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I am using Joi to validate a payload of a service in my node.js server using hapijs framework. It used to look like this (in my typescript code as well as after compiling to javascript):

payload: {
    para1: Joi.number().required(),
    para2: Joi.string()
}

Now I want to set default value of the two parameters. If the code is written in javascript, I can do this:

payload: {
    para1: Joi.number().required().default(1),
    para2: Joi.string().default("defaultstring")
}

I tested it in swagger and the default values actually became the values I set.

However, my project is written in typescript. I did the same thing and compiled typescript code. The result javascript looks like this: 

 payload: {
    para1: Joi.number().required()["default"](1),
    para2: Joi.string()["default"]("defaultstring")
 }

In swagger, the default values are not applied.

Here are my questions:

  1. why the code becomes different after compiling?
  2. what does ["default"]("defaultstring") mean and what does it do?
  3. how can I write typescript code to make sure it can compiled as Joi.string().default("defaultstring")

Update

According to @rsp's post, the format in question 2 is just different way to access object's property. I also get reference from here. But it doesn't explain if they have any difference. Does anyone have any idea?

Update2

Here is the difference between the two ways accessing JS property. It seems there is no negative effect using brackets way. However, in my case, the default values are not reflected on swagger. Will be doing research on it.

Community
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zhangjinzhou
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  • Finally figured it out. When I use typescript, the swagger notes are generated from typescript initial function, not from JS file. There is not issue with the code itself. – zhangjinzhou Mar 24 '17 at 22:06

4 Answers4

8

In JavaScript this:

required().default(1)

is the same as this:

required()["default"](1)

because you can access object properties either as:

object["propertyName"]

or:

object.propertyName

(with certain restrictions in the second case).

So it's strange that TypeScript would output the longer style if it doesn't have to, but it's also strange that the longer style doesn't work exactly the same as the shorter one.

I would try to manually change the compiled JavaScript to the shorter version and see if that helps. If it doesn't then the problem is somewhere else. My suspicion is that it will not help.

The .default() should work in TypeScript because it is defined in @types/joi - see:

But on the other hand there is this comment:

// TODO express type of Schema in a type-parameter (.default, .valid, .example etc)

Which may suggest that .default() implementation is not ready yet - see:

and also there's this issue: joi.d.ts out of date, missing types

rsp
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  • Thanks for the answer. It's good to know that there is another way to access object in JS. Probably I can't take you solution because changing compiled JavaScript is not a sustainable way to handle the issue. I use gulp when develop the software. Therefore every time I changed something all the files will be re-compiled. – zhangjinzhou Mar 24 '17 at 21:23
  • if the @types/joi is not well developed, why it was still compiled as the longer style JS? – zhangjinzhou Mar 24 '17 at 21:29
0

Use should use default() like in below code:

 validate: {
        payload: {
            para1: Joi.number().integer().min(1).max(100).default(10).required(),
            para2: Joi.string().min(1).max(100).default("TEST").required(),
        }
    }
Shubham Verma
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0

It still says key is required when we use Joi.boolean().required().default(true)

The below worked for me

JOI.boolean().default(false)
Sharan
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0
taxAmount: Joi.number().default(0),
totalAmount: Joi.number().default(0),

This works for me. Do not use .reqquired() keyword. The result will be in the

const validationResult = schema.validate(data);
validationResult.value
rbansal
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