<?php
$test = '{
    "100": {
        "name": "Sports",
        "contentID": "100"
    },
    "200": {
        "name": "Village",
        "contentID": "200"
    }
}';
$idWiseData = json_decode($test,true);


$test2 = '[
    {
        "contentID": "100",
        "contentStatus": "active"
    },
    {
        "contentID": "200",
        "contentStatus": "active"
    },
    {
        "contentID": "300",
        "contentStatus": "active"
    }
]';
$allTopics = json_decode($test2,true);

foreach ($allTopics as $key => &$topic) {
        $contentInfo = [
            'contentStatus'     => $topic['contentStatus']
        ];
$topic['contentName']  = isset($idWiseData[$topic['contentID']]['name']) ? $idWiseData[$topic['contentID']]['name'] : null;
}
echo "<pre>";
print_r($allTopics);
?>

上面的代码工作正常，我没有得到预期的输出。实际上$allTopics有 3 个对象（contentID 100 & 200 & 300）。$idWiseData具有对象（contentID 100 和 200）。我想从$idWiseDataname中获取值并根据 contentID 替换为$allTopics 。contentID 300 没有名字所以不应该出现这个对象。

预计输出

 Array
(
    [0] => Array
        (
            [contentID] => 100
            [contentStatus] => active
            [contentName] => Sports
        )

    [1] => Array
        (
            [contentID] => 200
            [contentStatus] => active
            [contentName] => Village
        )

)

我得到输出

 Array
(
    [0] => Array
        (
            [contentID] => 100
            [contentStatus] => active
            [contentName] => Sports
        )

    [1] => Array
        (
            [contentID] => 200
            [contentStatus] => active
            [contentName] => Village
        )

    [2] => Array
        (
            [contentID] => 300
            [contentStatus] => active
            [contentName] => 
        )

)

请任何人更新我的代码。

解决方案

据我了解，如果是，则需要unset()数组中的索引。这可以通过使用unset()来实现contentNamenull

foreach ($allTopics as $key => &$topic) {
    // your code here ....

    if (is_null($topic['contentName'])) {
        unset(allTopics[$key]);
    }
}