CODEWITHSUNDEEP
javascriptjsontypescript
MrTouya
MrTouya

Reputation: 656

How do I get JSON.Stringify to parse variable correctly?

I can't figure this out :-( My Client is doing an ajax httppost and I need the following to be sent over:

  private doSaveQuery(queryName: string): void {

    const info = {};
    info['userName'] = 'admin';
    info['userQueryName'] = queryName;
    ...

When it gets stringified it turns it into this:

{"userName":"admin",
 "userQueryName":{"queryName":"MyName"}
}

Why does it take my variable name "queryName" and adds it as a key to my value?????

How do I get to this?

{"userName":"admin",
  "userQueryName":"MyName"
}

I know this is something simple so please forgive my ignorance :-)

Thanks!

Upvotes: 0

Views: 59

Answers (2)

Javid Asgarov
Javid Asgarov

Reputation: 1532

You specified as a type string, and typescript probably should've caught that you're not passing a string. Because if you do, it would do what you want. But probably happens something like this.

function getMyJSON(queryName) {
   const info = {};
    info['userName'] = 'admin';
    info['userQueryName'] = queryName;
  return info;
}

const a = {queryName:"name"};

console.log(JSON.stringify(getMyJSON(a)));

Upvotes: 1

Jim Wright
Jim Wright

Reputation: 6058

When you are encoding the object it looks like queryName is actually an object that looks like {queryName: 'MyName'}.

To fix this you should make sure that queryName is just the string MyName.

info['userQueryName'] = 'MyName'

In order words, queryName is not what you think it is.

Example of your current implementation:

const queryName = {queryName: "MyName"}
const info = {
  userName: "admin",
  userQueryName: queryName
}
console.log(JSON.stringify(info))

You should change queryName to just the string:

const queryName = "MyName"
const info = {
  userName: "admin",
  userQueryName: queryName
}
console.log(JSON.stringify(info))

If you can't do then, then you could build info from the property:

const queryName = {queryName: "MyName"}
const info = {
  userName: "admin",
  userQueryName: queryName.queryName
}
console.log(JSON.stringify(info))

Upvotes: 1

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