如何在Java中解析字符串的示例

Question

我有以下字符串需要解析/提取其中的“ 20000”。

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

是否有建議這樣做的最佳方法？

這是更多有助於理解的代碼：

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);

困惑之后該怎么辦？

Answer 1

您可以使用此正則表達式從String提取單個非負整數

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

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

或者，如果您想保留單引號：

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

這將提取一個包含'='並在其后包含單個空格的模式。 它將打印一個String其中包含不帶'='的數字或空格。 因此，如果匹配，您會知道在'='之后有一個數字

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

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

根據您添加的代碼進行編輯 ，結果如下所示

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

嘗試這個：

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

浮點數和負數

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);

打印出來

20000
3.2