Why does JsonConvert.DeserializeObject return an error: Exception: Value cannot be null. (Parameter 'source')

Why does JsonConvert.DeserializeObject return an error: Exception: Value cannot be null. (Parameter 'source')

我有一个 public async Task GetCompaniesAsync() 编码如下:

public async Task GetCompaniesAsync()
{
    _getCompaniesSuccessful = false;

    var result = await _http.GetAsync($"https://somesite.com/api/companies");

    if (result.IsSuccessStatusCode)
    {
        // I split the code after the suggestion of @mjwills (thanks)
        var content = await result.Content.ReadAsStringAsync();
        var obj = JsonConvert.DeserializeObject<StockMarket>(content).Companies;

        // at this point, content.Length has a value (e.g. 24429)
        // but, obj is null (after JsonConvert)

        if (obj != null)
        {
            _companies = obj.ToList();
            _getCompaniesSuccessful = true;
        }
}

在其他一些 class 中,我使用以下代码:

await MyService.GetCompaniesAsync();
                
if (MyService.GetCompaniesSuccessful)
{
    foreach (var record in MyService.Companies)
    {
        await Context.Channel.SendMessageAsync($"{record.TickerSymbol,-10}\t{record.CompanyName,-10}");
    }
}

但是当我收到以下错误时:异常:值不能为空。 (参数'source') 难道它还没有取完数据?请帮忙...

顺便说一下,webapi url returns JSON 数据,它看起来像这样:

{
    "companies": [
        {
            "ticker": "HEY",
            "name": "Hey Corporation",
            "status": "open"
        },
        {
            "ticker": "PER",
            "name": "Pears Corporation",
            "status": "close"
        },
        {
            "ticker": "BRGR",
            "name": "Burger Inc.",
            "status": "open"
        },
    ]
}           

这些是 StockMarket 和 Company classes:

public class StockMarket
{
    public StockMarket()
    {
    }

    public ICollection<Company> Companies { get; set; }
}

public class Company
{
    public Company()
    {
        TickerSymbol = "";
        CompanyName = "";
        Status = "";
    }

    public string TickerSymbol { get; set; }
    public string CompanyName { get; set; }
    public string Status { get; set; }
}

查看示例,您的 JSON 有一个“名称”键,但您的公司 class 的 属性 名为“CompanyName”。

尝试将 属性 更改为简单的名称,或者用这个

装饰它

[JsonProperty("name", NullValueHandling = NullValueHandling.Ignore)]

此外,TickerSymbol 属性也是如此。