Question

您好我想删除＆＃34; news_list＆＃34;来自php json数组。

{
    "news_list": [
        {
            "id": 2,
            "group_id": 1,
            "news_title": "fbb",
            "news_description": "gfhgfh",
            "status": "Y",
            "created_at": "2017-05-11 16:04:26",
            "updated_at": "2017-05-11 16:04:26"
        },
        {
            "id": 3,
            "group_id": 1,
            "news_title": "ewrdf",
            "news_description": "dsfsdfdsfsdffffffffffffff",
            "status": "Y",
            "created_at": "2017-05-12 10:59:01",
            "updated_at": "2017-05-12 10:59:01"
        }
    ]
}

期望的输出：

[
    {
        "id": 2,
        "group_id": 1,
        "news_title": "fbb",
        "news_description": "gfhgfh",
        "status": "Y",
        "created_at": "2017-05-11 16:04:26",
        "updated_at": "2017-05-11 16:04:26"
    },
    {
        "id": 3,
        "group_id": 1,
        "news_title": "ewrdf",
        "news_description": "dsfsdfdsfsdffffffffffffff",
        "status": "Y",
        "created_at": "2017-05-12 10:59:01",
        "updated_at": "2017-05-12 10:59:01"
    }
]

是否有内置的PHP功能？如果我有任何内置功能，请转发给我。

Answer 1

试试这个......

$oldDataArray = json_decode('{"news_list":[{"id":2,"group_id":1,"news_title":"fbb","news_description":"gfhgfh","status":"Y","created_at":"2017-05-11 16:04:26","updated_at":"2017-05-11 16:04:26"},{"id":3,"group_id":1,"news_title":"ewrdf","news_description":"dsfsdfdsfsdffffffffffffff","status":"Y","created_at":"2017-05-12 10:59:01","updated_at":"2017-05-12 10:59:01"}]}');

$newDataArray = $oldDataArray->news_list;

echo '<pre>';
print_r($newDataArray);
echo '</pre>';

Answer 2

没有像直接函数来获取json的任何关键数据。您需要使用$selectall = sqlsrv_query($conn, "select * from Table where and Status = 2 or status = 6", $params, $options); while($fetchall = sqlsrv_fetch_array($selectall)) { $id = $fetchall['DATAID']; $dates = $fetchall['DATE']; if( 3 DAYS PAST ) { sqlsrv_query($conn, "UPDATE TABLE SET STATUS=5 WHERE DATAID=$ID") } }将json转换为数组或对象，然后执行或获取您要查找的内容。

json_decode

或者

$res = json_decode($data);   // return object
print_r($res->news_list);

删除数组名称的PHP数组转换

2 个答案: