Question

如何根据某些ajax调用的结果控制移动到下一步？ data.d返回bool值

$("#wizard").steps({
            onStepChanging: function (event, currentIndex, newIndex) {
                var move = false;
                if (currentIndex == 2) {
                    move = false;
                    $.ajax({
                        type: 'POST',
                        url: "Reservation.aspx/SomeFunction",
                        data: serializeData({  }),
                        contentType: "application/json",
                        dataType: 'json',
                        success: function (data) {
                            move = data.d;
                            return true;
                        },
                        error: ajaxLoadError
                    });
                }
                return move;
            },
            saveState: true

        });

Answer 1

$.ajax({
    type: 'POST',
    url: "Reservation.aspx/SomeFunction",
    async: false,
    contentType: "application/json",
    dataType: 'json',
    success: function (data) {
       move = data.d;
       return true;
    },
    error: ajaxLoadError
});

Answer 2

如果您不希望$ .ajax（）函数立即返回，请将async选项设置为false：

如果服务器没有响应你的ajax调用，那么为Ajax设置超时，那么它将继续进行下一个进程。

$("#wizard").steps({
            onStepChanging: function (event, currentIndex, newIndex) {
                var move = false;
                if (currentIndex == 2) {
                    move = false;
                    $.ajax({
                        type: 'POST',
                        url: "Reservation.aspx/SomeFunction",
                        data: serializeData({  }),
                        contentType: "application/json",
                        dataType: 'json',
                        async: false,
                        cache: false,
                        timeout: 30000,
                        success: function (data) {
                            move = data.d;
                            return true;
                        },
                        error: ajaxLoadError
                    });
                }
                return move;
            },
            saveState: true

        });

Answer 3

你可以使用Samy的同步ajax请求方法

$("#wizard").steps({
    onStepChanging: function (event, currentIndex, newIndex) {
        if (currentIndex == 2) {
            var requestResult = $.ajax({
                type: 'POST',
                url: "Reservation.aspx/SomeFunction",
                async: false,
                contentType: "application/json",
                dataType: 'json',
                error: ajaxLoadError
            });
            return requestResult.responseJSON.Result == "/*your expected value*/"
        }
    },
    saveState: true
});

Answer 4

我有同样的问题，我甚至想使用＆＃34; setStep＆＃34;在ajax加载后强制执行步骤，然后最新版本的jquery.steps取出了＆＃34; setStep＆＃34; ..

我最终使用＆＃34; next＆＃34;方法，并且必须使用全局触发器来停止onChanging事件的无限循环，简而言之，如果ajax返回有效数据，我强制向导转到下一步，否则，它将保持到当前步骤，这里是＃s; s代码

var $stopChanging = false; 

.... ....

onStepChanging: function (event, currentIndex, newIndex) {
      if ($stopChanging) {
        return true;
      } else {
        items = $.ajax({
        type: 'POST',
        url: "Reservation.aspx/SomeFunction",
        data: serializeData({  }),
        contentType: "application/json",
        dataType: 'json',
        success: function (data) {
            $stopChanging = true;
            wizard.steps("next");
        },
        error: ajaxLoadError
    });
   },
   onContentLoaded: function (event, currentIndex) {
       $stopChanging = false;
   }
}

逻辑如下：

点击＆＃34; next＆＃34;按钮触发onStepChanging
默认情况下，设置jquery.steps onStepChanging事件返回false，然后是$ .ajax调用，如果它返回有效数据（成功），则调用jquery.steps到转到下一步，onStepChanging再次触发，如果不是有效，什么都不做，保持现阶段

每次触发两个onStepChanging事件听起来不是一个好主意，但这就是我现在可以拥有的东西

您可能需要为不同的步骤索引行为添加更多条件

Answer 5

$("#wizard").steps({
        onStepChanging: function (event, currentIndex, newIndex) {
            var $out= false;
            if (currentIndex == 2) {
                $out= false;
                $.ajax({
                    type: 'POST',
                    url: "Reservation.aspx/SomeFunction",
                    data: serializeData({  }),
                    contentType: "application/json",
                    dataType: 'json',
                    success: function (data) {
                        move = data.d;

                        $out = true;
                    },
                    error: ajaxLoadError
                });
            }
            return $out;
        },
        saveState: true

    });

将全局变量$ out！

Answer 6

var items;

$("#wizard").steps({
onStepChanging: function (event, currentIndex, newIndex) {
    var move = false;
    if (currentIndex == 2) {
        move = false;

        items = $.ajax({
            type: 'POST',
            url: "Reservation.aspx/SomeFunction",
            data: serializeData({  }),
            contentType: "application/json",
            dataType: 'json',
            success: function (data) {
                move = data.d;
                return true;
            },
            error: ajaxLoadError
        });


    }
    return move;
},
saveState: true

});



items.success(function (data) {
//if can log in go to logged in page
if (data.success == true) {
    alert("Working");
    var move = data.d;
    return true;

} else {
    alert("didnt work");
}
// output of data
alert(JSON.stringify(data));
});

Answer 7

这是我经过多次尝试后才能开始工作的唯一方法，这就是@ joe-lu在上面的表现。你只想开始异步电话＆amp;返回false。这将使向导保持相同的步骤。然后在成功处理程序中，您以编程方式进入下一步。

$("#wizard").steps({
            onStepChanging: function (event, currentIndex, newIndex) {
                if (currentIndex == 2) {
                    //Would be a good idea to start a spinner here!
                    //would be a good idea to disable next button here!
                    $.ajax({
                        type: 'POST',
                        url: "Reservation.aspx/SomeFunction",
                        data: serializeData({  }),
                        contentType: "application/json",
                        dataType: 'json',
                        success: function (data) {
                            //stop spinner here!
                            //programmatically move to next step on success.
                            $("#wizard").steps("next");
                        },
                        error: ajaxLoadError
                    });
                }
                //Prevents natural movement to next step.
                //will be done programmatically
                return false;
            },
            saveState: true
        });

Answer 8

我遇到了类似的问题，但我使用parsleyjs进行验证。你可能会在我的代码中得到一个想法。

我的代码是这样的：

             onStepChanging: function (event, currentIndex, newIndex) {

                 // ======== Code that fails 

                 //var step = $wizard_advanced.find('.body.current').attr('data-step'),
                 //$current_step = $('.body[data-step=\"'+ step +'\"]');                        


                // check input fields for errors
                //$current_step.find('[data-parsley-id]').each(function() {
                    //this adds .md-input-danger to inputs if invalid
                    //there is remote validation occurring here via ajax
                    // async: false
                    //$(this).parsley().validate();
                //});

                // this is executed before ajax validation is finished 
                //return $current_step.find('.md-input-danger').length ? false : true;

                // ======== END of Code that fails 

                // FIX
                // waits on ajax validation to finish before returning
                if( $wizard_advanced_form.parsley().validate() ) {
                    return true;
                } else {
                    return false;
                }
                //FIX                    
            }

Answer 9

var is_async_step = false;
$("#wizard").steps({
        onStepChanging: function (event, currentIndex, newIndex) {
            //USED TO SEND USER TO NEXT STEP IS ASYNC REQUEST IS PRESENT - FOR AJAX CALL 
            if (is_async_step) {
                is_async_step = false;
                //ALLOW NEXT STEP
                return true;
            }

            if (currentIndex == 2) {                
                $.ajax({
                    type: 'POST',
                    url: "Reservation.aspx/SomeFunction",
                    data: serializeData({  }),
                    contentType: "application/json",
                    dataType: 'json',
                    success: function (data) {
                        move = data.d;

                        //Add below 2 lines for every Index(Steps).                            
                        is_async_step = true;
                        //This will move to next step.
                        $(form).steps("next");
                    },
                    error: ajaxLoadError
                });
            }
             //Return false to avoid to move to next step
             return false;
        },
        saveState: true
    });

Answer 10

我找到了另一个解决此问题的方法。 OnStepChanging仅支持boolean。有拉取请求#232，它增加了延迟对象的用法。（我在GitHub上也找到了如何使用延迟对象的方法）我将此修改后的版本包含在我的项目中，并在OnStepChanging中使用了它，如下所示：

    var def = $.Deferred();

    $.ajax({
        type: "POST",
        url: url,
        //async: false,
        beforeSend: function (xhr) {
            //ASP CORE Antiforgery token
            xhr.setRequestHeader("RequestVerificationToken",
                $('input:hidden[name="__RequestVerificationToken"]').val());
        },
        data: data,
        contentType: "application/json; charset=utf-8",
        dataType: "json",
        failure: function (xhr) {
            failureCallback(xhr);
        }
    }).done(function (response) {
        //Result of server validation
        var responseResult = response.result === "Success" ? true : false;
        // Check response
        def.resolve(responseResult);
        }).fail(function (response) {
            console.log(response);
            return false;
        });

    return def; // This is the Deferred that should be returned and NOT the one from jQuery Ajax

我希望这会对其他人有所帮助。：-）

jQuery向导步骤在ajax调用完成之前移动

10 个答案: