Question

我必须使用node.js应用程序中的express框架在DynamoDB表中插入以下数据

register.html

<input type = "text" placeholder="First Name" id = "txtFirstName"><br><br>
<input type = "text" placeholder="Last Name" id = "txtLastName"><br><br>        
<input type = "email" placeholder="Email" id = "txtEmail"><br><br>
<input type = "text" placeholder="Phone" id = "txtPhone"><br><br>
<input type = "text" placeholder="Zip Code" id = "txtZip"><br><br>

我理解我需要使用this post

中提到的快速body-parser

但我不清楚如何使用所描述的body-parser方法创建我需要在DynamoDB表中插入的JSON。我可以用jQuery来读取这些html项并创建一个JSON。例如，我需要创建一个如下所示的JSON：

var paramsInsert = {
                TableName:tableName,
                Item:{
                        "email": email,
                        "info":{
                                "FirstName": fName,
                                "LastName": lName,
                                "Phone": phone,
                                "ZipCode": zip
                        }
                }
        };

此paramsInsert最终传递给DynamoDB以插入如下表

insertAtTable(paramsInsert);

如何使用paramsInsert方法创建body-parser？

编辑：关注this link我写了下面的代码，但仍然没有得到输出

app.post('/register.html', function(req, res) {
  const { fname, lname, email, phone, zip } = req.body;
  console.log(fname)
}

Answer 1

您似乎错过了输入元素的name属性：

<input type = "text" placeholder="First Name" id = "txtFirstName" name="firstName">
<input type = "text" placeholder="Last Name" id = "txtLastName" name="lastName">      
<input type = "email" placeholder="Email" id = "txtEmail" name="email">
<input type = "text" placeholder="Phone" id = "txtPhone" name="phone">
<input type = "text" placeholder="Zip Code" id = "txtZip" name="zip">

指定name属性后，您现在应该可以执行以下操作：

app.post('/register.html', function(req, res) {
  const {
    firstName,
    lastName,
    email,
    phone,
    zip
  } = req.body

  const paramsInsert = {
    TableName: 'example',
    Item: {
      firstName,
      lastName,
      email,
      phone,
      zip
    }
  }
}

使用ES2018's object spread：

可以缩短时间

app.post('/register.html', function(req, res) {
  const paramsInsert = {
    TableName: 'example',
    Item: { ...req.body }
  }
}

Express JS：POST请求在DynamoDB中插入

1 个答案: