| User: editor -- 2011-01-28 << 707 709 >> |
| Hits: 2688 |
| Type: Text file parser |
| Search all Text file parser examples |
| Description: |
| How to extract all specified date format from a text file? Need to extract all format of date like 5:22 AM 11/08/2011, and remove all other contents. |
| Input Sample: |
| line1 ........ line2 . 5:22 AM 11/08/2001 line3 .......... line4 ............. 5:04 PM 1/28/2011 |
| Output Sample: |
| 5:22 AM 11/08/2001 5:04 PM 1/28/2011 |
| Answer: |
| Hint: You need to Download and install "Replace Pioneer" on windows platform to finish following steps. |
| 1. ctrl-o open source text file 2. ctrl-h open 'replace' dialog * in 'search for pattern', enter: * in 'replace with pattern', enter: * uncheck option of 'Print Unmatched Units' 3. click 'replace', done! 4. ctrl-s save to file. |
Screenshot 1: Replace_Window |
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Check Demo of Text file parser |
| Keywords: |
| ddd 110 contents date format unmatched date unmatch content specified print extract date from a text in all format date extract from text file extract date from text file specified format search and replace date format in remove specified replace date format format date |