Through bash shell script, i am trying to read yearmonth from file name -

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एनबी: इस सवाल का पालन

के माध्यम से bash shell script , मैं एक फ़ाइल नाम (FACT_PEOPLE_201401.txt) को पढ़ने की कोशिश कर रहा हूं। 

  फ़ाइल के लिए `एलएस -1 $ {OFSR_IN} / FACT_PEOPLE _ *। Txt में file_name = 'basename $ {file} `एक्स्ट_ युयर =` प्रतिध्वनि $ {File_name} | कट-सी 13-16` # YEAR Ext_Month के लिए = `प्रतिधारा $ {File_name} | कट-सी 17-18 '# महीने के लिए - एक्सचेंज को निकालने का प्रयास $ {EXT_Year} गूंज $ {EXT_Month} EXT_YEARMTH_1 = $ {Ext_Year} $ {var} $ {Ext_Month} # सम्मिलन वर्ष और amp; माह गूंज "\ $ \ $ एचआर_आइरमॉनथ = $ {EXT_YEARMTH_1}" & gt; & gt; $ {OFSR_CFG} / pmserver_OfsrHRSPeople05Loa.prm # यहां मूल्य को आउटपुट करना   
 मैं आउटपुट फ़ाइल config.prm में वर्ष महीना (EXT_YEARMTH_1) का मान प्रिंट करने का प्रयास कर रहा हूं  
 < कोड> गूंज "\ $ \ $ HR_YEARMONTH = $ {EXT_YEARMTH_1}" & gt; & gt; $ {OFSR_CFG} /pmserver_OfsrHRSPeople05Loa.prm   
 लेकिन मैं सही वर्ष 2014/01 जैसे प्रिंट नहीं कर पा रहा हूं। इसके बजाय मुझे आउटपुट के रूप में  
  $$ HR_YEARMONTH =   
 आउटपुट प्राप्त हो रहा है   
 कोड> $ इनपुट फ़ील_पेल = $ PMRootDir \ rlhrs \ data \ in \ FACT_PEOPLE_201306.txt $$ HR_YEARMONTH = 2014/01    
 
 
 यह:  
  फ़ाइल के लिए FACT_PEOPLE * .txt में है sed:। * _ \ ([0-9] [0-9] [0-9 ] [0- 9] \) \ ([0- 9] [0-9] \)। Txt: \ 1 / \ 2: '& lt; & lt; & lt; इनपुट के लिए "$ file" done   
  
  FACT_PEOPLE_201401.txt   
 प्रिंट होगा  
  2014/01   
 यदि आप वर्ष और महीना को वेरिएबल्स असाइन करना चाहते हैं, तो आप इसका उपयोग कर सकते हैं:  
  फ़ाइल के लिए FACT_PEOPLE _ *। Txt में पढ़ो- r वर्ष महीने & lt; & Lt; (sed:। * _ \ ([0-9] [0-9] [0-9] [0-9] \) \ ([0- 9] [0-9] \)। Txt : \ 1 \ 2: '& lt; & lt; & lt; "$ file") प्रतिलिपि YEAR: $ year गूंज MONTH: $ माह   
 क्या प्रिंट होगा  
  YEAR: 2014 माह: 01

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