由于脚本而不会插入mysql数据库

时间:2019-04-03 07:36:40

标签: javascript jquery sweetalert

我有一个脚本,询问是否确定要添加新数据。我的php没问题,因为当我想在禁用脚本的情况下插入数据时,它会正常插入。但是当我的脚本没有被禁用时,它将不再插入...我不知道为什么我没有任何错误(可能是因为有2个不同的文件)。

我的脚本

$(document).on("click", ".btnAddSubcat", function(e) {
    event.preventDefault();
    var subcatid = $('.CatEdit-select').val();
    var subcatvalue = $('.subCat').val();
    var subcatprocedure = $('.subProcedure').val();
    var url = "../service/functions/postActions.php";
    swal({
        title: 'Add a new sub-category?',
        text: "Are you sure to Add this sub-category?",
        type: 'warning',
        showCancelButton: true,
        confirmButtonColor: '#3085d6',
        cancelButtonColor: '#d33',
        confirmButtonText: 'Yes, Add it!'
    }).then(function () {
        $.ajax({
            type: 'POST',
            url : url,
            data: {
                'cat-select': subcatid,
                'subCat': subcatvalue,
                'Procedure': subcatprocedure
            },
            success: function (data) {
                swal({
                    title: 'Add!',
                    text: "You Add a sub-category!",
                    type: 'success',
                    confirmButtonColor: '#3085d6',
                    confirmButtonText: 'OK'
                }).then(function () {
                    window.location.reload();
                })
            }
        });
    })
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

如果您需要我的php插入功能或其他让我知道的信息

编辑

我的PHP函数

function AddSubCat() {
    $Catid = $_POST['cat-select'];
    $Subnaam = $_POST['subCat'];
    $procedure = $_POST['Procedure'];

    include '../../../include/dbConnection/dbcon.php';
    //set update query ready for database, with info from the Ajax POST method for update.
    $sqladdSubCat = 'INSERT INTO `*****`(`naam`,`Werkwijze`) VALUES ("'.$Subnaam.'","'.$procedure.'")';
    //if query is done right then 'Record updated successfully'
    if (mysqli_query($conn, $sqladdSubCat))
    {
        //get new colourId
        $newUpload_id = mysqli_insert_id($conn);
        //set update query hw_algoritmes thee cat id that may have change.
        $sqlalgotime = 'INSERT INTO `*****`(`categorie_id`, `subcategorie_id`) VALUES ("'.$Catid.'","'.$newUpload_id.'")';

        //if query is done right then 'Record updated successfully'
        if (mysqli_query($conn, $sqlalgotime)) {
            //check if there is a new front image add
            if ($_FILES['HWVideo']['name'] == "") {
                //Empty
            } else {
                uploadVideoHw($_FILES['HWVideo'], $newUpload_id);
            }
        } else {
            echo "Error updating record: " . mysqli_error($conn);
        }
        header('Location: /cms/service/index.php');
    } else {
        echo "Error updating record: " . mysqli_error($conn);
        // header('Location:../../Categorieen');
    }
}

0 个答案:

没有答案