Php财年报告，麻烦选择适当的年份 [英] Php fiscal year report, trouble with selecting proper year

问题描述

我正在使用php和codeigniter为客户创建报告。

此报告是一份财务报告，因此，我必须反映自财政年度。在这种情况下，7月31日。

我已经向用户查询报告的日期，但我如何让php知道要采取哪个会计年度？ p>

我有一个粗略的想法，沿着

行的东西

'如果月 - 日是在7月31日之前，使用current year-1
否则使用当前年度

但我不知道我将如何编码，或者如果它会工作，更优雅的做同样的事情。

解决方案

我认为你最好的打赌是dateTime类。

http://us3.php.net/manual/en/class.datetime。 php

我认为这样的东西是你需要的

  public function getCurrentYear DateTime $ dateToCheck）
 {
 $ today = new DateTime（）; 
 $ currentYear =（int）$ today-> format（'Y'）; 
 $ endFiscalYear = new DateTime（'31 July'）; // year left out，所以默认为今年
 if（$ dateToCheck< $ endFiscalYear）{//你需要PHP> = 5.2.2这个工作
 $ currentYear-- 
} 
 return $ currentYear; 
} 
  

您可以通过执行以下操作设置$ today： -

  $ today = new DateTime（'20 June 2011'）; 
 

在上面的链接中阅读更多

这是一个稍微不同的版本，应该是一个更健壮，因为它将返回任何日期的财政年度，而不只是当年的日期。

  function getFiscalYear（DateTime $ dateToCheck）
 {
 $ fiscalYearEnd = '31 July'; 
 $ year =（int）$ dateToCheck-> format（'Y'）; 
 $ fiscalyearEndDate = new DateTime（$ fiscalYearEnd。''。$ year）; 
 if（$ dateToCheck< = $ fiscalyearEndDate）{
 $ year--; 
} 
 return $ year; 
} 
  

使用： -

  $ dateToCheck = new DateTime（'1 jan 2009'）; //例如
 $ fiscalYear = getFiscalYear（$ dateToCheck）; 
  

这将返回2008

应该工作，如果你的PHP版本< 5.2

 函数getFiscalYear（$ timestamp）
 {
 $ year = '，$ timestamp）; 
 $ fiscalYearEndDate = strtotime（'31 July'。$ year）; 
 if（$ timestamp< $ fiscalYearEndDate）$ year--; 
 return $ year; 
} 
  

使用这样： -

$ b b

  $ date = strtotime（'1 Jan 2009'）; 
 fiscalYear = getFiscalYear（$ date）; 
  

将返回2008

I am building a report for a client using php and codeigniter.

This report is a financial report, as such, I must reflect money collected since the beginning of the fiscal year. In this case, July 31st.

I already query the user for the date of the report, but how would I get php to know which fiscal year to take?

I have a rough idea, something along the lines of

'If Month-Day is before July 31, use current year-1 Else use current year'

But I do not know how I would code that, or if it would work, or if there is a more elegant way of doing the same thing.

解决方案

I think your best bet is the dateTime class.
http://us3.php.net/manual/en/class.datetime.php
I think something like this is what you will need

public function getCurrentYear(DateTime $dateToCheck)
{
    $today = new DateTime();
    $currentYear = (int)$today->format('Y');
    $endFiscalYear = new DateTime('31 July'); //year left out, so will default to this year
    if($dateToCheck < $endFiscalYear){ //you need PHP >= 5.2.2 for this to work
        $currentYear--;
    }
    return $currentYear;
}

You can set $today by doing something like :-

$today = new DateTime('20 June 2011');

Read more in the link above

Here is a slightly different version which should be a bit more robust as it will return the fiscal year of any date you give it, not just dates in the current year.

function getFiscalYear(DateTime $dateToCheck)
{
    $fiscalYearEnd = '31 July';
    $year = (int)$dateToCheck->format('Y');
    $fiscalyearEndDate = new DateTime($fiscalYearEnd . ' ' . $year);
    if($dateToCheck <= $fiscalyearEndDate){
        $year--;
    }
    return $year;
}

use it like this :-

$dateToCheck = new DateTime('1 jan 2009'); // for example
$fiscalYear = getFiscalYear($dateToCheck);

This will return 2008

This version should work if your PHP version is < 5.2

function getFiscalYear($timestamp)
{
    $year = (int)date('Y', $timestamp);
    $fiscalYearEndDate = strtotime('31 July ' . $year);
    if($timestamp < $fiscalYearEndDate) $year--;
    return $year;
}

Use like this:-

$date = strtotime('1 Jan 2009');
fiscalYear = getFiscalYear($date);

Will return 2008

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