Call serviceName.functionName dynamically Angular

Question

I call a function

getData('serviceName', 'functionName')

That function looks like

function getData(service, functionName){
    service.functionName(request, $root.thing).then()
}

But I always get service.function is not a function, how do I call them properly?

fiddle

Have you returned functionName from your service?? Is a factory or service ? — Davide
– Davide, Commented Sep 27, 2016 at 12:23
@Davide I pass the proper parameters, if that's what you're asking. Also console.log(service) and console.log(functionName) give me proper values. Yea it's a factory. — strwoc
– strwoc, Commented Sep 27, 2016 at 12:24
What is .then ? Is a promise? Uhm are you sure that you return that function from your service? You are in controller when you call getData so? If you can provide a plunker we can help you better — Davide
– Davide, Commented Sep 27, 2016 at 12:26
I've edited the fiddle with the service and with the main controller part, I hope it helps. — strwoc
– strwoc, Commented Sep 27, 2016 at 12:36
myService is a string. It is not a real service. It doesn't have findData method. Please, provide full code and not the excerpts, so it would be possible to note what exactly should be fixed in your code. — Estus Flask
– Estus Flask, Commented Sep 27, 2016 at 12:39

taguenizy · Accepted Answer · 2016-09-27 14:06:24Z

2

If you use bracket notation you can achieve what you want

function getData(service, functionName){
    service[functionName](request, $root.thing).then()
}

answered Sep 27, 2016 at 14:06

taguenizy

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Comments

Ujjwal kaushik · Accepted Answer · 2016-09-27 13:15:32Z

0

function getData(service, functionName){
  var yourservice = $injector.get(service);
  yourservice.yourActualfunctionName(request, $root.thing).then() 
}

You can sure use your service as string ,Use it like this.Make sure you inject $injector wherever you are using it. not sure about function name because function is not in same file.

edited Sep 27, 2016 at 13:15

answered Sep 27, 2016 at 12:51

Ujjwal kaushik

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3 Comments

Matt Over a year ago

eval() is never the answer, especially since the language supports this with brackets: yourService[functionName](...)

Ujjwal kaushik Over a year ago

Yeah , you are right. My bad :( but this is the correct way to use service as string I think.

strwoc Over a year ago

The brackets did the thing, but I have to keep the service name static.

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