如何使用一个查询而不是多个

我有多个日期,我需要计算数据库中特定日期的记录数.我正在使用下面的代码,但是..问题是它会在循环内每次都有借口.我相信有更好的方法可以做到这一点.提前致谢.

  for ($i = 0; $i <= $total_difference; $i++){
     $date = strtotime("+$i day", strtotime($start_date));
     $check_date = date("Y-m-d", $date);

     // Here i have to check the record against $check_date
    $query = $this->db->select("SELECT COUNT(task_date) as COUNT FROM 
    time_management");
    $this->db->from('time_management');
    $this->db->where('task_date',$check_date);
    $result = $this->db->get()->row_array();
    $count = $result['COUNT'];
    echo "count for $check_date is $count";

 }

KIKO Softwar.. 5

您正在执行此查询:

SELECT COUNT(task_date) as COUNT 
FROM time_management 
WHERE task_date = ?

问号是日期的地方.

您需要做的是GROUP BY task_date设置开始日期和结束日期,如下所示:

SELECT task_date as TASKDATE,
       COUNT(task_date) as COUNT 
FROM time_management 
WHERE task_date BETWEEN ? AND ?
GROUP BY task_date

这两个问号是您想要计算的第一个和最后一个日期.

正如vivek_23所说,在结果中使用task_date很方便,所以我补充说.

我将让你把它翻译成Codeigniter使用的具体方法,这应该很容易(我不使用它).

如果您需要零计数结果(不会出现在此查询的结果中),则可以使用PHP添加它们.这比试图强制MySQL为不在数据库中的日期提供零计数更容易.像这样:

// suppose you start with these
$start_date       = '2018-05-04';
$total_difference = 6; 

// and suppose the result of your query looks like this
$counts = ['2018-05-04' => 5,
           '2018-05-06' => 3];

// then you can insert zero count results like this:
for ($days = 0; $days <= $total_difference; $days++) {
  $date = addDaysToDate($start_date,$days);
  if (!isset($counts[$date])) $counts[$date] = 0;
}

function addDaysToDate($date,$days)
{
  $timestamp = strtotime("+$days day",strtotime($date));
  return date("Y-m-d",$timestamp);
}

请注意,我定义了一个添加天数的函数,这似乎很有用.

您正在执行此查询:

SELECT COUNT(task_date) as COUNT 
FROM time_management 
WHERE task_date = ?

问号是日期的地方.

您需要做的是GROUP BY task_date设置开始日期和结束日期,如下所示:

SELECT task_date as TASKDATE,
       COUNT(task_date) as COUNT 
FROM time_management 
WHERE task_date BETWEEN ? AND ?
GROUP BY task_date

这两个问号是您想要计算的第一个和最后一个日期.

正如vivek_23所说,在结果中使用task_date很方便,所以我补充说.

我将让你把它翻译成Codeigniter使用的具体方法,这应该很容易(我不使用它).

如果您需要零计数结果(不会出现在此查询的结果中),则可以使用PHP添加它们.这比试图强制MySQL为不在数据库中的日期提供零计数更容易.像这样:

// suppose you start with these
$start_date       = '2018-05-04';
$total_difference = 6; 

// and suppose the result of your query looks like this
$counts = ['2018-05-04' => 5,
           '2018-05-06' => 3];

// then you can insert zero count results like this:
for ($days = 0; $days <= $total_difference; $days++) {
  $date = addDaysToDate($start_date,$days);
  if (!isset($counts[$date])) $counts[$date] = 0;
}

function addDaysToDate($date,$days)
{
  $timestamp = strtotime("+$days day",strtotime($date));
  return date("Y-m-d",$timestamp);
}

请注意,我定义了一个添加天数的函数,这似乎很有用.