Question

我想将所有从MySQL数据库返回的值解析为JSON文件。所以我这样写我的php：

foreach ($Posts->Show() as $Key => $Post) {
    $Data ['POST'][$Post['Post_Id']]["Id"]=$Post['Post_Id'];
    $Data ['POST'][$Post['Post_Id']]["Url"] = "https://eastcloud.ir/Blog?=" . $Post['Post_Url'];
    $Data ['POST'][$Post['Post_Id']]["Name"] = $Post['Post_Name'];
    $Data ['POST'][$Post['Post_Id']]["Date"] = $Post['Post_Date'];
    $Data ['POST'][$Post['Post_Id']]["Author"] = $Post['Post_Author'];
    $Data ['POST'][$Post['Post_Id']]["Image"] = "https://eastcloud.ir/" . $Post['Post_Image'];
    $Data ['POST'][$Post['Post_Id']]["Tags"] = "https://eastcloud.ir/" . $Post['Post_Image'];
    $ID++;
}

给定的json是这样的：

POST
17
{
    Id  "17"
    Url "https://eastcloud.ir/Blog?=Simplist-Documentations"
    Name    "Simplist Documentations"
    Date    "23 May, 2018"
    Author  "Milad"
    Image   "https://eastcloud.ir/Content/Shared/Posts/SimplistV2.png"
    Tags    "https://eastcloud.ir/Content/Shared/Posts/SimplistV2.png"
}

循环将继续获取值并将其解析为JSON（可使用here的JSON文件），但是如果我想在Json2Sharp上使用给定的JSON，它将告诉给定了无效的JSON文件。解决办法是什么？如何使用PHP从数据库值创建JSON？

Answer 1

使用json_encode创建json。这不是创建它的最佳方法手动。

要了解输出的问题，正确的数据JSON格式应如下所示：

{
    "Id":"17",
    "Url":"https://eastcloud.ir/Blog?=Simplist-Documentations",
    "Name":"Simplist Documentations",
    "Date":"23 May, 2018",
    "Author":"Milad",
    "Image":"https://eastcloud.ir/Content/Shared/Posts/SimplistV2.png",
    "Tags":"https://eastcloud.ir/Content/Shared/Posts/SimplistV2.png"
}

每行而不是最后一行之后，请仔细查看引号和双点和逗号。

Answer 2

您尝试使用json_encode()方法吗？

foreach ($Posts->Show() as $Key => $Post) {
    $Data['POST'][] = [
        $Post['Post_Id'] = [
            "Id" => $Post['Post_Id'],
            "Url" => "https://eastcloud.ir/Blog?=" . $Post['Post_Url'],
            "Name" => $Post['Post_Name'],
            "Date" => $Post['Post_Date'],
            "Author" => $Post['Post_Author'],
            "Image" => "https://eastcloud.ir/" . $Post['Post_Image'],
            "Tags" => "https://eastcloud.ir/" . $Post['Post_Image'],
        ],
    ];
    $ID++;
}

// here is your json from db fetching loop
$json = json_encode($Data);

更新1.0

请检查我根据您生成的JSON

{
    "POST" : [
        {
        "Id": "17",
        "Url": "https://eastcloud.ir/Blog?=Simplist-Documentations",
        "Name": "Simplist Documentations",
        "Date": "23 May, 2018",
        "Author": "Milad",
        "Image": "https://eastcloud.ir/Content/Shared/Posts/SimplistV2.png",
        "Tags": "https://eastcloud.ir/Content/Shared/Posts/SimplistV2.png"
        },
        {
        "Id": "18",
        "Url": "https://eastcloud.ir/Blog?=angular-2-vs-react",
        "Name": "انگیولار ۲ در مقابل ری‌اکت",
        "Date": "23 May, 2018",
        "Author": "Milad",
        "Image": "https://eastcloud.ir/Content/Shared/Posts/432644.png",
        "Tags": "https://eastcloud.ir/Content/Shared/Posts/432644.png"
        }
    ]
}

来自json2csharp的响应

public class POST
{
    public string Id { get; set; }
    public string Url { get; set; }
    public string Name { get; set; }
    public string Date { get; set; }
    public string Author { get; set; }
    public string Image { get; set; }
    public string Tags { get; set; }
}

public class RootObject
{
    public List<POST> POST { get; set; }
}

Answer 3

您可以使用数据库中的json_encode（）数据

如何为API调用创建json文件？

3 个答案: