PHP MYSQL Insert into with switch语句

首先，我很抱歉，因为我还是一个新手，想学习更多关于编码的知识

我现在正在做这个。 我的第一、第二和第三张桌子工作正常。我的问题是我的第四个表没有收到任何值。 我试图用switch语句在第四个表中插入数据，其中$eGrad等于Grade Level。 我想在我的st_grade表（第四个表）中插入基于等级的值，其中一些是常量值

我将感谢任何帮助。谢谢 这是我正在使用的代码

public function insert($stID, $eFullname, $eGrad, $secID, $eStatus, $eSY, $syID,  $rformOne, $rformTwo, $rPsa, $rGm, $rPic){

        $firstTB = "INSERT INTO st_enrolled (e_stid, e_fullname,e_grade,e_status,e_sy) VALUES('$stID', '$eFullname', '$eGrad',  '$eStatus', '$eSY')";
        $secondTB = "INSERT INTO st_sy (st_sy_stid, st_sy_syid) VALUES('$stID', '$syID')";
        $thirdTB = "INSERT INTO st_requirement (str_id, str_fullname, formOne, formTwo, psa, goodmoral, picture) VALUES ('$stID', '$eFullname',  '$rformOne', '$rformTwo', '$rPsa', '$rGm', '$rPic')";
        
        
        switch($eGrad){
            
            case "Grade1" :
                $fourthTB = "INSERT INTO `st_grade`(`st_id`,`subj_id`,`gr_id`,`sec_id`,`sy_id`,`first`,`second`,`third`,`fourth`,`ave`,`remarks`) VALUES 
                ($stID, 1, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 2, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 3, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 4, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 5, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 6, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 7, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),";
                break;
            case "Grade2" :
                $fourthTB = "INSERT INTO `st_grade`(`st_id`,`subj_id`,`gr_id`,`sec_id`,`sy_id`,`first`,`second`,`third`,`fourth`,`ave`,`remarks`) VALUES 
                ($stID, 8, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 9, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 10, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 11, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 12, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 13, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 14, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),";
                break;  
            default: 
                echo "error";
        }


        if ($this->conn->query($firstTB) && $this->conn->query($secondTB) && $this->conn->query($thirdTB) && $this->conn->query($fourthTB) ) {
            return true;
        } else {
            return false;
        }

    }

---

我已经删除了sql字符串中的最后一个逗号，现在应该可以了

public function insert($stID, $eFullname, $eGrad, $secID, $eStatus, $eSY, $syID,  $rformOne, $rformTwo, $rPsa, $rGm, $rPic){

        $firstTB = "INSERT INTO st_enrolled (e_stid, e_fullname,e_grade,e_status,e_sy) VALUES('$stID', '$eFullname', '$eGrad',  '$eStatus', '$eSY')";
        $secondTB = "INSERT INTO st_sy (st_sy_stid, st_sy_syid) VALUES('$stID', '$syID')";
        $thirdTB = "INSERT INTO st_requirement (str_id, str_fullname, formOne, formTwo, psa, goodmoral, picture) VALUES ('$stID', '$eFullname',  '$rformOne', '$rformTwo', '$rPsa', '$rGm', '$rPic')";
        
        
        switch($eGrad){
            
            case "Grade1" :
                $fourthTB = "INSERT INTO `st_grade`(`st_id`,`subj_id`,`gr_id`,`sec_id`,`sy_id`,`first`,`second`,`third`,`fourth`,`ave`,`remarks`) VALUES 
                ($stID, 1, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 2, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 3, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 4, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 5, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 6, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 7, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, '')";
                break;
            case "Grade2" :
                $fourthTB = "INSERT INTO `st_grade`(`st_id`,`subj_id`,`gr_id`,`sec_id`,`sy_id`,`first`,`second`,`third`,`fourth`,`ave`,`remarks`) VALUES 
                ($stID, 8, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 9, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 10, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 11, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 12, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 13, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),
                ($stID, 14, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, '')";
                break;  
            default: 
                echo "error";
        }


        if ($this->conn->query($firstTB) && $this->conn->query($secondTB) && $this->conn->query($thirdTB) && $this->conn->query($fourthTB) ) {
            return true;
        } else {
            return false;
        }

    }

---

结尾的逗号会导致问题

以前

($stID, 7, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),";
($stID, 14, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, ''),";

之后

---

在两种切换情况下，删除结尾处的逗号（，），这样做很好

打字问题不应回答，而应标记为该问题（

标记

问题下的链接>“需要改进”>“社区特定原因”>“不可复制或由打字造成”），除了留下一条评论，指示OP问题的原因。不应回答打字错误问题，应将其标记为该问题（

Flag

问题下的链接>“需要改进”>“社区特定原因”>“不可复制或由打字错误引起”），同时留下一条评论，说明问题的原因。到底是什么问题？你的问题到底是什么？检查如何

($stID, 7, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, '')";
($stID, 14, $eGrad, $secID, $syID, 0, 0, 0, 0, 0, '')";