Question

这是我的代码

$sql14 = ("SELECT count(`MarksID`) AS all FROM tbl_course_marks_result WHERE tbl_course_marks_result.`StudentID` = '$k' AND tbl_course_marks_result.`CourseID`='$courseID' AND tbl_course_marks_result.`SessionID`='$sessionID' AND tbl_course_marks_result.`TermID`='$termID' ");
    $result14 = mysqli_query($connect,$sql14);
if($result14 === FALSE) { 
    die(mysqli_error($connect));
}

此处显示此错误消息＆＃34;您的SQL语法有错误;检查与您的MariaDB服务器版本相对应的手册，以便在所有FROM tbl_course_marks_result附近使用正确的语法WHERE tbl_course_marks_result。StudentID =＆＃39; 17＆＃39;在第1行＆＃34;怎么解决这个???

Answer 1

所有都是保留关键字。使用不同的别名，如下所示

 $sql14 = ("SELECT count(`MarksID`) AS all_marksid FROM tbl_course_marks_result     WHERE tbl_course_marks_result.`StudentID` = '$k' AND  tbl_course_marks_result.`CourseID`='$courseID' AND tbl_course_marks_result.`SessionID`='$sessionID' AND tbl_course_marks_result.`TermID`='$termID' ");
$result14 = mysqli_query($connect,$sql14);
 if($result14 === FALSE) { 
die(mysqli_error($connect));
}

如何解决＆＃34;您的SQL语法中有错误。＆＃34;错误消息

1 个答案: