正則表達式模式用逗號分隔

Question

我有一個來自excel列的以下字符串

  "\"USE CODE \"\"Gef, sdf\"\" FROM 1/7/07\""

我想設置正則表達式模式來檢索整個字符串，以便我的結果將完全像

"USE CODE ""Gef, sdf"" FROM 1/7/07"

以下是我嘗試過的

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RegexMatches
{
    public static void main( String args[] ){

      // String to be scanned to find the pattern.
      String line = "\"USE CODE \"\"Gef, sdf\"\" FROM 1/7/07\", Delete , Hello , How are you ? , ";
      String line2 = "Test asda ds asd, tesat2 . test3";

      String dpattern = "(\"[^\"]*\")(?:,(\"[^\"]*\"))*,|([^,]+),";
      // Create a Pattern object
      Pattern d = Pattern.compile(dpattern);
      Matcher md = d.matcher(line2);

      Pattern r = Pattern.compile(dpattern);

      // Now create matcher object.
      Matcher m = r.matcher(line);
      if (m.find( )) {
         System.out.println("Found value: 0 " + m.group(0) );
       //  System.out.println("Found value: 1 " + m.group(1) );
         //System.out.println("Found value: 2 " + m.group(2) );
      } else {
         System.out.println("NO MATCH");
      }
   }
}

並且它的結果在，（逗號）之后中斷，因此輸出為

Found value: 0 "USE CODE ""Gef,

它應該是

Found value: 0 "USE CODE ""Gef sdf"" FROM 1/7/07",

對於第二行Matcher m = r.matcher(line2); 輸出應該是

Found value: 0 "Test asda ds asd",

Answer 1

您可以使用

(?:"[^"]*(?:""[^"]*)*"|[^,])+

見正則表達式演示

說明 ：

• " -前導報價
• [^"]* -除雙引號外的0+個字符
• (?:""[^"]*)* -0 +個""文本序列，后跟0 +字符（雙引號除外）
• " -尾隨報價

要么：

• [^,] -除逗號以外的任何字符

整個模式被匹配1次或多次，因為它用(?:...)+括起來，而+匹配1次或多次。

IDEONE演示 ：

String line = "\"USE CODE \"\"Gef, sdf\"\" FROM 1/7/07\", Delete , Hello , How are you ? , ";
String line2 = "Test asda ds asd, tesat2 . test3";
Pattern pattern = Pattern.compile("(?:\"[^\"]*(?:\"\"[^\"]*)*\"|[^,])+");
Matcher matcher = pattern.matcher(line);
if (matcher.find()){                        // if is used to get the 1st match only
    System.out.println(matcher.group(0)); 
}
Matcher matcher2 = pattern.matcher(line2); 
if (matcher2.find()){
    System.out.println(matcher2.group(0)); 
}