Question

我希望从列表中获取最长日期＆lt;＆gt;。

但我不知道如何将x.time转换为DateTime。

它总是在异常中显示如下错误： String was not recognized as a valid DateTime。

这是我的代码：

List<DTOSaveFromFile> lst = Load();

public static List<DTOSaveFromFile> Load()
{
   string[] data = File.ReadAllLines(dataPath);
   return (from i in data select new DTOSaveFromFile(i)).ToList<DTOSaveFromFile>();
}

foreach (var rows in lst)
{
    DateTime biggest = lst
        Select(x => DateTime.ParseExact(x.time, "d/M/yyyy", System.Globalization.CultureInfo.InvariantCulture)) //to get the list of `string` of `time` from your List<DTOSaveFromFile> + //To convert it to DateTime 
        .Max(); //Get the max among the DateTime
}

来自班级的

x.time：

public class DTOSaveFromFile
{
    public string reportName { get; set; }
    public string eMail { get; set; }
    public string time { get; set; }
}

如何解决此问题？感谢。

Answer 1

添加DateTime.ParseExact的格式以处理05-01-2016案例

string[] formats = new string[] {"d/M/yyyy", "d-M-yyyy"}; //notice the dash

然后你的查询就像：

string[] formats = new string[] {"d/M/yyyy", "d-M-yyyy"}; //notice the dash
DateTime biggest = lst
        .Select(x => DateTime.ParseExact(x.time, formats, //now use formats here
                    System.Globalization.CultureInfo.InvariantCulture, 
                    System.Globalization.DateTimeStyles.AssumeLocal)) //to get the list of `string` of `time` from your List<DTOSaveFromFile> + //To convert it to DateTime 
        .Max(); //Get the max among the DateTime

请注意，DateTime.ParseExact中的所有参数都可能需要System.Globalization.DateTimeStyles.AssumeLocal，format。

此外，每当您因新format而发现错误时，只需在formats数组中添加<?php header('Content-type: application/json'); $status = array( 'type'=>'success', 'message'=>'Thank you for contact us. As early as possible we will contact you ' ); $name = @trim(stripslashes($_POST['name'])); $email = @trim(stripslashes($_POST['email'])); $subject = @trim(stripslashes($_POST['subject'])); $message = @trim(stripslashes($_POST['message'])); $email_from = $email; $email_to = 'reflexivezhcet@gmail.com';//replace with your email $body = 'Name: ' . $name . "\n\n" . 'Email: ' . $email . "\n\n" . 'Subject: ' . $subject . "\n\n" . 'Message: ' . $message; $success = @mail($email_to, $subject, $body, 'From: <'.$email_from.'>'); echo json_encode($status); die;

无法将字符串转换为DateTime

1 个答案: