if let variable - Use of unresolved identifier - ios

Question

I am using SwiftyJSON to call some APIs and fetch some data.
When I use:
if let variable = json["response"]["fieldname"] {
} else {
println("error")
}
I am not able to use the variable later on, for example to append the value to an array.
For example:
if let variable1 = json["response"]["fieldname1"] {
} else {
println("error")
}
if let variable2 = json["response"]["fieldname2"] {
} else {
println("error")
}
var currentRecord = structure(variable1, variable2) ---> This line returns an error (use of unresolved identifier variable1) as not able to find variable1 or variable2
myArray.append(currentRecord)
How can I solve this?

Answer 1

The scope of an if let is inside the brackets immediately following it:
if let jo = joseph {
// Here, jo is in scope
} else {
// Here, not in scope
}
// also not in scope
// So, any code I have here that relies on jo will not work
In Swift 2, a new statement, guard was added, that seems to have exactly the kind of behaviour you want:
guard let jo = joseph else { // do something here }
// jo is in scope
If you're stuck in Swift 1, though, an easy way for you to unwrap those variables without a pyramid of doom is:
if let variable1 = json["response"]["fieldname1"], variable2 = json["response"]["fieldname2"] {
var currentRecord = structure(variable1, variable2)
myArray.append(currentRecord)
} else {
println("error")
}

Answer 2

#oisdk already explained that the scope of a variable defined by if let is only inside the braces of that statement.
That's what you want, because if it if let statement fails, the variable is undefined. The whole point of if let is to unwrap your optional safely, so that inside the braces, you can be sure the variable is valid.
Another solution to your problem (in Swift 1.2) is to use multiple if let statements:
if let variable1 = json["response"]["fieldname1"],
let variable2 = json["response"]["fieldname2"]
{
//This code will only run if both variable1 and variable 2 are valid.
var currentRecord = structure(variable1, variable2)
myArray.append(currentRecord)}
else
{
println("error")
}

Answer 3

Your code checks variable2 always even variable 1 fails. But that causes (edited!) not the error.
You can check and assign both variables in the same line. The "true" branch will be executed only if both variables are not nil
let response = json["response"]
if let variable1 = response["fieldname1"], variable2 = response["fieldname2"] {
let currentRecord = structure(variable1, variable2)
myArray.append(currentRecord)
} else {
println("error")
}