Question

当我从下拉列表中选择项目时，如果用户再次添加该项目，它将添加到表格中，它应显示已添加项目的错误。

<select id="dropdownUsers"></select>
<table border="1" id="tableshow">
    <thead>
        <tr>
            <td>ID</td>
            <td>Name</td>
            <td>unit Price</td>
            <td>qunatity</td>
            <td>total</td>
            <td>Actions</td>
        </tr>
    </thead>

<tbody id="tabledata"></tbody>
</table>

ajax调用绑定表

 var val = $("#dropdownUsers").val();
                             $.ajax({
                                 dataType: 'json',
                                 type: 'POST',
                                 contentType: "application/Json; charset=utf-8",
                                 url: '/Home/fillDropdownbyids',
                                 data: JSON.stringify({ id: val }),
                                 success: function (response) {
                                     $("#id").append(response.user.iUserInfoId + " ")
                                     var tr = '';


                                     tr += '<tr class="row"><td class="rowid' + val + '">' + response.user.iUserInfoId + '</td><td>' + response.user.strFirstName + '</td><td><input type="text" id="unittextbox' + val + '"></td><td><input type="text" id="quantextbox' + val + '" onfocusout="get(' + val + ')"></td><td><input type="text" id="totaltextbox' + val + '" ></td><td><input type="button" value="Delete" onclick="deleteRow(this)"></td></tr>';


                                     $('#tabledata').append(tr);
                                 },
                             });

这将从表中获取第一个td的行，它将返回id

数组的列表

 $('#tableshow > tbody  > tr ').each(function (index, td) {
                     arr = $(this).find('td');
                     arrlist[index] = arr[0].innerHTML;
                     index++;
                 });

我想比较这个td值和下拉选择值如果匹配那么它应该显示记录存在，否则它允许将数据插入表

Answer 1

试试这个：

var val = $("#dropdownUsers").val();
var isAvailiable = myFunction(val);
if(isAvailiable){
$.ajax({
        dataType: 'json',
        type: 'POST',
        contentType: "application/Json; charset=utf-8",
        url: '/Home/fillDropdownbyids',
        data: JSON.stringify({ id: val }),
        success: function (response) {
                        $("#id").append(response.user.iUserInfoId + " ")
                         var tr = '';
                         tr += '<tr class="row"><td class="rowid' + val + '">' + response.user.iUserInfoId + '</td><td>' + response.user.strFirstName + '</td><td><input type="text" id="unittextbox' + val + '"></td><td><input type="text" id="quantextbox' + val + '" onfocusout="get(' + val + ')"></td><td><input type="text" id="totaltextbox' + val + '" ></td><td><input type="button" value="Delete" onclick="deleteRow(this)"></td></tr>';
                         $('#tabledata').append(tr);
                           },
    })
} else {
    //Show error message
}

//Call this before calling myFuction();
$('#tableshow > tbody  > tr ').each(function (index, td) {
    arr = $(this).find('td');
    arrlist[index] = arr[0].innerHTML;
    index++;
});

//Function to check that value is present array or not
function myFunction(val){
    var isAvailiable = false
    for(i=0;i<arrlist.length;i++){
        if(arrlist[i] === val){
            isAvailiable = true;
        }
    }
    return isAvailiable;
}

不应允许重复记录

1 个答案: