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To display the date column containing "19/"

Query

SELECT username,DATE_FORMAT(dialdate,GET_FORMAT(DATE,'EUR')) as new,DATE_FORMAT(dialtime,'%r') as dialtime,price FROM processeddata_table WHERE DialDate between '2008/07/04' and '2013/07/26' and dialdate like '%19/% 00:00:00'
Posted

Simply ;)
Use LEFT[^] function.

SQL
SELECT ...
FROM ....
WHERE ... LEFT(dialdate,2) = '19'


More about String functions[^]

Note: Is this statement true: (year>=2008 And year <=2013) AND (year>=1900 AND year<=1999)?
 
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v2
ur,what is your problem ?

SQL
SELECT username,DATE_FORMAT(dialdate,GET_FORMAT(DATE,'EUR')) as new,DATE_FORMAT(dialtime,'%r') as dialtime,price FROM processeddata_table WHERE DialDate between '2008/07/04' and '2013/07/26' and dialdate like '%19%' 
 
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Comments
Roopa 10064853 26-Jul-13 2:21am    
I want to display dialdate containing 19/
Chui PuiKwan 26-Jul-13 2:24am    
1.like '%19/%'
2.
INSTR(dialdate,'19/')>0
Roopa 10064853 26-Jul-13 2:41am    
thank u but not getting any output
Chui PuiKwan 26-Jul-13 2:48am    
If you could give me a demo data, I can help you to debug it .
Roopa 10064853 26-Jul-13 2:52am    
2013-12-12

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