循环遍历多个目录并读取html文件的状态

Question

我需要从多个report.html grep“status”并制作合并报告，这些报告保存在以下目录中。用**突出显示的所有目录对于每个报告都不同。我怎么能实现这一点。 C：\ Program Files（x86）\ Jenkins \ jobs ** E2E_Sanity ** \ jobs ** ABC_E2E_Sanity ** \ builds ** 41 ** \ archive \ performanceTestsReports ** pcRun106821 ** \ Report

Answer 1

例如使用python和glob模块：

import glob

files = r"C:\Program Files (x86)\Jenkins\jobs*\jobs*\builds*\archive\performanceTestsReports*\Report"
l = glob.glob(files)
for f in l:
    print (f)

Answer 2

找一个可以的小型常规程序：

• 提取修复行号（如果状态位于固定行）

• 搜索单词 - 状态并提取信息

  int lineNo = 1
  //if the row number is fixed you can extracted by minLine and maxLine
  int minLine = 1
  int maxLine = 20
  def line
  def status
  def statusRegex
  
  
  def folder = new File("C:\\Users\\user\\Desktop\\ero")
  
  folder.eachFile{it->
  println "File: ${it.absolutePath}"
  
  it.withReader { reader->
      while ((line = reader.readLine()) != null & lineNo <= maxLine) {
  
          if (lineNo >= minLine) {
              //        println "${lineNo}. ${line}" //if you need specific line numbers
          }
          lineNo++
          //search for status and print the line
          status = line.find("status")
          //search for status by regex and extract all up to <
          statusRegex = line.find(/(?s)status (.*?)\</)
  
          if(status){
              println '    full line' + line
          }
          if(statusRegex){
              println '    by regex' + statusRegex
          }
      }
  }
  }