Question

我正在开发Web API，我在返回数据时遇到错误;

无法隐式转换类型

我尝试修复错误但失败了，让我得到像你这样的伟大开发者的帮助。我会很感激。以下是代码：

var a = ['','one ','two ','three ','four ', 'five ','six ','seven ','eight ','nine ','ten ','eleven ','twelve ','thirteen ','fourteen ','fifteen ','sixteen ','seventeen ','eighteen ','nineteen '];
var b = ['', '', 'twenty','thirty','forty','fifty', 'sixty','seventy','eighty','ninety'];

function inWords (num) {
    if ((num = num.toString()).length > 9) return 'overflow';
    n = ('000000000' + num).substr(-9).match(/^(\d{2})(\d{2})(\d{2})(\d{1})(\d{2})$/);
    if (!n) return; var str = '';
    str += (n[1] != 0) ? (a[Number(n[1])] || b[n[1][0]] + '-' + a[n[1][1]]) + 'million ' : '';
    str += (n[2] != 0) ? (a[Number(n[2])] || b[n[2][0]] + '-' + a[n[2][1]]) + 'hundred ' : '';
    str += (n[3] != 0) ? (a[Number(n[3])] || b[n[3][0]] + '-' + a[n[3][1]]) + 'thousand ' : '';
    str += (n[4] != 0) ? (a[Number(n[4])] || b[n[4][0]] + '-' + a[n[4][1]]) + 'hundred ' : '';
    str += (n[5] != 0) ? ((str != '') ? 'and ' : '') + (a[Number(n[5])] || b[n[5][0]] + '-' + a[n[5][1]]) + 'only ' : '';
    return str;
}

console.log(inWords(999999999));
console.log(inWords(1095));

此声明：WHERE creation_date BETWEEN CONVERT(VARCHAR(10), @DateFrom, 103) AND CONVERT(VARCHAR(10), @DateTo, 103)给出错误，其他声明没问题。实际上它只是抱怨public IEnumerable<TABLE1> LoadSubIndustryByID(int id) { foreach (var subIndustry in Db.TABLE1.Where((u) => u.us_user_id == id)) { var childIndustry = Db.TABLE2.Where((c) => c.sci_cat_id == subIndustry.sci_cat_id); yield return childIndustry; } }。

Answer 1

当您使用TABLE2类型时，您的函数正在返回TABLE1的IEnumerable。

您应该更改功能以返回IEnumerable<TABLE2>

Answer 2

public IEnumerable<TABLE2> LoadSubIndustryByID(int id)
{
    foreach (var subIndustry in Db.TABLE1.Where((u) => u.us_user_id == id))
    {

        var childIndustry = Db.TABLE2.Where((c) => c.sci_cat_id == subIndustry.sci_cat_id);

        yield return childIndustry;
    }
}

这样可行。

显式转换错误

2 个答案: