Swift - Construct URL with filter parameters

Question

I am trying to construct an URL with filter parameters, like so:

let url = URL(string: "http://myapi.com/api/customers?filter={\"where\":{\"name\":\"myName\"}}")

But this returns nil. Is there a way to have an URL object of this format? Or should I use another method to construct my request?

Thanks in advance !

Use URLComponents, see for example https://stackoverflow.com/a/24888789/1187415 or https://stackoverflow.com/a/43668198/1187415. — Martin R, Sep 27 '18 at 12:40

score 2 · Accepted Answer · answered Sep 27 '18 at 12:39

2

var url = URLComponents(string: "http://myapi.com/api/customer")!

url.queryItems = [
    URLQueryItem(name: "filter", value: "{\"where\":{\"name\":\"myName\"}}")
]

answered Sep 27 '18 at 12:39

meggar

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That did the work, thank you! – M. Garrigues Sep 27 '18 at 12:57

ielyamani · Answer 2 · 2018-09-27T12:47:51.293

1

You have to add percent encoding to the string if it doesn't starts with http://:

let string = "myapi.com/api/customers?filter={\"where\":{\"name\":\"myName\"}}"

let escaped = string.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!

let url = URL(string: escaped)

edited Sep 27 '18 at 12:47

answered Sep 27 '18 at 12:28

ielyamani

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5

That won't work because it will escape the `://` and `=` parts of the URL. You need to escape the parts first, then assemble them. – Duncan C Sep 27 '18 at 12:31
That worked too, thanks! Though I think I should still use URLComponents as it's its purpose? – M. Garrigues Sep 27 '18 at 12:56
1

@M.Garrigues yes, that is the recommended way of adding query items – ielyamani Sep 27 '18 at 12:57

Swift - Construct URL with filter parameters

2 Answers2