Question

数据存储在数据库中，如下所示，下午，晚上是列名，逗号分隔的字符串是它们的值。

morning = 'Fri,Sun';   
afternoon = 'Thrs,Fri,Sat,Sun';
evening = 'Thrs,Fri,Sat,Sun';

此查询有效：

AND (    FIND_IN_SET( 'Sat', posts.afternoon )
      OR FIND_IN_SET( 'Sat', posts.evening ) )

但我想要这样的东西，但我知道不能：

AND (    FIND_IN_SET( 'Fri','Sat', posts.afternoon )
      OR FIND_IN_SET( 'Thrs','Fri','Sat', posts.evening ) )

所以我至少要像下面一样单独搜索我已经测试过的结果：

AND ( FIND_IN_SET ('Fri', posts.afternoon) OR FIND_IN_SET ('Sat', posts.afternoon) OR FIND_IN_SET ('Sat', posts.evening))

问题是如何修改我现有的PHP脚本以匹配它，好吗？以下是我的完整代码：

 <?php
    JSON formatted data received via ajax
    $return = '{"avail":["Wed-2,Thrs-2","Thrs-3"]}';

    In the above -1 means for monrning, -2 for afternoon and -3 for evening.I'm categorising them below.'
    $avail = $data['avail'];
    $days = array();
    $cols = array();

    $size = sizeof($avail);
    if(($avail != "")&&($size > 1)){
    $periods = array();
    foreach($avail as $v){
        list($day, $column) = explode("-", $v); // make sure you validated the data to avoid errors
       $periods[$column][] = "'" . mysql_escape_string($day) . "'"; //strtolower// PHP would automatically create an empty array if $periods[$column] was not defined
    }
    $intToCol = array(1 => "morning", 2 => "afternoon", 3 => "evening");
    // $periods should now be identical to ["2" => ["'sun'", "'mon'"], "3" => ["'sun'"]]

    $conditions = array();
    foreach($periods as $int => $days){
        $dayString = implode(",", $days);
        $conditions[] = " FIND_IN_SET ($dayString,  posts." . $intToCol[$int].")" ;
    }
    $add_here = "AND (" . implode(" OR ", $conditions) . ")";
    }else if(($avail != "")&&($size == 1))
    {
        foreach($avail as $k=>$v)
            {
                 $v;

                $array = explode('-', $v);
                $day =$array[0]; // Wed
                $column =  $array[1]; // 2

                if($column == 1)
                {
                $col = "morning";

                }
                if($column == 2)
                {
                    $col = "afternoon";
                }
                if($column == 3)
                {
                    $col = "evening";
                }

            }

        $add_here = " posts.".$col." = '".$day."' ";

        $sql = "SELECT * , (3956 * 2 * ASIN(SQRT( POWER(SIN(('$lat' - lat) *  pi()/180 / 2), 2) +COS('$lat' * pi()/180) * COS(lat * pi()/180) * POWER(SIN(('$lon' - lon) * pi()/180 / 2), 2) ))) as distance  from posts WHERE posts.subname LIKE '%$subject%' AND posts.pricing <= '$rate' ".$add_here."".$IsTutionCentre." GROUP BY posts.UUID having  distance <= '$distance' order by distance";

        $stmt =connection::$pdo->prepare($sql);
            $stmt->execute();
            $place=array();
            while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) { 
                   $place[] = $row;
                   }
                   $_SESSION['subject'] = $place;

        //send back to ajax call  made page
        echo json_encode($place);
    }
    ?>

Answer 1

这个MySQL函数find_in_set（）只能搜索一组字符串中的一个字符串。

第一个参数是一个字符串，所以没有办法让它将逗号分隔的字符串解析成字符串（你根本不能在SET元素中使用逗号！）。第二个参数是一个SET，它反过来用逗号分隔的字符串表示，因此你希望find_in_set（'p，q，r'，'p，q，r，s'）工作正常，但它肯定可以'根据定义，在任何SET中找到字符串'p，q，r' - 它包含逗号。

您可以使用如下命令：

where setcolumn like '%p,q%'

要在FIND_IN_SET

1 个答案: