Question

如何将复选框值更新到数据库

$fmcourse = $_POST['fm_courses'];
foreach($fmcourse as $fmc) {
    $studentid = mysqli_real_escape_string($con, $_POST["id"]);
    $sql = "UPDATE enrollments SET course_fk='$fmc' WHERE student_fk = '$studentid'";
}

var_dump(fmc)输出此

string(2) "18" string(2) "20" string(2) "22" string(2) "24" string(2) "26"

当我更新course_fk时，在所有行上重复相同的id

+-----+------------+-----------+
| eid | student_fk | course_fk |
+-----+------------+-----------+
|  1  |          1 |        17 |
|  2  |          1 |        17 |
|  3  |          1 |        17 |
+-----+------------+-----------+

我正在寻找这个

+-----+------------+-----------+
| eid | student_fk | course_fk |
+-----+------------+-----------+
|  1  |          1 |        17 |
|  2  |          1 |        20 |
|  3  |          1 |        22 |
+-----+------------+-----------+

Answer 1

下面的代码使用函数 record_exists（）检查，如果记录在enrollments表中有重复项，如果有，则代码只会相应地更新重复项。如果没有重复项，则会为学生创建新的注册记录。

更新了代码。

$fmcourse = $_POST['fm_courses'];
    $student_id = $_POST["id"];
    $batch_id = $_POST["name"];
    delete_enrollments( $student_id, $batch_id );

    foreach($fmcourse as $fmc) {
        $course_id = $fmc;

        if( record_exists( $student_id, $course_id, $batch_id ) == FALSE ) {
            $stmt = mysqli_prepare($conn, "INSERT INTO enrollments (enrollment_id, student_id, course_id, batch_id, is_deleted, joining_date) VALUES( NULL, ?, ?, ?, 0, NOW() );");
            $is_binded = mysqli_stmt_bind_param($stmt, "iii", $student_id, $course_id, $batch_id);
            $is_exec = mysqli_stmt_execute($stmt);
            mysqli_stmt_close($stmt);
        }
        else {
            undelete_enrollment($student_id,  $course_id, $batch_id);
        }
    }

    function undelete_enrollment($student_id,  $course_id, $batch_id) {
        global $conn;
        $stmt = mysqli_prepare($conn, "UPDATE enrollments SET is_deleted = 0, joining_date = NOW() WHERE student_id= ? AND course_id = ? AND batch_id = ?");
        $is_binded = mysqli_stmt_bind_param($stmt, "iii", $student_id, $course_id,$batch_id);
        $is_exec = mysqli_stmt_execute($stmt);
        mysqli_stmt_close($stmt);
        return $is_exec;
    }

    function record_exists( $student_id,  $course_id, $batch_id ) {
        global $conn;

        $stmt = mysqli_prepare($conn, "SELECT COUNT(enrollment_id) as total FROM enrollments WHERE student_id= ? AND course_id = ? AND batch_id = ?");
        $is_binded = mysqli_stmt_bind_param($stmt, "iii", $student_id, $course_id, $batch_id);
        $is_exec = mysqli_stmt_execute($stmt);
        $result = mysqli_stmt_get_result($stmt);
        $record = mysqli_fetch_assoc($result);
        mysqli_stmt_close($stmt);
        return ( isset( $record['total'] ) AND $record['total'] > 0 );
    }

    function delete_enrollments( $student_id, $batch_id ) {
        global $conn;
        $stmt = mysqli_prepare($conn, "UPDATE enrollments SET is_deleted = 1 WHERE student_id = ? AND batch_id =?;");
        $is_binded = mysqli_stmt_bind_param($stmt, "ii", $student_id, $batch_id);
        $is_exec = mysqli_stmt_execute($stmt);
        mysqli_stmt_close($stmt);
        return $is_exec;
    }

Answer 2

您需要根据eid而不是student_fk

更新您的记录

您还需要eid。随着您的student_fk值重复出现。它总是更新最后一个值

因此您的where条件存在问题。您需要添加eid而不是student_fk的条件。同时确保每行$eid不同。

$sql = "UPDATE enrollments SET course_fk ='$fmc' WHERE eid = '$eid'";

此外，您的代码对SQL注入是开放的。为防止这种情况，请使用预备声明

将复选框更新到mysql数据库

2 个答案: