如何在控制台中打印每行特定数量的字符？

Question

我想在控制台中每行打印80个字符。我可以每行打印80个字符，但是当我在字符串中剩余80个字符时，它们不会被打印。我怎么可以补救这个？

public void output(String plainText, String cipherText, String iv, String key, String filename) { 
     System.out.println("CBC Vigenere by "); 
     System.out.println("Plaintext file name: " + filename); 
     System.out.println("Vigenere Keyword: " + key); 
     System.out.println("Initialization vector: " + iv + "\n"); 
     //This is the code for printing 80 char per line 
     System.out.println("Clear Text: \n"); 
     int j = 0; 
     String output = ""; 
     for (int i = 0; i < plainText.length(); i++) { 
      j++; 
      output = output + plainText.charAt(i); 
      if (j == 80) { 
       System.out.println(output); 
       output = ""; 
       j=0; 
      }else{ 
       //an attempt to print the remaining characters 
       // System.out.println(output); 
      } 


     } 
     //same process as above. 
     System.out.println(" \n Cipher Text:\n "); 
     j = 0; 
     output = ""; 
     for (int i = 0; i < cipherText.length(); i++) { 
      j++; 
      output = output + cipherText.charAt(i); 
      if (j == 80) { 
       System.out.println(output); 
       output = ""; 
       j = 0; 

      } else if ((cipherText.length() - i) <= 80) { 
       if (j == 80) { 
        if (j == (cipherText.length() - i)) { 
         //if (j == 80) { 
         System.out.println(output); 
         output = ""; 
         j = 0; 
         System.out.println((output)); 

         j = 0; 
        } 
       } 
      } 
     } 
     System.out.println("\n\nNumber of characters in clean plaintext file: " + (plainText.length() - padding)); 
     System.out.println("Block Size: " + key.length()); 
     System.out.println("Number of pad Characters added: " + padding); 

    } 

输出：

Clear Text: 

csthesciencethatdealswiththetheoryandmethodsofprocessinginformationindigitalcomp 
utersthedesignofcomputerhardwareandsoftwareandtheapplicationsofcomputersitthedev 
elopmentimplementationandmaintenanceofcomputerhardwareandsoftwaresystemstoorgani 
zeandcommunicateinformationelectronicallyabbreviationitcomputersaremanmadetoolst 
hataidusinsolvingotherproblemsabiologististryingtofigureouthowlifeworksphysicist 
sandchemistsaretryingtofigureouthowitemsreactinouruniversemathematiciansaretryin 
gtofigureoutrulesformanmadesystemsanyresearchproblemthatmayimproveacomputerscapa 
bilityofhelpingsolveaproblemoranyresearchproblemthatshedslightaboutanewwaytodoso 
methingwithacomputerispartofcsmostexcitingresearchmedicalapplicationsexpertsyste 
msfordiagnosisremotesurgerynanodeviceswithcomputingpowertodelivermedicineetcwene 
edhelptryingtocreateacomprehensiveemraccessibletotherightpeopleonlycarsthatcandr 
ivethemselvesseemslikethebestwayweknowhowtosolvelotsofproblemsisbythrowinglotsof 
computingpowerattheminsteadoflookingforelegantsolutionsthisdoesntsoundexcitingbu 
titwillbeexcitingwhentheresultsareachievediewatsoncsstudentstendtofindjobswheret 
heyprogramatleastsomeintheprocesstheyaresolvingproblemschallengesitsimpossibleto 
teachallthenewlanguagestoysultimatelywejustneedtoteachourstudentshowtothinksotha 
ttheycanpickupnewthingsontheirownourbiggestchallengeisgettingthemtobuyintothatet 

Cipher Text: 

qiqxbxknvieuejlkvwcpbtquevshswdsbyitsndorkbkwxswngygxwrkgfsbqvxwmsgiatpheilbuxnu 
ovzuyvmhwtvjiedrmnfkglrkfsddglexlsvsqfqjxiiqcbcqnyjvblktkjfeeaqklehghsfsutuiftqw 
bpwmpyjzytumfgqbqjqzluspcjjzdaovvqeczpilbicjnibqzbcjavhqrqmdmhtgegjgjscscoxnkvel 
kxrsbcwwpyjoswsihdtzrtyunqhczoipglabjjbcgkotmfovvwkseieszmvovataazltsolwviveqxpy 
dgdosztziuhqztoobkbzmyixdgvsosvbkxctpjokdkdjizcncjozbbznupuepihytkwguytvhbgrphoi 
nqhbqqbkqkqkxjikgqvlimdzsntmeifjtfwvwhphntudutoppoqjfmqlpxifiutoappzehyhfgqhondi 
qjlwakbjmfngliduchxgqokirtcththfafhamzxzihgruoskadlxlmsnlvqljihejqqaluhwhifioqky 
sokekkfxtjlgabnbiukgrdpzxsimdlkgpfaflnjxxuvqdjodosvtuiftemtahbokjmjudggknqnmxiwe 
pqpfearegfundjwgxhigxgtorpmcvcoynzyehuhkbyzkywfvokhwrmauscozgppxpnkijunryqyqdqdc 
wciygjvrimdmsmqyjujgiigwrvwqdlywjzyjjjmzmovkohpiedramexdxzhlzbbnzhkszcoxzjztcfwr 
hcvonlnksamomkvjgrgdnnihegpvaqzgbsmmkuteeedulnworpfbdrqgqdbzxsiochhroqnrbzbjiesd 
hgshgtztkmmylywkpkhaymybjkukuobuegvbqfebhcccxvzhrdrdmqipdyvkokdklqlfjbmrgfybybto 
ckvhwildyateedleltwnlikjvrfnrgmcdsxgjchozamigjffxhxxssrtfjeuzgybybtdnoythjhzbnjp 
jfjjycqlbxfhuijdwhkteusgqspuhvgdeplzzpksvdkpnxnxisrsqtreczczclwwwsiictspapdeodze 
nqsnqjdikcrdjpqpqpljwzqaemalvvgvfwtiyqwcvynbcwptnpvigcfmtulnwijuikikxetasahjzvzr 
zdgnuvmomdxyicakuyvgishtczbbtciwpxaqfuqlrlrgcsmthuscbrwtutuvehyixxsbsxsmhhyhtihq 
hrvdanhidymqnnhzkayqseralbznomjqdmelmbzpkpkzyhqabvohdhqtqufpsqgczdumvsknkjkyrzca 

我缺少字符串中的最后20个字符。什么是添加它们的实用方法？字符串的大小会有所不同，所以它并不总是相同的大小。

Answer 1

如果output在for循环完成后不为空，请将其打印出来。

Answer 2

试试这个。

for (int i = 0, size = cipherText.length(); i < size; i += 80) 
    System.out.println(cipherText.substring(i, Math.min(i + 80, size))); 

Answer 3

由于您连接了如此多的字符串，因此您的代码效率极低。我建议将它分成长度为80的子字符串，然后打印出来。

final int lineLength = 80; 
final int numLines = Math.ceil(plainText.length/lineLength); 
String[] lines = new String[num_lines]; 
int charend; 
for(int i=0, charini=0; i < numLines; i++) { 
    charend = Math.min(charini + 80, plainText.lenght()); 
    lines[i] = plainText.substring(charini,charend); 
    charini += lineLength; 
} 

// The print function 
for(int i=0; i<lines.length; i++) { 
    System.out.println(lines[i]); 
} 

此外，我建议你将代码拆分成方法，而不是复制两次。