Question

利用和从Facebook的开发商中选取,我收到以下代码。

    //build fql string
    $fql = "SELECT name, birthday_date FROM user 
              WHERE uid IN
                (SELECT uid2 FROM friend WHERE uid1 = me())
              AND strlen(birthday_date) != 0 AND substr(birthday_date,0,3) 
            IN ( {$currentMonth}/ ,  {$nextMonth}/ )";
    $fql.=$orderBy;
    $fql.=" LIMIT " . $limit;

    //query facebook
    $params  =   array(
                  method     =>  fql.query ,       
                  query      => $fql,
                  callback   =>        
                  );
    $birthdays  =  $this->facebook_obj->api($params);

。

这是“工作”,但我要说,今天是11月29日,我有20位朋友,11月份有出生日,而<>条码规定为10。在此情况下,在11月29日前,10个出生日都很有可能返回。

我真的想做的是,从今天开始,本月和下个月,即将到来。我怎么能改变这一 que,以实现这一目标?

感谢!

* <>。 Edit, see comment thread between Shawn and I for possible Alternative Solutions. 尚未接受任何答复,因为没有直接涉及原法典/问题。

Answer 1

这是几个月后,但也许会帮助面临同样问题的其他人。基本上,FQLIN的操作者没有检查该数值是否属于某一范围,但该数值是否与所提供的任何不同数值相符。举例来说,如果价值为09和10岁,那么9月和10月份的所有出生日都将返回,因为这些出生月数与规定的月份数相符。

What you can do is calculate the following values using PHP or whatever other language you are using and use them in your FQL (today is 17 March and say we want 17 April is end date):

Today s day of the month (17),
Today s month of the year (03)
End date day of the month (17)
End date month of the year (04)

保证日数和月数分别采用d和MM(如果是单一数字,则增加零)。之后,贵方论坛希望:

      SELECT name, birthday_date
      FROM user 
      WHERE uid IN (SELECT uid2 FROM friend WHERE uid1 = me())
      AND strlen(birthday_date) != 0
      AND ((substr(birthday_date, 0, 2) =  03 
      AND substr(birthday_date, 3, 5) >=  17 )
      OR (substr(birthday_date, 0, 2) =  04 
      AND substr(birthday_date, 3, 5) <  17 ))
      ORDER BY birthday_date

之后,3月17日至4月17日期间,亲爱的朋友将返回。

Answer 2

AND created_time > strtotime( now ) AND created_time < $endofnext

http://www.php.net/manual/en/Function.strtotime.php#97025"rel=“nofollow” http://www.php.net/manual/en/Function.strtotime.php#97025

使用“现在”将回去目前这一天。

这项职能将在下个月的最后一天恢复,无论时间是哪个月。

<?php 
/* $now can be set to any strtotime(), yesterday, last month, last year etc; */
$now = "now"; 
function EndOfNext($now){
    function lastDayOfMonth($month =   , $year =   ) 
    { 
       if (empty($month)) { 
          $month = date( m ); 
       } 
       if (empty($year)) { 
          $year = date( Y ); 
       } 
       $result = strtotime("{$year}-{$month}-01"); 
       $result = strtotime( -1 second , strtotime( +1 month , $result)); 
       return date( m/d/y , $result); 
    } 
    function addRealMonth($timeStamp) 
    { 
        $tempMonth = date( m , $timeStamp); 
        $tempYear  = date( Y , $timeStamp); 
        if($tempMonth == "12") 
        { 
            $tempMonth = 1; 
            $tempYear++; 
        } 
        else 
            $tempMonth++; 

        $newDate = lastDayOfMonth($tempMonth, $tempYear); 
        return strtotime($newDate); 
    }
    $newChargeDate = strtotime($now); 
    $newChargeDate = addRealMonth($newChargeDate); 
    return date("m/d/y", $newChargeDate); 
}
$endofnext = EndOfNext($now);
echo $endofnext;
?>

Answer 3

你们还可以让联邦劳工局为你们做一些工作,例如:

SELECT uid, name, birthday_date, strtotime(substr(birthday_date, 0, 5)) > now()
FROM user
WHERE uid in (SELECT uid2 FROM #freunde)
AND birthday_date <>   
AND strpos(birthday_date, 06 ) = 0
ORDER BY substr(birthday_date, 0, 5)

returning a boolean type "anon" field telling if you have to include this result in the display list (true) or if the birthday for this user lies already in the past (false). Unfortunately I wasn t able to have this proposition work as a WHERE clause - keep getting the error under, but would love to know if someone was luckier than I.

  "error": {
    "message": "Call to a member function on a non-object", 
    "type": "BadMethodCallException"
  }

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