首页 > 解决方案 > 为什么 JsonConvert.DeserializeObject 返回错误:异常:值不能为空。(参数“来源”)

问题描述

我有一个public async Task GetCompaniesAsync()这样编码的:

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;
        }
}

在其他一些课程中,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') 会不会是数据还没有完成呢?请帮忙...

顺便说一句,webapi url 返回 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 类:

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; }
}

标签: c#.net-corejson.netdiscord.net

解决方案


查看示例,您的 JSON 有一个“名称”键,但您的公司类的属性称为“公司名称”。

尝试将属性更改为简单的名称,或者用这个来装饰它

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

此外,TickerSymbol 属性也是如此。


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