我正在使用Jquery创建评论,并希望将其添加到页面而不重新加载页面。我有一个commentConfirm原型函数在发布之前捕获它。我不能在该项目上使用remote:true,所以代码如下:
$(function createComment() {
$("#new_comment").on("submit", function(e) {
const values = {
description: $('#comment_description').val(),
rating: $('#comment_rating').val()
};
const newComment = new Comment(values);
newComment.commentConfirm();
});
});
function Comment(comment) {
this.description = comment.description;
this.rating = comment.rating;
}
Comment.prototype.commentConfirm = function(e) {
let doIt = confirm(`You are about to comment: "${this.description}" and give a rating of: ${this.rating} stars`);
if(!doIt)
return;
let params = {
'comment[description]': this.description,
'comment[rating]': this.rating
};
$.post(this.action, params).success(function(response) {
$('div.comments_container').append('<div class="new_comment_' + `${response.id}` + '"> </div>')
$('div.new_comment_'+ `${response.id}`).append('<h3 class="cheading">' + `${response.user.name}` + ' gives ' + `${response.rating}` + ' out of 5 stars! </h3>')
$('div.new_comment_'+ `${response.id}`).append('<p class="cdescription">' + `${response.description}` + '</p>')
$('div.new_comment_'+ `${response.id}`).append('<a class="ecomment" href="/recipes/' + `${response.recipe_id}` + '/comments/' + `${response.id}` + '/edit">Edit</a>' + " ")
$('div.new_comment_'+ `${response.id}`).append('<a class="dcomment" rel="nofollow" data-method="delete" href="/comments/' + `${response.id}` + '">Delete</a>')
$('form#new_comment')[0].reset();
});
};
不确定这是否引起问题,但这是我在控制器中的create函数:
def create
if logged_in?
comment = Comment.new(comment_params)
comment.recipe = find_by_recipe_id
comment.user = current_user
if comment.description.empty? || comment.rating == nil
redirect_to recipe_path(comment.recipe), alert: "Please fill out all fields"
else
comment.save
render json: comment.to_json(only: [:rating, :description, :id, :recipe_id],
include: [user: { only: [:name]}])
end
else
redirect_to login_path, alert: "You must be logged in to comment"
end
end
任何帮助他解决问题的人,将不胜感激!
如果可以帮助回答其他一些问题,请访问以下仓库:https://github.com/Bartekswistak/fun_guy_chef/tree/jquery
答案 0 :(得分:0)
我无法确认您的其余代码是否正常工作,但是您将需要手动处理submit
事件以防止页面重新加载...
$("#new_comment").on("submit", function(e) {
e.preventDefault();
....
答案 1 :(得分:0)
submit event重新加载页面。要阻止它重新加载页面,请prevent its default action。因此,您的第一个功能将如下所示:
$(function createComment() {
$("#new_comment").on("submit", function(e) {
e.preventDefault();
const values = {
description: $('#comment_description').val(),
rating: $('#comment_rating').val()
};
const newComment = new Comment(values);
newComment.commentConfirm();
});
});