Question

我正在使用jquery和ajax在用户从下拉框中选择值时运行MySQL查询。我想将第一个和第二个下拉框中的值作为帖子传递到下一页，查询可以检查该值是否存在。

我很容易就能像这样传递一个值：

       <script type="text/javascript">
            $(function() {
//alert('Document is ready');

                $('#block').change(function() {
                    var sel_stud2 = $('#username_select').val();
//alert('You picked: ' + sel_stud);

                    $.ajax({
                        type: "POST",
                        url: "include/block_user.php",
                        data: 'theOption2=' + sel_stud2,
                        success: function(whatigot) {
//alert('Server-side response: ' + whatigot);
                            $('#LaDIV2').html(whatigot);
                            $('#theButton').click(function() {
                                alert('You clicked the button');
                            });
                        } //END success fn
                    }); //END $.ajax
                }); //END dropdown change event
            }); //END document.ready
        </script>

然而，当我尝试传递我的选择下拉框值时，我得到两者的索引未定义错误。这是我尝试过的。请有人告诉我我哪里出错了，谢谢

<script type="text/javascript">
            $(function() {
//alert('Document is ready');

                $('#block').change(function() {
                    var sel_stud2 = $('#username_select').val();
                    var sel_stud3 = $('#block').val();
//alert('You picked: ' + sel_stud);

                    $.ajax({
                        type: "POST",
                        url: "include/block_user.php",
                        data: "{'theOption2=': + sel_stud2,'theOption3=': + sel_stud3}",  
                        success: function(whatigot) {
//alert('Server-side response: ' + whatigot);
                            $('#LaDIV2').html(whatigot);
                            $('#theButton').click(function() {
                                alert('You clicked the button');
                            });
                        } //END success fn
                    }); //END $.ajax
                }); //END dropdown change event
            }); //END document.ready
        </script>

Answer 1

@A. Wolff是正确的。您正在引用您的对象数据（并且您的注释表明您仍然留下外部双引号）：

请改为尝试：

     data: {theOption2: sel_stud2,
            theOption3: sel_stud3},

在ajax jquery中组合多个数据？

1 个答案: