php - 如何将查询附加到先前选择的查询？

Question

public function gettopfilteredcourse($filtertype, $topfiltername, $filterid){
    $sql="SELECT * FROM course c
          INNER JOIN category s
          ON c.categoryId=s.categoryId
          INNER JOIN users u
          ON c.userId=u.userId";

  if($topfiltername == 'trash'){
     $sql .= " WHERE isCompleted='3'";
  } else if($topfiltername == 'forapproval'){
     $sql .= " WHERE isCompleted='1'";
  } else if($topfiltername == 'drafts'){
     $sql .= " WHERE isDraft='1'";
  } else {
     $sql .= " WHERE isCompleted='2'";
  }

  if($filtertype == 'free'){
     $sql .= " AND c.fee=''";
  } else if($filtertype == 'paid'){
     $sql .= " AND c.fee!=''";
  } else {
     $sql .= " AND c.categoryId='$filterid' ";
  } 
   $query=$this->db->query($sql);
   return $query;
}

在第一个if条件$topfiltername == 'forapproval'被选中时，它正确执行。当谈到第二个 if 条件 $filtertype == 'paid' 时，我想将其附加到 $topfiltername == 'forapproval'，但它总是附加第一个 if 条件 else 语句 $sql .= "WHERE isCompleted='2'";。怎么办请帮帮我。

Answer 1

如果您使用的是codeigniter。请使用数据库。codeigniter.com/userguide3/database/query_builder.html