在Java中如何降低多层嵌套的复杂度_嵌套语法拆分技巧

if (user != null) {
    if (user.isActive()) {
        if (user.hasPermission()) {
            // 主逻辑
        } else {
            throw new SecurityException();
        }
    } else {
        throw new IllegalStateException();
    }
} else {
    throw new IllegalArgumentException();
}

if (user == null) {
    throw new IllegalArgumentException();
}
if (!user.isActive()) {
    throw new IllegalStateException();
}
if (!user.hasPermission()) {
    throw new SecurityException();
}
// 主逻辑，无需嵌套

private boolean canAccessResource(User user, Resource resource) {
    return user != null 
           && user.isActive() 
           && user.hasPermission(resource);
}

if (!canAccessResource(user, resource)) {
    throw new SecurityException();
}
// 执行操作

if ("ALI_PAY".equals(type)) {
    // 支付宝逻辑
} else if ("WECHAT_PAY".equals(type)) {
    // 微信逻辑
}

PaymentStrategy strategy = strategyMap.get(type);
strategy.pay(amount);

if (user != null) {
    Address addr = user.getAddress();
    if (addr != null) {
        String city = addr.getCity();
    }
}

String city = Optional.ofNullable(user)
    .map(User::getAddress)
    .map(Address::getCity)
    .orElse("Unknown");