Question

我试图将日期格式从TumblingEventTimeWindows.of(Time.minutes(5), Time.minutes(3))更改为08-28-2018 06:33 PM，以便将其存储在mysql时间戳中，但无法在PHP中完成

Answer 1

这是您的答案

echo date("Y-m-d H:i:s", strtotime(str_replace('-', '/', "08-28-2018 06:33 PM")));

注意： m / d / y或d-m-y格式的日期可通过以下方式消除歧义在各个组件之间的分隔符处：如果分隔符是斜线（/），则假定为美国m / d / y；而如果分隔符是破折号（-）或点（。），然后是欧洲d-m-y格式假设。但是，如果年份以两位数格式给出，并且分隔符是破折号（-，日期字符串被解析为y-m-d。

来源Link。

这里在工作code。

Answer 2

使用DateTime类：

// Create a PublisherService to send WCF requests to the Publisher Actor's mailbox
let publisherService = PublisherService(fun request -> publisherActor <! request)

let rec initHost () =                        
    // Create a new WCF ServiceHost for the Publisher Service
    let host = new ServiceHost(publisherService, [| config.PublisherServiceUri |])
    host.AddServiceEndpoint(typeof<IPublisherService>.FullName, NetTcpBinding(SecurityMode.None), config.PublisherServiceUri) |> ignore

    // Log any WCF Service Faulted events and try to start a new ServiceHost
    host.Faulted |> Event.add (fun _ -> 
        log.errorf "Publisher Service '%s' faulted. Restarting in 10 seconds..." config.PublisherServiceUri
        // BUG: Always fails with CommunicationObjectFaulted exception -- probably shouldn't be here
        try host.Close ()
        with | ex -> log.errorx ex "Error closing faulted Service Host"
        // TODO: Do something smarter than just sleeping for 10 seconds before restarting the service host
        Async.Sleep 10000 |> Async.RunSynchronously
        // BUG:  This isn't tail-recursive, will probably blow the stack eventually
        initHost () |> ignore)

    // Start the WCF Service Host
    host.Open()
    log.debugf "Publisher Service '%s' started." config.PublisherServiceUri

initHost()

格式为// Creating from actual format $date = DateTime::createFromFormat('m-d-Y h:i A', '08-28-2018 06:33 PM'); // Output: 2018-08-28 18:33:00 echo $date->format('Y-m-d H:i:s');的是大写AM / PM（see date parameters）。

Answer 3

引用：

DateTime :: createFromFormat ：http://php.net/manual/en/datetime.createfromformat.php

DateTime :: format ：http://php.net/manual/en/datetime.format.php

PHP代码：

<?php
$datestring = "08-28-2018 06:33 PM";
$input_format = "m-d-Y h:i A";
$output_format = 'Y-m-d G:i';
$date =  DateTime::createFromFormat($input_format, $datestring);

echo  $date->format($output_format). "\n"; //output 2018-08-28 18:33

输出：https://3v4l.org/ajfJk

Answer 4

要将08 / 28-2018 06:33 PM更改为2018-07-23 18:33:00这是由于php内置日期功能而导致的小型便捷解决方案。

date('Y-m-d H:i:s',strtotime('08-28-2018 06:33'));

结果将显示您这样

 2018-07-23 18:33:00

PHP中日期格式为AM和PM的字符串

4 个答案: