title: grep or awk, how to convert data to multiple columns in csv file
body: <p>I have following log file data</p>
  
  <pre><code>2/1/1/2/tasdfn.c:
  
  LOG:
          backslash-newline should be deleted before tokenizing
      No diagnostics line
  --
  2/1/1/2/tlsdf.c:
  
  LOG:
  +++ stderr ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
  tlsdf.c:15:5: error: use of undeclared identifier 'wchar_t'
  --
  2/2/4/1/tasdf.gen(0):
  
  LOG:
      511 external identifiers in one source file
      Compilation failed ungracefully
  --
  2/2/4/1/tiasdf.gen(0):
  
  LOG:
      8 nesting levels of #include files
      Compilation failed ungracefully
  --
  </code></pre>
  
  <p>And wanted to covert above data to following: where <code>coulmn1</code> should <code>testname</code>(i.e 2/1/1/2/testname.c) and <code>coulmn2</code> should <code>LOG</code> details:</p>
  
  <p>expected Output in csv file:</p>
  
  <pre><code>2/1/1/2/tasdfn.c:           LOG:
                              backslash-newline should be deleted before tokenizing
                              No diagnostics line
  
  
  2/1/1/2/tlsdf.c:            LOG:
                              +++ stderr ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
                              tlsdf.c:15:5: error: use of undeclared identifier 't'
  
  
  
  2/2/4/1/tasdf.gen(0):       LOG:
                              511 external identifiers in one source file
                              Compilation failed ungracefully
  </code></pre>
  
  <p>Not able to get the desired output from <code>grep</code> and <code>awk</code>. 
  Using grep i can get testname i.e and Log details as separate output</p>
  
  <pre><code>cat test.log | grep -Ea '*.c:$|*.gen\([0-9]\):' //prints testname
  cat test.log | grep -Ea '*.c:$|*.gen\([0-9]\):' -A4 //prints testname with Log details
  </code></pre>
  
  <p>how to get data in mentioned format where .csv file as output??</p>