        $v['class'] = 'added';
    }
}

print_r($all_tags);

输出：

Array
(
    [0] => Array
        (
            [id] => 1
            [title] => title 1
            [description] => description 1
            [class] => added
        )

    [1] => Array
        (
            [id] => 2
            [title] => title 2
            [description] => description 2
            [class] => added
        )

    [2] => Array
        (
            [id] => 3
            [title] => title 3
            [description] => description 3
        )

    [3] => Array
        (
            [id] => 4
            [title] => title 4
            [description] => description 4
        )

    [4] => Array
        (
            [id] => 5
            [title] => title 5
            [description] => description 5
            [class] => added
        )
)

Answer 2

我的建议是首先获取$list_tags ID，然后仅在$all_tags上进行迭代。

当数组不完全相同但id匹配时，此方法将起作用。

$list_tags_ids = array_column($list_tags, 'id');

foreach($all_tags as &$val){
    if(in_array($val['id'], $list_tags_ids)) {
        $val['class'] = 'added';
    }
}

Answer 3

我有这个方法来做所需的输出。我也总是对其他方式开放=）

$all_tags = array();

$all_tags[] = array(
    'id' => 1,
    'title' => 'title 1',
    'description' => 'description 1'
);
$all_tags[] = array(
    'id' => 2,
    'title' => 'title 2',
    'description' => 'description 2'
);
$all_tags[] = array(
    'id' => 3,
    'title' => 'title 3',
    'description' => 'description 3'
);
$all_tags[] = array(
    'id' => 4,
    'title' => 'title 4',
    'description' => 'description 4'
);
$all_tags[] = array(
    'id' => 5,
    'title' => 'title 5',
    'description' => 'description 5'
);



$list_tags = array();
 $list_tags[] = array(
    'id' => 1,
    'title' => 'title 1',
    'description' => 'description 1'
);
$list_tags[] = array(
    'id' => 2,
    'title' => 'title 2',
    'description' => 'description 2'
);
$list_tags[] = array(
    'id' => 5,
    'title' => 'title 5',
    'description' => 'description 5'
);






foreach ($all_tags as $allTagKey => $allTagElement) {
    foreach ($list_tags as $listTagKey => $listTagElement) {
        if ($allTagElement['id'] == $listTagElement['id']) {
            $all_tags[$allTagKey]['class'] = 'added';
            break; // for performance and no extra looping
        }
    }
}

echo '<pre>';
print_r($all_tags);

如果第二个数组与PHP具有匹配的ID键/值，则将数据添加到数组

3 个答案: