情况:我正在检查文件名,文件名存储在名为String
的{{1}}变量中,并且根据我在str
语句中检查的条件设置名为if
的变量的值。
mailType
问题:有没有更好的方法在java中缩短我的代码并且对内存友好?
答案 0 :(得分:20)
使用LinkedHashMap<String, String>
:
LinkedHashMap<String, String> mapping = new LinkedHashMap<>();
mapping.put("unsupported", "unsupported");
mapping.put("final_result", "final_result");
// ... etc
然后迭代地图,直到找到匹配的键:
for (Map.Entry<String, String> entry : mapping.entrySet()) {
if (str.contains(entry.getKey()) {
mailType = entry.getValue();
break;
}
}
这里的关键点是LinkedHashMap
保留了插入顺序(与HashMap
不同),因此您可以实际指定要测试匹配的顺序(其他地图实现也会这样做,例如Guava的ImmutableMap
; LinkedHashMap
只是您开箱即用的内容之一。)
如果你需要为外壳嵌套这个,你可以简单地应用相同的模式:
LinkedHashMap<String, LinkedHashMap<String, String>> outerMapping =
new LinkedHashMap<>();
outerMapping.put("template", mapping);
outerMapping.put("properties", someOtherMapping);
然后以相同的方式迭代键:
for (Map.Entry<String, LinkedHashMap<String, String>> outerEntry : outerMapping.entrySet()) {
if (str.contains(outerEntry.getKey()) {
// Apply the iteration above, using outerEntry.getValue().
}
}
答案 1 :(得分:2)
你说你发现Andy's answer(这就是我要做的)太复杂了。他的original comment建议else if
可能适合您:
if (str.contains("template")) {
if (str.contains("unsupported"))
mailType = "unsupported";
else if (str.contains("final_result"))
mailType = "final_result";
else if (str.contains("process_success"))
mailType = "Process Success";
else if (str.contains("receive"))
mailType = "Receive";
else if (str.contains("sen"))
mailType = "sent";
else if (str.contains("welcome"))
mailType = "welcome";
else if (str.contains("manual"))
mailType = "Manual";
} else if (str.contains("properties")) {
if (str.contains("unsupported"))
mailType = "unsupported";
else if (str.contains("final_result"))
mailType = "final_result";
else if (str.contains("process_success"))
mailType = "Process Success";
else if (str.contains("receive"))
mailType = "Receive";
else if (str.contains("sen"))
mailType = "sent";
else if (str.contains("welcome"))
mailType = "welcome";
else if (str.contains("manual"))
mailType = "Manual";
}
或更好,一直使用{}:
if (str.contains("template")) {
if (str.contains("unsupported")) {
mailType = "unsupported";
} else if (str.contains("final_result")) {
mailType = "final_result";
} else if (str.contains("process_success")) {
mailType = "Process Success";
} else if (str.contains("receive")) {
mailType = "Receive";
} else if (str.contains("sen")) {
mailType = "sent";
} else if (str.contains("welcome")) {
mailType = "welcome";
} else if (str.contains("manual")) {
mailType = "Manual";
}
} else if (str.contains("properties")) {
if (str.contains("unsupported")) {
mailType = "unsupported";
} else if (str.contains("final_result")) {
mailType = "final_result";
} else if (str.contains("process_success")) {
mailType = "Process Success";
} else if (str.contains("receive")) {
mailType = "Receive";
} else if (str.contains("sen")) {
mailType = "sent";
} else if (str.contains("welcome")) {
mailType = "welcome";
} else if (str.contains("manual")) {
mailType = "Manual";
}
}
答案 2 :(得分:1)
试试这个
static String containsAndValue(Collection<String> collection, String oldValue, String... strings) {
if (strings.length % 2 != 0)
throw new IllegalArgumentException("strings");
for (int i = 0; i < strings.length; i += 2)
if (collection.contains(strings[i]))
return strings[i + 1];
return oldValue;
}
和
if (str.contains("template")) {
mailType = containsAndValue(str, mailType,
"unsupported", "unsupported",
"final_result", "final_result",
"process_success", "Process Success",
"receive", "Receive",
"sen", "sent",
"welcome", "welcome",
"manual", "Manual");
} else if (str.contains("properties")) {
mailType = containsAndValue(str, mailType,
"unsupported", "unsupported",
"final_result", "final_result",
"process_success", "Process Success",
"receive", "Receive",
"sen", "sent",
"welcome", "welcome",
"manual", "Manual");
}
答案 3 :(得分:1)
这不那么复杂吗?
public enum MailType {
UNSUPPORTED("unsupported", "unsupported"),
FINAL_RESULT("final_result", "final_result"),
PROCESS_SUCCESS("process_success", "Process Success"),
RECEIVE("receive", "Receive"),
SENT("sen", "sent"),
WELCOME("welcome", "welcome"),
MANUAL("manual", "Manual");
private final String query;
private final String value;
MailType(String query, String value) {
this.query = query;
this.value = value;
}
public static String find(String text){
for(MailType mailType: values()){
if(text.contains(mailType.query)) {
return mailType.value;
}
}
return null;
}
}
并像这样使用它:
String mailType = MailType.find(str);
答案 4 :(得分:0)
首先,如果您使用byte[] bytes = Encoding.UTF8.GetBytes(publicKeyString);
IBuffer keyBuffer = CryptographicBuffer.DecodeFromBase64String(publicKeyString);
AsymmetricKeyAlgorithmProvider provider = AsymmetricKeyAlgorithmProvider.OpenAlgorithm(AsymmetricAlgorithmNames.RsaPkcs1);
CryptographicKey publicKey = provider.ImportPublicKey(keyBuffer,CryptographicPublicKeyBlobType.Pkcs1RsaPublicKey);`
而非简单else if
,则缩进的次数会减少。
else
在您的特定情况下,我建议尝试使用正则表达式来简化您的工作:
if(str.contains("unsupported"))
mailType="unsupported";
else if(str.contains("final_result"))
mailType="final_result";
else if(str.contains("process_success"))
mailType="Process Success";
else if(str.contains("receive"))
mailType="Receive";
else if(str.contains("sen"))
mailType="sent";
else if(str.contains("welcome"))
mailType="welcome";
else if(str.contains("manual"))
mailType="Manual";