Is there any function to replace NULL or empty space with special character in hive? when I execute the below statement it returns a semantic exception stating trim works only on string/nvarchar
CASE
WHEN TRIM(q.address) = '' OR q.address IS NULL THEN '?'
ELSE q.address END as address
Please help.
Use LENGTH() to check the length of the column value. It returns > 0, if there is some value else return 0 for empty or NULL value.
Also frame the column value in CASE WHEN ... END block
The final query may look like:
SELECT CASE WHEN LENGTH(address) > 0 THEN address ELSE '?' END AS address
FROM table_name;
Please refer Replace the empty or NULL value with specific value in HIVE query result
Hope this help you!!!
In order to replace nulls you can use Coalesce
Coalesce( q.address, '?')
But it seems your field adress is not of the proper type to use trim, can you show us the type of that field?
Related
I'm trying to replace any cell that has a date in it with "null" so that I can fill values in PowerQuery. I've used formula's that can can perform the replacement, but the non-date values that I want to remain as text are replaced with "Error". Is there a way to do this?
Something like this might work
#"RemoveDates" = Table.TransformColumns(#"PriorStepNameGoesHere",{{"Date", each try if Number.From(Date.From(_))>0 then null else _ otherwise _, type text}})
x = Table.TransformColumns(Source,{{"Column1", (a)=> try Date.FromText(a) otherwise null , type text}})
CASE WHEN (r.code_value4 = 0 AND LENGTH(ltrim(rtrim(xx.AFFILIATE_CODE))) > 0) AND
r_intercompany.code_value1 is not null AND
ltrim(rtrim(xx.AFFILIATE_CODE)) <> (CASE WHEN xx.COMPANY_CODE_JE_EXCEPTION_FLAG = 1
THEN r.code_value3 ELSE r.code_value1 END)
THEN r_intercompany.code_value1
ELSE NVL(r_mga_acct.code_value2, xx.ACCOUNT_NUMBER) END
I have view in which the above part of the sql is being used in select statement and as well as to compare with a field while joining. When I run the view after making this change, it is throwing the error ORA-01722 invalid number.
Please let me know on how to correct this.
Thanks
" it is throwing the error ORA-01722 invalid number."
It is likely somewhere you are comparing a numeric column with a string column . Oracle is implicitly casting the string to a number but the column contains values which cannot be converted, so it hurls.
"Please let me know on how to correct this."
Don't rely on implicit data conversion. Go through your code, check the data types of all the columns. Where you find a varchar2 column being compared to a number you need to cast the number to a string. For literals that means quoting them - '1' instead of 1 - and for columns that means wrapping them in to_char() calls.
My Suspicion is either r.code_value4 field or xx.COMPANY_CODE_JE_EXCEPTION_FLAG field is String type hence try the below
CASE WHEN (r.code_value4 = '0' AND LENGTH(ltrim(rtrim(xx.AFFILIATE_CODE))) > 0) AND
r_intercompany.code_value1 is not null AND
ltrim(rtrim(xx.AFFILIATE_CODE)) <> (CASE WHEN xx.COMPANY_CODE_JE_EXCEPTION_FLAG = '1'
THEN r.code_value3 ELSE r.code_value1 END)
THEN r_intercompany.code_value1
ELSE NVL(r_mga_acct.code_value2, xx.ACCOUNT_NUMBER) END
I am trying to see if a string contains a dot.
Set Root_Currency = Root_TaxDataSummary.SlvObject("Currency")
curr_val = InStr(Root_Currency,".")
If curr_val.exist Then
pass
else
fail
Is there anything wrong with the way I am going about this?
InStr returns an integer representing the position the searched text can be found in the string.
curr_val.exist won't work because the integer type doesn't have an exist method. Instead:
If curr_val > 0 Then
Or (if this is the only use of that variable):
If InStr(Root_Currency,".") > 0 Then
Lastly, because 0 is treated as False in VBScript, you don't need to include the equality. Either a position is found for the character or you get back a 0/false:
If InStr(Root_Currency,".") Then
InStr returns a 'simple' number (1 based index/position of needle in haystack, or 0 meaning 'not found', or Null meaning Null argument) not an object. So change your code to:
If curr_val Then
' found
Else
' not found
End If
In Oracle I want to check whether the string has "=' sign at the end. could you please let me know how to check it. If it has '=' sign at the end of string, I need to trailing that '=' sign.
for eg,
varStr VARCHAR2(20);
varStr = 'abcdef='; --needs to trailing '=' sign
I don't think you need "pattern matching" here. Just check if the last character is the =
where substr(varstr, -1, 1) = '='
substr when called with a negative position will work from the end of the string, so substr(varstr,-1,1) extracts the last character of the given string.
Use the REGEX_EXP function. I'm putting a sql command since you didn't specify on your question.:
select *
from someTable
where regexp_like( someField, '=$' );
The pattern $ means that the precedent character should be at the end of the string.
see it here on sql fiddle: http://sqlfiddle.com/#!4/d8afd/3
It seems that substr is the way to go, at lease with my sample data of about 400K address lines this returns 1043 entries that end in 'r' in an average of 0.2 seconds.
select count(*) from addrline where substr(text, -1, 1) = 'r';
On the other hand, the following returns the same results but takes 1.1 seconds.
select count(*) from addrline where regexp_like(text, 'r$' );
I have a table which has city column having few records with state values as well-separated by comma.
There are other records without, as well. I want to take the state values for those present into a separate field called state.
How to do that? I tried the code below and it is saying "missing right parenthesis":
SELECT DECODE(ORA_CITY,
INSTR(ORA_CITY,',') > 0,
SUBSTR(ORA_CITY, INSTR(ORA_CITY, ','), LENGTH(ORA_CITY) ) ,
NULL) AS STATE
from ADDRESS
I don't know if you still need it but use CASE:
SELECT CASE
WHEN INSTR(ORA_CITY, '5') > 0 THEN
SUBSTR(ORA_CITY, INSTR(ORA_CITY, '5'), LENGTH(ORA_CITY))
ELSE
NULL
END STATE
FROM ADDRESS
Clearly you have not understood decode syntax.
Try the following:
SELECT DECODE(INSTR(ORA_CITY,','),
0,
NULL,
SUBSTR(ORA_CITY, INSTR(ORA_CITY, ','), LENGTH(ORA_CITY) )) AS STATE
FROM ADDRESS
The correct syntax is:
DECODE( expression , search , result [, search , result]... [,
default] ), where
expression is the value to compare.
search is the value that is compared against expression.
result is the value returned, if expression is equal to search.
default is optional. If no matches are found, the DECODE function will
return default. If default is omitted, then the DECODE function will
return null (if no matches are found).
Examples here and here
SELECT REGEX_REPLACE(ORA_CITY, '.*, *', '') AS STATE
FROM ADDRESS
WHERE ORA_CITY LIKE '%,%'
This uses regular expression to replace all upto the comma, and then maybe spaces with nothing. A WHERE included.