Question

我正在使用jQuery-step库和一个组合formvalidation.js的表单。我想延迟低于方法响应，直到ajax完成。

我希望在ajax完成时发送返回 true / false 。

onStepChanging: function (e, currentIndex, newIndex) {

    var fv = $stepForm.data('formValidation'), // FormValidation instance
        // The current step container
        $container = $stepForm.find('section[data-step="' + currentIndex + '"]');

    // Validate the container
    fv.validateContainer($container);
    var isValidStep = fv.isValidContainer($container);
    if (isValidStep === false || isValidStep === null) {
        // Do not jump to the next step
        return false;
    }
    if (isValidStep === true) {
        //call ajax here
        //wait for ajax to complete then move to next or keep 
    }
    return true; //-> i want to delay this 

},

Answer 1

我通过在javascript window中设置属性然后在setTimeout函数中包装返回来完成

第1步：

$(document).ready(function(){ window.flag = ''; });

第2步

function submitForm($stepForm) { 
    $.ajax({
        type: 'POST',
        url: 'http://' + location.host + "/ajax",
        dataType: 'JSON',
        data: {'method': 'submitApplication', 'formdata': $stepForm.serialize()},
        success: function (data) {
          window.flag = true;
        }
}

第3步：

onStepChanging: function (e, currentIndex, newIndex) {
  if (isValidStep === true) {
                submitForm($stepForm);
  }
 .... code
  setTimeout(function () {
     return window.flag;
  }, 2000);
}

Answer 2

在你的情况下：

function submitForm($stepForm) {
    return new Promise((resolve, reject) =>{
      $.ajax({
        type: 'POST',
        url: 'http://' + location.host + "/ajax",
        dataType: 'JSON',
        data: {'method': 'submitApplication', 'formdata': $stepForm.serialize()},
        /* Resolving true on success, you could also return a json object if you need the response */ 
        success: () => { resolve(true); },
        /* Rejecting the promise in case of an error */
        error: () => { reject(false) } 
     })
  })
}

然后：

onStepChanging: function (e, currentIndex, newIndex) {
  if (isValidStep === true) {
       submitForm($stepForm)
       .then( /*...Do whatever you want ..*/)
       .catch( console.log );
  }

如何延迟jquery步骤函数响应

2 个答案: