關於如何解析java中的字符串的示例

Question

我有下面的字符串，我需要解析/提取'20000'。

"where f_id = '20000' and (flag is true or flag is null)" 

任何建議最好的方法來做到這一點？

這裏有更多的代碼來幫助理解：

List<ReportDto> reportDtoList = new ArrayList<ReportDto>(); 
for (Report report : reportList) { 
List<ReportDetailsDto> ReportDetailsDtoList = new ArrayList<ReportDetailsDto>(); 

ReportDto reportDto = new ReportDto(); 
reportDto.setReportId(report.getReportId()); 
reportDto.setReportName(report.getName()); 

Pattern p = Pattern.compile("=\\s'[0-9]+'"); 
String whereClause = report.getWhereClause(); 
Matcher m = p.matcher(whereClause); 

困惑在這之後該怎麼辦？

Answer 1

您可以使用此正則表達式來從你的String

Pattern p = Pattern.compile("[0-9]+"); 
Matcher m = p.matcher(text); 

if (m.find()) { 
    System.out.println(m.group()); 
} 

提取單個非負整數，或者，如果你想保留單引號：

Pattern p = Pattern.compile("['0-9]+"); 

這將提取包括'='模式之後有一個空間。它將打印一個String，其中包含沒有'='或空格的數字。所以，如果這符合你知道有根據您加入這個代碼'='

Pattern p = Pattern.compile("=\\s'[0-9]+"); 
Matcher m = p.matcher(text); 

if (m.find()) { 
    System.out.println(m.group().substring(3)); 
} 

編輯 後一個數字它會是什麼樣子

List<ReportDto> reportDtoList = new ArrayList<ReportDto>(); 
Pattern p = Pattern.compile("=\\s'[0-9]+"); 
for (Report report : reportList) { 
    List<ReportDetailsDto> ReportDetailsDtoList = new ArrayList<ReportDetailsDto>(); 

    ReportDto reportDto = new ReportDto(); 
    reportDto.setReportId(report.getReportId()); 
    reportDto.setReportName(report.getName()); 

    String whereClause = report.getWhereClause(); 
    Matcher m = p.matcher(whereClause); 
    if (m.find()) { 
     String foundThis = m.group().substring(3); 
     // do something with foundThis 
    } else { 
     // didn't find a number or = 
    } 
} 

Answer 2

試試這個：

Pattern p = Pattern.compile("-?\\d+"); 
String s = "your string here"; 
Matcher m = p.matcher(s); 
List<String> extracted = new ArrayList<String>(); 
while (m.find()) { 
    extracted.add(m.group()); 
} 

Answer 3

浮動和底片

個

Pattern p = Pattern.compile("(-?\\d+)(\\.\\d+)?"); 
     String s = "where f_id = '20000' 3.2 and (flag is true or flag is null)"; 
     Matcher m = p.matcher(s); 
     List<String> extracted = new ArrayList<String>(); 
     while (m.find()) { 
      extracted.add(m.group()); 
     } 

     for (String g : extracted) 

      System.out.println(g); 

打印出

20000 
3.2