rxjs: rxjs/webSocket SyntaxError: Unexpected token s in JSON at position 0

Bug Report

Current Behavior using string urlConfigOrSource generates SyntaxError: Unexpected token s in JSON at position 0 with WebSocketSubjectConfig and defined deserializer function error does not appear.

Reproduction

    this.pubsubSubject = webSocket('ws://localhost:8080');
    this.pubsubSubject.subscribe(
      (msg) => console.log('message received: ' + msg),
      (err) => console.log(err),
      () => console.log('complete')
    );

throws error

SyntaxError: Unexpected token s in JSON at position 0
    at JSON.parse (<anonymous>)
    at deserializer (WebSocketSubject.js:12)
    at tryCatcher (tryCatch.js:6)
    at WebSocket.socket.onmessage [as __zone_symbol__ON_PROPERTYmessage] (WebSocketSubject.js:174)
    at WebSocket.wrapFn (zone.js:1188)
    at ZoneDelegate.push../node_modules/zone.js/dist/zone.js.ZoneDelegate.invokeTask (zone.js:421)
    at Object.onInvokeTask (core.js:3811)
    at ZoneDelegate.push../node_modules/zone.js/dist/zone.js.ZoneDelegate.invokeTask (zone.js:420)
    at Zone.push../node_modules/zone.js/dist/zone.js.Zone.runTask (zone.js:188)
    at ZoneTask.push../node_modules/zone.js/dist/zone.js.ZoneTask.invokeTask [as invoke] (zone.js:496)

to fix this we need use:

this.pubsubSubject = webSocket({
    url: 'ws://localhost:8080',
    deserializer: () => {}
});

after that this.pubsubSubjectsubscribe() doesn’t throw error

Environment

Runtime: @angular 6.1.8 chrome 69.0.3497.100
RxJS version: 6.3.2
angular-cli 6.2.3,

About this issue

Original URL
State: closed
Created 6 years ago
Reactions: 1
Comments: 16 (10 by maintainers)

Most upvoted comments

By default webSocketSubject is trying to deserialize the message by applying JSON.parse(). If you message is not an string is going to fail (the error you are getting now). Now when you tell to webSocket to use this deserializer () => {}, you are telling it, just give me back an undefined value (Since it is returning nothing).

Here you have two options,

use JSON.stringify on the server or any other tool to make the message an string.
use a custom deserializer but not forget to get your data back. 😄

Example: Here you can apply whatever transformation you want. I just not applying anything but avoiding the default behaviour (JSON.parse()). Since JSON.parse()attempts to transform an string to a JSON object and surely you are not expecting that type of data (an string coming from your server).

const WebSocketSubject = webSocket({
    url: 'ws://localhost:3200',
    deserializer: msg => msg
});

+11

luillyfe on Oct 16, 2018

This seems like it was just a debugging session. Closing.

benlesh on Feb 9, 2021

@luillyfe it works in this way, thanks.

iamcco on Oct 24, 2018