As of ABAP 7.40 SP5 I can use the CONCAT function to combine two fields in my CDS view select list. It's limited to two parameters but I can work around that by chaining this function to combine multiple fields or build larger strings. What I can't do this way is combine two fields with a space separating them. When I do this:
define view Z...
as select from but000 as bp
{
concat( concat( bp.name_first, ' '), bp.name_last )
}
The space ' ' is silently trimmed from the resulting string. How can I separate fields with a space?
ABAP 7.50
ABAP 7.50 will include the CONCAT_WITH_SPACE function that addresses this problem. With that function the example above can be written simply as:
CONCAT_WITH_SPACE( bp.name_first, bp.name_last, 1 )
With the 1 referring to the number of spaces that are to be inserted between the two arguments.
7.50 also introduces other string functions like INSTR, LEFT, LENGTH, LTRIM, RIGHT, RPAD and RTRIM. 7.51 looks set to add LOWER and UPPER to that list.
ABAP 7.40
There is no clean way to accomplish the same in this release. The only way seems to be to concatenate the two fields with a dummy string that encloses the space in a character group that won't appear in the fields that are being selected. After combining you can then remove those characters from the result, leaving just the space. I've taken this approach from Christian Seitel on the SAP forums.
REPLACE(CONCAT( CONCAT( bp.name_first, '|-| |-|'), bp.name_last),'|-|', '')
This works because it will process this string as follows:
name_first|-| |-|
name_first|-| |-|name_last
name_first name_last
If you are still struggling with a <7.50 version. You can try:
concat(concat("first_string",(' ')), "second_string")
I hope it helps.
Related
I am trying to split a field by delimiter in LookML. This field either follows the format of:
Managers (AE)
Managers (AE - MM)
I was able to split to first case using this
sql: case
when rlike (${user_role_name}, '^.*[\\(\\)].*$') then split_part(${user_role_name}, ' ', -1)
However, I haven't been able to get the 2nd case to do the same. It's in a case statement so I am going to add another when statement, but am not able to figure out the regex for parentheses that contains spaces.
Thanks in advance for the help!
By "split" the string, I think you mean you want to extract the part in parentheses, right?
I would do this using a regex substring method. You didn't mention what warehouse you're using, and the syntax will vary a little, but on snowflake that would look like:
regexp_substr(${user_role_name}, '\\([^)]*\\)')
So, for example, with the inputs you gave:
select regexp_substr('Managers (AE)', '\\([^)]*\\)')
union all
select regexp_substr('Managers (AE - MM)', '\\([^)]*\\)')
result
(AE)
(AE - MM)
The environment I am currently working in is Snowflake.
As a matter of data sensitivty, I will be using pseudonyms for my following question.
I have a specific field in one of my tables called FIELD_1. The data in this field is structured as such:
I am trying to figure out how to automatically extract from my FIELD_1 the output I have in FIELD_2.
Does anyone have any idea what kind of query I would need to achieve this? Any help would be GREATLYappreciated! I am really quite stuck on this problem.
Thank you!
You seem to want everything up to the first four numbers. Then to replace the underscores with spaces. If so:
select replace(regexp_substr(field_1, '^[^0-9]*[0-9]{4}'), '_', ' ')
Or alternatively, if you want the first three components separated by underscores:
select replace(regexp_substr(field_1, '^[^_]+_[^_]+_[0-9]{4}'), '_', ' ')
If the data is as simplistic in reality as you've described here, you can use a variable-length LEFT() function in conjunction with REPLACE() to get the desired output:
SELECT FIELD_1, REPLACE(LEFT(FIELD_1, LEN(FIELD_1)-10),'_',' ') AS FIELD_2
FROM table_name
See also:
SELECT - Snowflake Documentation
LEFT - Snowflake Documentation
REPLACE - Snowflake Documentation
LENGTH, LEN - Snowflake Documentation
I'm attempting to isolate eight digits from a cell that contains other numbers as well as text and no rhyme or reason to where it is placed. An example return would look something like this:
will deliver 11/07 in USA at 12:30 with conf# 12345678
I need the conf# only, but it could be at the end, beginning, middle of the string and I don't know how to isolate it. I'm working in DB2 so I can't use functions such as PATINDEX or CHARINDEX, so what are my other option for pulling out only "12345678" regardless of where it is located?
While DB2 doesn't have PATINDEX or CHARINDEX, it does have LOCATE.
If your DB2 version supportx pureXML, you can use the regular expression support in XQuery, something like:
select xmlcast(
xmlquery(
' if (fn:matches( $YOURCOLUMN, "(^|.*[^\d])(\d{8})([^\d].*$|$)")) then fn:replace( $YOURCOLUMN,"(^|.*[^\d])(\d{8})([^\d].*$|$)","$2") else "" '
)
as varchar(20)
)
from YOURTABLE
This assumes that 8-digit sequence appears only once in the column. You may need to tweak the regex to support some border cases.
I require a select query that adds a space to the data based on the placement of the capital letters i.e. 'HelpMe' using this query would be displayed as 'Help Me' . Note i cannot use a stored function to do this the it must be done in the query itself. The Data is of variable length and query must be in SQL. Any Help will be appreciated.
Thanks
You need to use user defined function for this until MS give us support for regular expressions. Solution would be something like:
SELECT col1, dbo.RegExReplace(col1, '([A-Z])',' \1') FROM Table
Aldo this would produce leading space that you can remove with TRIM.
Replace regular expresion function:
http://connect.microsoft.com/SQLServer/feedback/details/378520
About dbo.RegexReplace you can read at:
TSQL Replace all non a-z/A-Z characters with an empty string
Assume if you are using Oracle RDBMS, you use the following,
REGEX_REPLACE
SELECT REGEXP_REPLACE('ILikeToWatchCSIMiami',
'([A-Z.])', ' \1')
AS RX_REPLACE
FROM dual
;
Managed to get this output: * SQLFIDDLE
But as you see it doesn't treat well on words such as CSI though.
jkdfhdjfhjh&name=ijkjkjkjkjkjk&id=kdjkjkjkjkjkjjjd&class=kdfjjfjdhfjhf
The above string has some characters starting with & and ending with =
for example we have &name= and I just need this from the above string.
similarly I need &id=, &class=
I need the output under a single column.
Final Extract
----------------------
&id=, &class=, &name=
can anyone help me out in writing a query for this.
You could try this :
select regexp_replace('jkdfhdjfhjh&name=ijkjkjkjkjkjk&id=kdjkjkjkjkjkjjjd&class=kdfjjfjdhfjhf', '\\w*?(&.*?=)\\w+((?=&)|$)', '\\1, ', 'g');
result:
regexp_replace
-------------------------
&name=, &id=, &class=,
Then it's up to you to remove the last ,.
The regexp_replace function is available in version 8.1 and after.
If you want the values along with each variable, I would implement this by splitting on "&" into an array and then taking a slice of the desired elements:
SELECT (string_to_array('jkdfhdjfhjh&name=ijkjkjkjkjkjk&id=kdjkjkjkjkjkjjjd&class=kdfjjfjdhfjhf','&'))[2:4];
Output in PostgreSQL 8.4 (array type):
{name=ijkjkjkjkjkjk,id=kdjkjkjkjkjkjjjd,class=kdfjjfjdhfjhf}
The example string is very wide so here's the general form to show the array slicing more clearly:
SELECT ((string_to_array(input_field,'&'))[2:4];
NOTE: You must have the extra parentheses around the string_to_array() call in order for the array slicing to work--you'll get an error otherwise.