On Thu, 8 Oct 2009, Rich Shepard wrote:
Date: Thu, 8 Oct 2009 13:43:06 -0700 (PDT) From: Rich Shepard <[email protected]> To: [email protected] Subject: [PLUG] Date Reformatting in Database TableI downloaded a database from an agency's Web site, and the date formats are not that used by SQL (YYYY-MM-DD). Instead, they are in the form of D/M/YYYY (or DD/MM/YYYY). Not only that, but the column they are in is the second to last of 11 columns; a representative sample is: 1993-1|Water Quality|WVR|Yamhill, City of|Yamhill|Hamlin|Holt|Npv|NPDES-Waste Discharge Limits|7/6/1993|01993-1|Water Quality|WVR|Yamhill, City of|Yamhill|Hamlin|Holt|Npv|NPDES-Waste Discharge Limits|7/6/1993|0 Is there a practical way to apply sed and/or awk to convert the date column above from 7/6/1993 to 1993-06-07? Rich
You did say »practical«. A perl script would be the most practical way to deal with this file. I'd want to examine one line of text at a time, parsing each line on the | characters. The data field would always be the 19th element of the resulting array (counting from 0). The date would then be parsed on the / characters, the elements rejuggled, and padded with 0s if neccesary. Then the whole line would be reassembled with the new date inserted into position 19, and written to a new file. I'd give you a sample script here, but that would take all the fun out if for you. Carlos
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