查询构建SQL功能

Question

我正在基于变量查询数据库，但对于如何获取我的函数不起作用有些困惑。 如果SQL条件数组中只有一个字符串，我希望它说WHERE，如果有多个，则我希望第一个是WHERE，其余的是AND。 当前，如果我有一个字符串，并且rets为空白，则查询如下所示：SELECT * FROM tb_visitors AND'2017-08-01'和'2017-08-31'之间的日期。 我希望首先在哪里，是否有更好的方法可以做到这一点？

        $sites =  $data['sites'];
        $attendence =  $data['filter'];
        $time = $data['time'];
        $department = $data['department'];
        $staff = $data['staff'];

        $sql = 'SELECT * FROM tb_visitors' ;
        $sql_criteria= "";

        if(is_numeric($department) == true){
           $sql_criteria[] = "dept_id = "."'".$department."'"; 
        } else if($staff !== 'Staff Member') {
             $sql_criteria[] = 'staff='.'"'.$staff.'"'; 
        } 

        if($department == 'Department' || $staff == 'Staff Member' ){
           $sql_criteria[] = ""; 
        }

        // Time
        if($time == 'Date Range' || $time == 'Custom' || $time == 'Today'){
           $sql_criteria[] = ""; 
        } else if($time == 'Today'){
           $today_date = date("Y-m-d"); 
           $sql_criteria[] = "date ="."'".$today_date."'";  
        } else if($time == 'Tomorrow'){
           $d= strtotime("tomorrow");
           $date =  date("Y-m-d", $d);
           $sql_criteria[] = "date ="."'".$date."'"; 
        }  else if($time == 'Yesterday'){
           $date = date('Y-m-d',strtotime("-1 days"));
           $sql_criteria[] = "date ="."'".$date."'"; 
        } else if($time == 'Last Month'){
           $start_date =  date("Y-n-j", strtotime("first day of previous month"));
            $end_date =  date("Y-n-j", strtotime("last day of previous month"));
           $sql_criteria[] = "date BETWEEN '".$start_date."' AND '".$end_date."'"; 
        } else if($time == 'Next Month'){
           $start_date =  date("Y-n-j", strtotime("first day of next month"));
           $end_date =  date("Y-n-j", strtotime("last day of next month"));
           $sql_criteria[] = "date BETWEEN '".$start_date."' AND '".$end_date."'"; 
        } else if($time == 'This Week'){
           $start_date = date("Y-m-d", strtotime('monday this week'));
           $end_date =  date("Y-m-d", strtotime('sunday this week'));
           $sql_criteria[] = "date BETWEEN '".$start_date."' AND '".$end_date."'";  
        } else if($time == 'This Month'){
           $start_date = date('Y-m-01');
            $end_date = date('Y-m-t');
            $sql_criteria[] = "date BETWEEN '".$start_date."' AND '".$end_date."'";
        } else if($time == 'Last Week'){
           $start_date = date("Y-m-d", strtotime("last week monday"));
           $end_date =  date("Y-m-d", strtotime("last week sunday"));
           $sql_criteria[] = "date BETWEEN '".$start_date."' AND '".$end_date."'"; 
        } else if($time == 'Next Week'){
           $start_date = date("Y-m-d", strtotime("next monday"));
           $end_date =    date("Y-m-d", strtotime('+1 week sunday'));
           $sql_criteria[] = "date BETWEEN '".$start_date."' AND '".$end_date."'";  

        } 

        // Attendence 
        if($attendence == 'Status' || $attendence == 'All Visits'){
            $sql_criteria[] = ""; 
        } else if($attendence == 'No Show'){
            $sql_criteria[] = "no_show = 'Y'";  
        } else if($attendence == 'Completed Visits'){
            $sql_criteria[] = "seen IS NOT NULL";  
        } else if($attendence == 'Upcoming Visits'){
           $today_date = date("Y-m-d"); 
           $sql_criteria[] = "date > '".$today_date."'"; 
        } 

        //Sites

        if($sites == 'All Sites' || $sites == 'Sites' ){
            $sql_criteria[] = ""; 
        } else if($sites == 'Twickenham'){
            $sql_criteria[] = "dept_id NOT IN ('28',  '29',  '30',  '32',  '41',  '44')";
        } else if($sites == 'Hammersmith'){
            $sql_criteria[] = "dept_id IN ('36')"; 
        } else if($sites == 'Heatherwood'){
           $sql_criteria[] = "dept_id IN ('37')";  
        } else if($sites == 'Hillingdon'){
          $sql_criteria[] = "dept_id IN ('38')";   
        } else if($sites == 'Nuffeild'){
            $sql_criteria[] = "dept_id IN ('39')"; 
        } else if($sites == 'St Georges'){
            $sql_criteria[] = "dept_id IN ('41')"; 
        } else if($sites == 'Queen Victoria'){
            $sql_criteria[] = "dept_id IN ('40')"; 
        } else if($sites == 'Stoke Mandeville'){
           $sql_criteria[] = "dept_id IN ('42')";  
        }


    $i=0;
    foreach ($sql_criteria as $a => $b){

       if($b == ""){
            $join = "";   
        } else if($i == 0){
            $join = "WHERE";
        } else {
            $join = "AND";  
        }


    $sql_final = $join." ".$b;
    $i++;
    $sql = $sql." ".$sql_final;
    }
   $result = $db->db_num("$sql");
   echo $sql;

Answer 1

由于您已经在数组中where了where子句，因此可以使用count和implode 。

$sql = 'SELECT * FROM tb_visitors';
$sql_criteria = array();

/*
* Build the $sql_criteria array
* [...]
*/

if (count($sql_criteria)) // Should the where clause be built?
{
    $sql .= " WHERE " . implode(" AND ", $sql_criteria);
}

注意：正如评论者已经指出的那样，您容易受到SQL注入的攻击。 考虑到这一点很重要，因为注入可能是“偶然”发生的，并不总是由于有针对性的攻击而引起的。 例如，如果您的$staff的值为“ John O'Sullivan”，则名称中的单引号会打断您的查询。