在java中獲取兩個字符串之間的字符串，從前兩行開始

Question

我需要從文件中獲取一個字符串，介於起始字符串和結束字符串之間，而且在返回的字符串中，我需要在起始字符串之前有 2 行。

例如，下面是我的字符串塊，我需要找到以“:25:RO68BTRL04701202281399XX”開頭並以“-}”結尾的行，但還要返回起始搜索字符串之前的前兩行：

{1:F01BTRLRO22AXXX1111111111}{2:I940BTRLRO22XXXXN}{3:{108:047MSOG16103001F}}{4: <br />
:20:047RONCRT0028139901<br />
:25:RO68BTRL04701202281399XX<br />
:28C:00697/00001<br />
:60F:C160411RON580868,35<br />
:61:1604110411C537,91NTRFC31EIOP161020001<br />
:86:PLATI CARTELE  AP 1351271   CALGIRA  FCT.12138114 S I FCT.12145656 CALGIRA TR ANSILVANIA SRL RO81BTRL RO NCRT0278617701 36 BTRL RO2 2 Incasare OP   canal ele ctronic<br />
:61:1604110411C279,44NTRF801EIOP161020101
:86:12135020 12144465 AP11309 84 FANTASI STILL SRL FANT ASI STILL SRL RO56BTRL 001 01202N50350XX 16 BTRL RO22 Incasare OP   canal elec tronic<br />
:61:1604110411C1149,41NTRF020EIOP161020002<br />
:86:CV F 12143919 12135600 EN IGMA COM SRL RO08BTRL 0200 1202D01120XX 159 BTRL RO22 Incasare OP   canal elec tronic<br />
:61:1604110411C53,52NTRF325EIOP161020051<br />
:86: AP1383577 cv fact 1213519 0/ 31.03.20 16 reincarcari electronice GLISSANDO CO STEA SNC RO26BTRL 02501202 121834XX 20 BTRL RO22 Inca sare OP   canal electronic<br />
:61:1604110411D695,15NTRF047EPOP161020204<br />
:86:PLATA PP F 2479 PRAM CADE S SRL 7227 BTRL RO22 Plata OP intra   canal electronic<br />
:61:1604110411C223,01NTRF010EIOP161020085<br />
:86: AP 1353368 F 12134396 121 40760 COM TURISM MARKET S RL Comturism Market SRL C OMTURISM MARKET SRL RO21B TRL01001202M68296XX 94 BT RLRO22 Incasare OP   cana l electronic<br />
:62F:C160411RON1162675,04 <br />
:64:C160411RON1162675,04 <br />
-}

我在 Java 中的代碼是這樣的（正在工作，只是它沒有返回我想要的，這意味着 startString 減去 2 行和 endString 之間的字符串）：

 import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; public class bt { public static void main(String[] args) { // TODO code application logic here File fileName = new File("bta.txt"); try { String startToken = ":25:RO68BTRL04701202281399XX"; String endToken = "-}"; boolean output = false; int nrLinie=0; Scanner scan = new Scanner(fileName); while (scan.hasNextLine()) { String line = scan.nextLine(); if (output==false && line.indexOf(startToken) > -1) { output = true; //line = line.substring(line.indexOf(startToken)+startToken.length()); line = line.substring(line.indexOf(startToken)+1,line.indexOf(endToken)+endToken.length()); } else if (output && line.indexOf(endToken) > -1) { //output = false; //System.out.println(line.substring(0, line.indexOf(endToken))); } if (output) { System.out.println(line); } } } catch (FileNotFoundException e) { e.printStackTrace(); } } }

謝謝，因為任何幫助都非常感謝。

Answer 1

這個正則表達式可以做到。 有關演示，請參閱regex101 。

((?:[^\r\n]*\r?\n){2}):25:RO68BTRL04701202281399XX(.*?)-\}