【发布时间】:2020-01-25 23:07:10
【问题描述】:
我的网站在单击“喜欢”按钮后挂起。我的 mongo db 更新正常,如果我在单击后重定向站点以加载新页面,那很好。有没有办法在不重新加载页面的情况下做到这一点?
JS
app.post("/", async function(req) {
console.log("/ works as post route");
const requestedPostID = Object.keys(req.body)
console.log(Object.keys(req.body));
// Load the document
const doc = await Post.findOne({ _id: requestedPostID });
const update = { likeCount: 5 };
await doc.updateOne(update);
const updatedDoc = await Post.findOne({ likeCount: 5 });
updatedDoc.likeCount;
});
HTML
<form class="" action="/" method="post">
<div class="field is-grouped">
<button type="submit" name="<%=post._id%>" class="card-footer-item"><i class="fas fa-thumbs-up"></i></button>
<button type="submit" name="dislikeButton" class="card-footer-item"><i class="fas fa-thumbs-down"></i></button>
</div>
</form>
【问题讨论】:
标签: node.js mongodb mongoose ejs