[英]Efficiently parse the huge string response
我有一個服務,它返回以下格式的數據。 我已將其縮短以便理解,但總的來說這是一個非常大的反應。 格式總是一樣的。
process=true
version=2
DataCenter=dc2
Total:2
prime:{0=1, 1=2, 2=3, 3=4, 4=1, 5=2}
obvious:{0=6, 1=7, 2=8, 3=5, 4=6}
mapping:{3=machineA.dc2.com, 2=machineB.dc2.com}
Machine:[machineA.dc2.com, machineB.dc2.com]
DataCenter=dc1
Total:2
prime:{0=1, 1=2, 2=3, 3=4, 4=1, 5=2, 6=3}
obvious:{0=6, 1=7, 2=8, 3=5, 4=6, 5=7}
mapping:{3=machineP.dc1.com, 2=machineQ.dc1.com}
Machine:[machineP.dc1.com, machineQ.dc1.com]
DataCenter=dc3
Total:2
prime:{0=1, 1=2, 2=3, 3=4, 4=1, 5=2}
obvious:{0=6, 1=7, 2=8, 3=5, 4=6}
mapping:{3=machineO.dc3.com, 2=machineR.dc3.com}
Machine:[machineO.dc3.com, machineR.dc3.com]
我試圖解析上面的數據並將其存儲在三個不同的地圖中。
Map<String, Map<Integer, Integer>> prime = new HashMap<String, Map<Integer, Integer>>();
Map<String, Map<Integer, Integer>> obvious = new HashMap<String, Map<Integer, Integer>>();
Map<String, Map<Integer, String>> mapping = new HashMap<String, Map<Integer, String>>();
以下是描述:
dc2
,值為{0=1, 1=2, 2=3, 3=4, 4=1, 5=2}
。 dc2
,值為{0=6, 1=7, 2=8, 3=5, 4=6}
dc2
{0=6, 1=7, 2=8, 3=5, 4=6}
dc2
{0=6, 1=7, 2=8, 3=5, 4=6}
dc2
{0=6, 1=7, 2=8, 3=5, 4=6}
dc2
{0=6, 1=7, 2=8, 3=5, 4=6}
。 dc2
,值為{3=machineA.dc2.com, 2=machineB.dc2.com}
。 同樣對於其他數據中心也是如此。
解析上述字符串響應的最佳方法是什么? 我應該在這里使用正則表達式還是簡單的字符串解析?
public class DataParser {
public static void main(String[] args) {
String response = getDataFromURL();
// here response will contain above string
parseResponse(response);
}
private void parseResponse(final String response) {
// what is the best way to parse the response?
}
}
任何例子都會有很大的幫助。
您可以像ShellFish一樣建議並按'\\ n'拆分響應,然后處理每一行。
一個正則表達式的方法如下(它不完整,但足以讓你開始):
public static void main(String[] args) throws Exception {
String response = "process=true\n" +
"version=2\n" +
"DataCenter=dc2\n" +
" Total:2\n" +
" prime:{0=1, 1=2, 2=3, 3=4, 4=1, 5=2}\n" +
" obvious:{0=6, 1=7, 2=8, 3=5, 4=6}\n" +
" mapping:{3=machineA.dc2.com, 2=machineB.dc2.com}\n" +
" Machine:[machineA.dc2.com, machineB.dc2.com]\n" +
"DataCenter=dc1\n" +
" Total:2\n" +
" prime:{0=1, 1=2, 2=3, 3=4, 4=1, 5=2, 6=3}\n" +
" obvious:{0=6, 1=7, 2=8, 3=5, 4=6, 5=7}\n" +
" mapping:{3=machineP.dc1.com, 2=machineQ.dc1.com}\n" +
" Machine:[machineP.dc1.com, machineQ.dc1.com]\n" +
"DataCenter=dc3\n" +
" Total:2\n" +
" prime:{0=1, 1=2, 2=3, 3=4, 4=1, 5=2}\n" +
" obvious:{0=6, 1=7, 2=8, 3=5, 4=6}\n" +
" mapping:{3=machineO.dc3.com, 2=machineR.dc3.com}\n" +
" Machine:[machineO.dc3.com, machineR.dc3.com]";
Map<String, Map<Integer, Integer>> prime = new HashMap();
Map<String, Map<Integer, Integer>> obvious = new HashMap();
Map<String, Map<Integer, String>> mapping = new HashMap();
String outerMapKey = "";
int findCount = 0;
Matcher matcher = Pattern.compile("(?<=DataCenter=)(.*)|(?<=prime:)(.*)|(?<=obvious:)(.*)|(?<=mapping:)(.*)").matcher(response);
while(matcher.find()) {
switch (findCount) {
case 0:
outerMapKey = matcher.group();
break;
case 1:
prime.put(outerMapKey, new HashMap());
String group = matcher.group().replaceAll("[\\{\\}]", "").replaceAll(", ", ",");
String[] groupPieces = group.split(",");
for (String groupPiece : groupPieces) {
String[] keyValue = groupPiece.split("=");
prime.get(outerMapKey).put(Integer.parseInt(keyValue[0]), Integer.parseInt(keyValue[0]));
}
break;
// Add additional cases for obvious and mapping
}
findCount++;
if (findCount == 4) {
findCount = 0;
}
}
System.out.println("Primes:");
prime.keySet().stream().forEach(k -> System.out.printf("Key: %s Value: %s\n", k, prime.get(k)));
// Add additional outputs for obvious and mapping
}
結果:
Primes:
Key: dc2 Value: {0=0, 1=1, 2=2, 3=3, 4=4, 5=5}
Key: dc1 Value: {0=0, 1=1, 2=2, 3=3, 4=4, 5=5, 6=6}
Key: dc3 Value: {0=0, 1=1, 2=2, 3=3, 4=4, 5=5}
參考解釋正則表達式模式: http : //docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html
答案取決於您對格式的確定程度和確切程度。 一個非常簡單的方法解析字符串並進行最小字符串比較以確定鍵值:
private static final String DATA_CENTER = "DataCenter=";
private static final int DATA_CENTER_LEN = DATA_CENTER.length();
private static final String PRIME = " prime:";
private static final int PRIME_LEN = PRIME.length();
// etc.
Map<String, Map<Integer, Integer>> prime = new HashMap<>();
// etc.
String response = "...";
Scanner scanner = new Scanner( response );
while(scanner.hasNextLine()){
String line = scanner.nextLine();
if( line.startsWith( DATA_CENTER ) ){
String dc = line.substring( DATA_CENTER_LEN );
line = scanner.nextLine(); // skip Total
prime.put( dc, str2map(scanner.nextLine().substring(PRIME_LEN)) );
obvious.put( dc, str2map(scanner.nextLine().substring(OBVIOUS_LEN)) );
mapping.put( dc, str2mapis(scanner.nextLine().substring(MAPPING_LEN)) );
}
}
更明確的nextLine()調用甚至可以避免對“DataCenter”的測試。
這里有幾個幾乎相同的方法來分割大括號並創建一個地圖:
private static Map<Integer,Integer> str2map( String str ){
Map<Integer,Integer> map = new HashMap<>();
str = str.substring( 1, str.length()-1 );
String[] pairs = str.split( ", " );
for( String pair: pairs ){
String[] kv = pair.split( "=" );
map.put( Integer.parseInt(kv[0]),Integer.parseInt(kv[1]) );
}
return map;
}
private static Map<Integer,String> str2mapis( String str ){
Map<Integer,String> map = new HashMap<>();
//...
map.put( Integer.parseInt(kv[0]),kv[1] );
}
return map;
}
如果白色空間有可能發生變化,您可以保持安全,使用
private static final String PRIME = "prime:";
// ...
prime.put( dc, str2map(scanner.nextLine().trim().substring( PRIME_LEN )) );
如果無法保證線路的順序或完整性,則可能需要進行測試:
line = scanner.nextLine().trim();
if( line.startsWith( PRIME ) ){
prime.put( dc, str2map(scanner.nextLine().substring( PRIME_LEN )) );
}
通過更少的穩定性/信任,可以指示正則表達式解析。
在這種情況下,我會做簡單的字符串解析,為每一行應用正則表達式 。 在偽代碼中,這樣的事情:
for line in response
if line matches /^DataCenter/
key = datacenter name
else if line matches / *prime/
prime.put(key, prime value)
else if line matches / *obvious/
obvious.put(key, obvious value)
else if line matches / *mapping/
mapping.put(key, mapping value)
else
getline
您可以通過首先檢查該行的第一個字符來優化此處。 如果它是除了空格或D
之外的任何東西,你可以轉到下一行。 如果格式始終相同,您甚至可以對要解析的行進行硬編碼。 在您提供的示例中,您可以:
skip 2 lines
repeat
extract datacenter name
skip 1 line
extract prime
extract obvious
extract mapping
add above stuff to the maps
skip 1 line
until EOF
這將快得多,但如果格式改變則會失敗。
您可以使用諸如ANTLR之類的解析器生成器,或者您可以手動編寫解析器代碼。 根據您需要處理的輸出量和頻率,您可能會發現遇到這樣的麻煩並不值得,並且只需遍歷每一行並手動解析它(例如,正則表達式或indexOf)就足夠了足夠。
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