数据库Integer类型数据为何在iOS端变为String？

This isn't exactly a "bug" in either your PHP or Swift code—it's a combination of how JSON serialization/deserialization works and how you're handling the data in Swift. Let's break this down:

1. What's happening on the PHP side #

Your MySQL id and emailConfirmed are Integer types, and when you fetch them with mysqli_fetch_assoc(MYSQLI_ASSOC), PHP correctly returns them as integer values. When you run json_encode($returnArray), PHP by default serializes integers as JSON numeric types (e.g., "id": 123, "emailConfirmed": 1), so the JSON response is correct here.

2. The issue in your Swift code #

When Swift's JSONSerialization.jsonObject parses the JSON, numeric values are converted to NSNumber (the Objective-C/Swift wrapper for numeric types). The problem comes in how you're accessing these values:

let confirmedUser = parsedJSON["emailConfirmed"] as! String

You're force-casting the NSNumber directly to a String, which causes Swift to implicitly convert the numeric value to its string representation. Similarly, when you save the entire parsedJSON (an NSDictionary) to UserDefaults, any NSNumber values will be treated as strings if you later access them with a string cast.

Fixes you can apply #

Option 1: Correct the type casting in Swift #

Instead of casting to String, cast the values to Int directly to preserve their numeric type:

// Handle emailConfirmed correctly
guard let confirmedUser = parsedJSON["emailConfirmed"] as? Int else {
    print("Failed to parse email confirmation status")
    return
}
if confirmedUser == 1 {
    // Proceed with login flow
    UserDefaults.standard.set(parsedJSON, forKey: "parsedJSON")
    userInfo = UserDefaults.standard.object(forKey: "parsedJSON") as? NSDictionary
    self.activityIndicator.stopAnimating()
    self.performSegue(withIdentifier: "goToHomePage", sender: self)
} else {
    // Show unconfirmed alert
    self.activityIndicator.stopAnimating()
    self.showAlert(alertTitle: "sorry", alertMessage: "please complete your registration process first, please check your email", actionTitle: "OK")
}

// For id, use the same approach when accessing it later
if let userId = userInfo?["id"] as? Int {
    // Use the integer ID as needed
}

Option 2: Explicitly enforce numeric types in PHP (optional) #

While not strictly necessary for your case, you can add the JSON_NUMERIC_CHECK flag to json_encode to eliminate any edge cases where PHP might accidentally serialize numbers as strings:

echo json_encode($returnArray, JSON_NUMERIC_CHECK);

Bonus: Avoid force casting in Swift #

Force casting (as!) can lead to crashes if the JSON structure changes unexpectedly. Using optional binding (like the guard let example above) makes your code more robust and easier to debug.

内容的提问来源于stack exchange，提问作者Agung