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Find and replace?

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# 1  
Old 06-05-2013
Find and replace?
Hi All,

I have data like below
Code:
1 CREATE TABLE temp123 
2 (
3 col1 INTEGER,
4 col2 CHAR(3) CHARACTER SET LATIN NOT CASESPECIFIC,
5 col3 DECIMAL(19,0),
6 col4 VARCHAR(80) CHARACTER SET LATIN NOT CASESPECIFIC,
7 start_dt DAte FORMAT 'YY/MM/DD',
8 end_dt DATE FORMAT 'YY/MM/DD', 
9 datecol1 date FORMAT 'YY/MM/DD',
10 change_dt TIMESTAMP(6),
11 tempdatecolumn VARCHAR(100)      
12 )

here i have to replace wherever "DATE" is coming i have to replace the entire string in that line with some other value

like start_dt DAte FORMAT 'YY/MM/DD', should be replcaed with start_dt varchar(100),

(it has to match exactly). in my case the line 7,8 & 9 should match whereas 12 should not match.
Last edited by Franklin52; 06-05-2013 at 03:06 AM.. Reason: Please use code tags
# 2  
Old 06-05-2013
This is not easy.
1. You should use code tags
2. You have DAte, DATE, date, that can be solved with [Dd][Aa][Tt][Ee]
3. Line 9 is not in same format as 7 and 8, it does not have _dt
4. 12 will not match, it contains only ) ??
# 3  
Old 06-05-2013
Could this help you ?
Code:
 
 perl -pe 's/\s+date\s+(.*)/ varchar2\(100\),/i' filename

This User Gave Thanks to pravin27 For This Post:
baranisachin
# 4  
Old 06-05-2013
Hi jotne,
The find and replace should start only when we encounter "DATE", as a whole. so the string before it doesnt matter, ony the string from "DATE" and after that till the end of line is what matters.
Thanks.
# 5  
Old 06-05-2013
Give example output on how it should look like when done.
This since 9 is different then 7 and 8
# 6  
Old 06-05-2013
Thanks pravin27 :-) Works like a charm Smilie

---------- Post updated at 02:03 AM ---------- Previous update was at 02:02 AM ----------

pravin27,
Is there anyway we can do it without using PERL?
# 7  
Old 06-05-2013
GNU sed does support a case insensitive search using I flag:
Code:
sed "s#date format 'yy/mm/dd'#varchar2(100)#gI" file

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