Convert.ToDateTime函数出错，错误为“”；字符串未被识别为有效的DateTime“；

本文关键字：识别 DateTime 有效 字符串 函数 ToDateTime 出错 错误 Convert | 更新日期: 2023-09-27 17:57:26

这是我的代码：

private void GenerateExcelDataToday()
{
    try
    {
        // Setting path for the Excel File
        string path = System.IO.Path.GetFullPath(Server.MapPath("~/Closure.xlsx"));
        if (Path.GetExtension(path) == ".xls")
        {
            oledbConn = new OleDbConnection("Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" + path + ";Extended Properties='"Excel 8.0;HDR=Yes;IMEX=2'"");
        }
        else if (Path.GetExtension(path) == ".xlsx")
        {
            oledbConn = new OleDbConnection(@"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=" + path + ";Extended Properties='Excel 12.0;HDR=YES;IMEX=1;';");
        }
        oledbConn.Open();
    }
    catch (Exception ex)
    {
        lblError.Text = ex.ToString();
    }
    finally
    {
        oledbConn.Close();
    }
    ClosureReport();
}
private void ClosureReport()
{
    OleDbCommand cmd = new OleDbCommand();
    OleDbDataAdapter oleda = new OleDbDataAdapter();
    DataSet ds = new DataSet();
    cmd.Connection = oledbConn;
    cmd.CommandType = CommandType.Text;
    cmd.CommandText = "SELECT * FROM [Data$]";
    oleda = new OleDbDataAdapter(cmd);
    oleda.Fill(ds);
    grvData.DataSource = ds.Tables[0].DefaultView;
    grvData.DataBind();
    grvData.Visible = false;
    Day();
}
private void Day()
{
    DateTime Date = DateTime.Today;
    double G;
    for (int i = 0; i < grvData.Rows.Count; i++)
    {
        if (grvData.Rows[i].Cells[22].Text == "AU")
        {
           G = (Date - Convert.ToDateTime(grvData.Rows[i].Cells[3].Text)).TotalDays;
           AUDay1.Text = Convert.ToString(G);
        }
    }
}

单元格3数据似乎有错误。我想将字符串值转换为DateTime，但它显示错误：String was not recognized as a valid DateTime.

excel数据的格式如下：4/1/2014上午1:16:32

我该如何解决这个问题？

Convert.ToDateTime函数出错，错误为“”；字符串未被识别为有效的DateTime“；

由于您的grvData.Rows[i].Cells[3].Text是4/1/2014 1:16:32 AM，因此它没有标准的日期和时间格式。这就是Convert.ToDateTime方法抛出FormatException的原因。

可以使用DateTime.TryParseExact或DateTime.ParseExact方法解析自定义格式的字符串。

string s = "4/1/2014 1:16:32 AM";
DateTime date;
if (DateTime.TryParseExact(s, "M/d/yyyy h:mm:ss tt",
                               CultureInfo.InvariantCulture,
                               DateTimeStyles.None, out date))
{
     Console.WriteLine(date);
}
else
{
     Console.WriteLine("Your string is not valid.");
}

输出将为；

4/1/2014 1:16:32 AM

此处为demonstration。

有关更多信息，请查看；

  • 自定义日期和时间格式字符串
 G = (Date - DateTime.ParseExact(grvData.Rows[i].Cells[3].Text).ToString().Trim(), "M/d/yyyy h:mm:ss tt", CultureInfo.InvariantCulture).TotalDays;

你可以试试这个：

string strDate = "4/1/2014 1:16:32 AM";
DateTime datDate;
if (DateTime.TryParseExact(strDate, new string[] { "M/d/yyyy h:m:s tt" },
                        System.Globalization.CultureInfo.InvariantCulture,
                        System.Globalization.DateTimeStyles.None, out datDate))
{
    Console.WriteLine(datDate);
}

根据您的代码：

 ..................................................
    for (int i = 0; i < grvData.Rows.Count; i++)
    {
     if (grvData.Rows[i].Cells[22].Text == "AU")
     {
     string strDate = grvData.Rows[i].Cells[3].Text;
     DateTime presenetDate;           
     if (DateTime.TryParseExact(strDate, new string[] { "MM/d/yyyy h:m:s tt" },
                            System.Globalization.CultureInfo.InvariantCulture,
                            System.Globalization.DateTimeStyles.None, out datDate))
     {
       G = (Date - presenetDate).TotalDays;
           AUDay1.Text = Convert.ToString(G);
     }
     else
     {
       AUDay1.Text = "Not a Valid date";
     }
   .......................................
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