我被要求处理这个后端预定作业，该作业将一些客户数据（从电子商务数据库）导出到自定义格式文本文件。接下来的代码就是我找到的。
我只是想删除它，但我不能。我是否有可能在不改变它的情况下改进它？
public class AConverter implements CustomerConverter {
protected final Logger LOG = LoggerFactory.getLogger(AConverter.class);
private final static String SEPARATOR = ";";
private final static String CR = "\n";
public String create(Customer customer) {
if (customer == null)
return null;
LOG.info("Exporting customer, uidpk: {}, userid: {}", customer.getUidPk(), customer.getUserId());
StringBuilder buf = new StringBuilder();
buf.append("<HEAD>");
buf.append(SEPARATOR);
buf.append(String.valueOf(customer.getUidPk()));
buf.append(SEPARATOR);
byte[] fullName = null;
try {
fullName = customer.getFullName().getBytes("UTF-8");
} catch (UnsupportedEncodingException e1) {
fullName = customer.getFullName().getBytes();
}
String name = null;
try {
name = new String(fullName, "UTF-8");
} catch (UnsupportedEncodingException e) {
name = customer.getFullName();
}
buf.append(limitString(name, 40));
buf.append(SEPARATOR);
final CustomerAddress preferredShippingAddress = customer.getPreferredShippingAddress();
if (preferredShippingAddress != null) {
final String street1 = preferredShippingAddress.getStreet1();
if (street1 != null) {
buf.append(limitString(street1, 40));
}
} else {
buf.append(" ");
}
buf.append(SEPARATOR);
final String addressStr = buildAddressString(customer);
buf.append(limitString(addressStr, 40));
buf.append(SEPARATOR);
buf.append(limitString(customer.getEmail(), 80));
buf.append(SEPARATOR);
if (preferredShippingAddress!=null && preferredShippingAddress.getStreet2() != null) {
buf.append(limitString(preferredShippingAddress.getStreet2(), 40));
} else {
buf.append(" ");
}
buf.append(SEPARATOR);
buf.append(limitString(customer.getPhoneNumber(), 25));
buf.append(SEPARATOR);
if (preferredShippingAddress!=null) {
if(preferredShippingAddress.getCountry()!=null) {
buf.append(preferredShippingAddress.getCountry());
} else {
buf.append(" ");
}
} else {
buf.append(" ");
}
buf.append(SEPARATOR);
if (preferredShippingAddress!=null) {
if(preferredShippingAddress.getCountry()!=null) {
buf.append(preferredShippingAddress.getCountry());
} else {
buf.append(" ");
}
} else {
buf.append(" ");
}
buf.append(SEPARATOR);
String fodselsnummer = " ";
try {
Map<String, AttributeValue> profileValueMap = customer.getProfileValueMap();
AttributeValue attributeValue = profileValueMap.get("CODE");
fodselsnummer = attributeValue.getStringValue();
} catch (Exception e) {
}
buf.append(fodselsnummer);
buf.append(CR);
final String string = buf.toString();
return string;
}
private String buildAddressString(Customer customer) {
final CustomerAddress preferredShippingAddress = customer.getPreferredShippingAddress();
if (preferredShippingAddress != null) {
final String zipOrPostalCode = preferredShippingAddress.getZipOrPostalCode();
final String city = preferredShippingAddress.getCity();
if (zipOrPostalCode != null && city != null) {
return zipOrPostalCode + " " + city;
} else if(zipOrPostalCode == null && city != null) {
return city;
} else if(zipOrPostalCode != null && city == null) {
return zipOrPostalCode;
}
}
return " ";
}
private String limitString(String value, int numOfChars) {
if (value != null && value.length() > numOfChars)
return value.substring(0, numOfChars);
else
return value;
}
}
答案 0 :(得分：0)
你说你想改进它，你想删除它，但你不能。我不确定你为什么不能。我也不明白你为什么要删除它。但在我看到Martin Fowler的 Refactoring 之前，这听起来像我曾经拥有的态度。如果你还没有，我强烈建议你阅读那本书。
当然可以在不重写所有内容的情况下改进此代码（或任何代码）。最明显的改进是通过创建一些实用程序方法来消除create方法中的一些重复代码，然后将create方法拆分为几个较小的方法 - 模板方法。
此外，create方法中存在一些可疑的代码，它将客户的名称转换为UTF-8字节流，然后再转换为字符串。我无法想象这是为了什么。最后，如果customer为null，则返回null。这不太可能是必要的或明智的。
为了好玩，我决定对这段代码进行一些重构。 （请注意，正确的重构涉及单元测试;我没有对此代码进行任何测试，甚至没有编译下面的代码，更不用说测试它了。）以下是一种可以重写此代码的方法：
public class AConverter implements CustomerConverter {
protected final Logger LOG = LoggerFactory.getLogger(AConverter.class);
private final static String SEPARATOR = ";";
private final static String CR = "\n";
public String create(Customer customer) {
if (customer == null) throw new IllegalArgumentException("no cust");
LOG.info("Exporting customer, uidpk: {}, userid: {}",
customer.getUidPk(), customer.getUserId());
StringBuilder buf = new StringBuilder();
doHead(buf, customer);
doAddress(buf, customer);
doTail(buf, customer);
return buf.toString();
}
private void doHead(StringBuilder buf, Customer customer) {
append(buf, "<HEAD>");
append(buf, String.valueOf(customer.getUidPk()));
append(buf, limitTo(40, customer.getFullName()));
}
private void doAddress(StringBuilder buf, Customer customer) {
append(buf, limitTo(40, street1of(customer)));
append(buf, limitTo(40, addressOf(customer)));
append(buf, limitTo(80, customer.getEmail()));
append(buf, limitTo(40, street2of(customer)));
append(buf, limitTo(25, customer.getPhoneNumber()));
append(buf, countryOf(customer));
append(buf, countryOf(customer));
}
private void doTail(StringBuilder buf, Customer customer) {
buf.append(fodselsnummerOf(customer));
buf.append(CR);
}
private void append(StringBuilder buf, String s) {
buf.append(s).append(SEPARATOR);
}
private String street1of(Customer customer) {
final CustomerAddress shipto = customer.getPreferredShippingAddress();
if (shipto == null) return " ";
if (shipto.getStreet1() != null) return shipto.getStreet1();
return " ";
}
private String street2of(Customer customer) {
final CustomerAddress shipto = customer.getPreferredShippingAddress();
if (shipto == null) return " ";
if (shipto.getStreet2() != null) return shipto.getStreet2();
return " ";
}
private String addressOf(Customer customer) {
final CustomerAddress shipto = customer.getPreferredShippingAddress();
if (shipto == null) return " ";
final String post = preferredShippingAddress.getZipOrPostalCode();
final String city = preferredShippingAddress.getCity();
if (post != null && city != null) return post + " " + city;
if (post == null && city != null) return city;
if (post != null && city == null) return post;
return " ";
}
private String countryOf(Customer customer) {
final CustomerAddress shipto = customer.getPreferredShippingAddress();
if (shipto == null) return " ";
if (shipto.getCountry() != null) return shipto.getCountry();
return " ";
}
private String limitTo(int numOfChars, String value) {
if (value != null && value.length() > numOfChars)
return value.substring(0, numOfChars);
return value;
}
private String fodelsnummerOf(Customer customer) {
try {
Map<String, AttributeValue> profileValueMap =
customer.getProfileValueMap();
AttributeValue attributeValue = profileValueMap.get("CODE");
return attributeValue.getStringValue();
} catch (Exception e) {
return " ";
}
}
}
我还注意到，如果客户数据的任何字段（例如电子邮件地址）中碰巧有分号，则自定义格式文本文件的格式存在问题，因为这是您的分隔符字符。我相信这是一个众所周知的问题？