Php 如何基于当前日期导入和打印CSV值？

我有一个CSV文件，它是一个带有日期、小时和姓名的轮班时间表。知道如何根据当前日期/时间提取姓名吗？CSV如下所示：

...
14 Oct 2015, 02, 12:00 - 18:00, 6, "", "John Doe", "Joe Smith"
14 Oct 2015, 03, 18:00 - 00:00, 6, "Jenny Roe", "", "Henry Smith"
15 Oct 2015, 01, 00:00 - 06:00, 6, "Jake Blake", "Bob Ford", ""
...

我需要的是运行代码并打印计划下一班的人员的姓名，如：

Jenny Roe
Henry Smith

我知道我可以像这样以数组的形式加载文件：

<?php
$csv = array();
$lines = file('schedule.csv', FILE_IGNORE_NEW_LINES);

foreach ($lines as $key => $value)
{
    $csv[$key] = str_getcsv($value);
}

echo '<pre>';
print_r($csv);
?>

$results[] = $data

...
01 Oct 2015 ,65 ,07:00 - 15:00 ,8 ,"","John Doe","Joe Smith","Martin Taylor","Henry Smith","Mike Miller"
01 Oct 2015 ,22 ,15:00 - 23:00 ,8 ,"","Bob Ford","Sarah Smith","Jack Williams","",""
01 Oct 2015 ,11 ,23:00 - 7:00 ,8 ,"","","Jenny Roe","Adam Davis","Jake Blake",""
02 Oct 2015 ,21 ,07:00 - 19:00 ,12 "Antonio Garcia","John Doe","Joe Smith","","Henry Smith","Mike Miller"
02 Oct 2015 ,22 ,19:00 - 07:00 ,12 ,"","Bob Ford","Sarah Smith","Jack Williams","",""
02 Oct 2015 ,11 ,07:00 - 15:00 ,8 ,"","","Jenny Roe","Adam Davis","Jake Blake",""
...

或将其打印为文本：

$currentDate = new DateTime(); //we create an object with the current time

$file = fopen("schedule.csv","r");

$results = array();

//we loop through the file, each line will be an array containing each fields
while (($data = fgetcsv($file, 0, ",")) !== FALSE)
{
    //we create an array that has three elements: the start time, the "-" and the endtime of the shifts
    $shiftTime = explode(" ", $data[2]); 
    //We create two datetime objects that mark the beginning of the shift and the end of the shift
    $startShift = dateTime::createFromFormat("d M Y H:i", $data[0] . " " . $shiftTime[0]);
    $endShift = dateTime::createFromFormat("d M Y H:i", $data[0] . " " . $shiftTime[2]);

    //We compare the two previously created objects to the current date and if 
    //the current date is between the start of the shift and the end of the shift, 
    //we know that person is scheduled to work.
    if($currentDate >= $startShift && $currentDate <= $endShift)
        $results[] = $data;
}

---

现在，我想知道从哪种方法开始会更好？如果有人能帮我解决这个问题，我保证我最终会学习PHP（doh！），并在这里发布回复作为证明。

我会尝试这样做：

<?php
$csv = array();
$lines = file('schedule.csv', FILE_IGNORE_NEW_LINES);

foreach ($lines as $key => $value)
{
    $csv[$key] = str_getcsv($value);
}

echo '<pre>';
print_r($csv);
?>

$results[] = $data

...
01 Oct 2015 ,65 ,07:00 - 15:00 ,8 ,"","John Doe","Joe Smith","Martin Taylor","Henry Smith","Mike Miller"
01 Oct 2015 ,22 ,15:00 - 23:00 ,8 ,"","Bob Ford","Sarah Smith","Jack Williams","",""
01 Oct 2015 ,11 ,23:00 - 7:00 ,8 ,"","","Jenny Roe","Adam Davis","Jake Blake",""
02 Oct 2015 ,21 ,07:00 - 19:00 ,12 "Antonio Garcia","John Doe","Joe Smith","","Henry Smith","Mike Miller"
02 Oct 2015 ,22 ,19:00 - 07:00 ,12 ,"","Bob Ford","Sarah Smith","Jack Williams","",""
02 Oct 2015 ,11 ,07:00 - 15:00 ,8 ,"","","Jenny Roe","Adam Davis","Jake Blake",""
...

它实际上将$data的内容附加到$results数组的末尾。最后，你会有这样的结果：

<?php
$csv = array();
$lines = file('schedule.csv', FILE_IGNORE_NEW_LINES);

foreach ($lines as $key => $value)
{
    $csv[$key] = str_getcsv($value);
}

echo '<pre>';
print_r($csv);
?>

$results[] = $data

...
01 Oct 2015 ,65 ,07:00 - 15:00 ,8 ,"","John Doe","Joe Smith","Martin Taylor","Henry Smith","Mike Miller"
01 Oct 2015 ,22 ,15:00 - 23:00 ,8 ,"","Bob Ford","Sarah Smith","Jack Williams","",""
01 Oct 2015 ,11 ,23:00 - 7:00 ,8 ,"","","Jenny Roe","Adam Davis","Jake Blake",""
02 Oct 2015 ,21 ,07:00 - 19:00 ,12 "Antonio Garcia","John Doe","Joe Smith","","Henry Smith","Mike Miller"
02 Oct 2015 ,22 ,19:00 - 07:00 ,12 ,"","Bob Ford","Sarah Smith","Jack Williams","",""
02 Oct 2015 ,11 ,07:00 - 15:00 ,8 ,"","","Jenny Roe","Adam Davis","Jake Blake",""
...

---

好了。我希望这能把事情弄清楚。

谢谢你对Osuwariboy的所有帮助！这感觉像是一场艰苦的斗争，但最终还是奏效了。多亏了你

我将尝试用解决方案全面描述该案例：

我的schedule.csv文件格式如下（日期、未使用的编号、轮班时间、轮班长度、名称（最多6个）：

这里的主要问题是，我要找的不是当前计划的人，而是下一班的人。班次是8小时（常规）或12小时（周末和一些随机的日子）。基于此，我不能简单地看8小时或12小时，因为这会产生各种问题，如：

• 在12小时班次开始时向前看8小时（显示相同的班次）

• 在8小时轮班结束时展望前方12小时（跳过下一班）

轮班日期也存在问题，例如2015年10月1日23:00-7:00开始时为10月1日，结束时为10月2日

根据Osuwariboy的提示，我提出了以下解决方案：

---

也许它不是最漂亮、最干净的代码，但是，嘿，它似乎在任何情况下都能工作。我希望有人也会觉得它有用。

谢谢！这看起来很好。我可以理解你的操作方式，但我不太清楚的是开始时的

$results=array（）；

（它代表什么？）在最后的

if

-（

$result[]=$data；

）至于如何打印本例中的输出？我在文章中添加了一些解释。如果事情仍然不清楚，请不要犹豫再次询问。正如我预期的那样，但仍然让我感到困惑。谢谢！问题是我在填充

$results

时遇到问题。

schedule.csv

包含有效条目，

$startShift

和

$endShift

填充正常，但在

if

语句中，无法读取任何内容

$results

-只有一个空的

数组（）

。当

$currentData

是Unix时间戳，而其他值类似于DateTime对象时，您确定我可以比较时间吗？我添加了

date\u default\u timezone\u set（'Europe/London'）；

以确保生成的时间正确，并且我尝试了

$currentDate=date（'Y-m-d H:I:s:u'））

-不行。你说得对，我可能弄错了$currentDate。你试过了吗？$currentDate=new DateTime（）；，这应该会创建一个值为“now（）”的DateTime对象（与系统的当前日期和时间相同）。这将允许你在我的最终“if”中使用比较结果。