Question

我需要编写一个方法，允许我计算用户填充的行数。例如：

 User    Name    Age    Country    Gender    Height
    1       Mike    34     USA        Male      6
    2       Bill    23     CA                   5
    3       Jane    31     USA

在上面的例子中，我想查询数据库并返回一个反映用户记录完成程度的值。如：

User 1 = 100% complete
User 2 = 80% complete
User 3 = 60% complete

我正在使用Codeigniter 3.你能帮助我吗？

Answer 1

您可以尝试这样::

$result = mysql_query('SELECT * FROM `MyTable`');
while($row = mysql_fetch_row($result)){
$empty_count = 0;
$count = count($row);
for($i = 0; $i < $count; $i++)
    if($row[$i] === '' || $row[$i] === 'NULL')
        $empty_count++;
echo 'User '.$row[0].' = '.((int)(100*(1-$empty_count/($count-1)))).'% complete';
}

希望它能奏效。

Answer 2

使方法传递方法

中的参数

  function profileCompleted($user){
         $percentage = 0;
         $sql ="SELECT * FROM user  WHERE user=$user"
       $query = $this->db->query($sql);

      if ($query->num_rows() > 0)
          { 
            $notEmpty =   0;
            $totalField =5;
            foreach ($query->result() as $row)
              {
               $notEmpty +=  ($row->Name != '') ? 1 : 0;
                //do with all field
              }
          $percentage = $notEmpty/$totalField *100;
          }
        return $percentage.'%';
    }

Answer 3

我认为它也可以通过SQL查询来完成：

SELECT id,
(
    CASE name WHEN NULL THEN 0 ELSE 1 END
    +
    CASE WHEN age IS NULL THEN 0 ELSE 1 END
    +
    CASE country WHEN '' THEN 0 ELSE 1 END
    +
    CASE gender WHEN '' THEN 0 ELSE 1 END
    +
    CASE WHEN height IS NULL THEN 0 ELSE 1 END
) * 100 / number_of_fields AS complete
from your_table;

计算Codeigniter中连续填充的MySQL字段数

3 个答案: