Parsing JSON from PHP in Swift 4

Question

I'm getting an error when loading JSON from a PHP output page using Swift 4.

Note: The PHP script has been checked and is working fine

let myUrl = URL(string: "http://localhost/swift/signin_up/login.php");
var request = URLRequest(url:myUrl!)
request.httpMethod = "POST"// Compose a query string
request.addValue("application/json", forHTTPHeaderField: "Content-Type")
DispatchQueue.main.async {
    let task = URLSession.shared.dataTask(with: request) { (data: Data?, response: URLResponse?, error: Error?) in
        if error != nil {
            print("error=\(error!)")
            return
        }

        // You can print out response object
        print("response = \(response!)")
        print("==================================")
        //Let's convert response sent from a server side script to a NSDictionary object:
        do {
            let json = try JSONSerialization.jsonObject(with: data!, options: .mutableContainers) as? NSDictionary

            if let parseJSON = json {
                print(parseJSON)
            }
        } catch {
            print(" error:\(error)")
        }
    }
    task.resume()

I get this error message from Xcode:

error: Error Domain=NSCocoaErrorDomain Code=3840 "JSON text did not start with array or object and option to allow fragments not set." UserInfo={NSDebugDescription=JSON text did not start with array or object and option to allow fragments not set.}

When I changed options to:

options: [.mutableContainers, .allowFragments]

I get this error message:

error: Error Domain=NSCocoaErrorDomain Code=3840 "Garbage at end." UserInfo={NSDebugDescription=Garbage at end.}

This is the PHP output I tested:

{"id":"1","email":"aa","password":"aa"}

When you get some error, please include the error message at least. — OOPer
– OOPer, Commented Aug 14, 2018 at 21:51
I guess your php server is returning some sort of error, what do you get if you put print(String(data: data!, encoding: .utf8)) before the line let json = ... ? — OOPer
– OOPer, Commented Aug 14, 2018 at 22:38
i replace a code with above and get message error{error:Error Domain=NSCocoaErrorDomain Code=3840 "Invalid value around character 0." UserInfo={NSDebugDescription=Invalid value around character 0.} @OOPer — M_Zakaria
– M_Zakaria, Commented Aug 14, 2018 at 22:49
print(String(data: data!, encoding: .utf8)) would never output such message and I never suggested any sort of replace. Are you reading my comment? — OOPer
– OOPer, Commented Aug 14, 2018 at 22:51

M_Zakaria · Accepted Answer · 2018-08-15 05:22:50Z

1

Be careful when you set character_set in database server choose (utf8_unicode_ci) to get correct json format when using

json_encode(//json variable);

in php

answered Aug 15, 2018 at 5:22

M_Zakaria

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