My oracle table is as follows ( Address column having multiple lines):
ID Address
--------------------
1456897 No 61
11th Street
Tatabad Coimbatore - 641012
How to get the desired result as (with Address column as a single line) ?
ID Address
-------------------------
1456897 No 61 , 11th Street, Tatabad Coimbatore - 641012
I don't know if your database has its newlines as \x0a or \x0d or \x0d\x0a. I therefore propose a the following solution that handles all three kind of new lines. It will however replace mutliple newlines with one ,. This might be what you want, or it might not.
select
id,
regexp_replace(
address,
'('||chr(10)||'|'||chr(13)||')+',
', ') as address,
....
from
....
remove new line character in the column - something like
SELECT REPLACE(Address_column, '\n', ' ') -- \n might be also \r\n or even \r
FROM table_name
Related
I have a pipe delimited file which has to be loaded via SQL*Loader in Oracle.
My control file looks like this:
LOAD DATA
REPLACE
INTO TABLE1
FIELDS TERMINATED BY '|'
TRAILING NULLCOLS
(
ID "TRIM(:ID)",
TEXT "NVL(TRIM(:TEXT),' ')"
)
The TEXT column in the data file can contain text with "|"- i.e., delimiter too.
How can I accept pipe in the TEXT column?
You can't escape the delimiter; but if you want everything up to the first pipe to be the ID and everything after the first pipe to be TEXT, you could treat the record in the data file as a single field and split it using SQL functions, e.g.:
LOAD DATA
INFILE ...
REPLACE
INTO TABLE TABLE1
TRAILING NULLCOLS
(
ID CHAR(4000) "regexp_replace(:ID, '^(.*?)(\\|(.*))?$', '\\1')",
TEXT EXPRESSION "regexp_replace(:ID, '^(.*?)(\\|(.*))?$', '\\3')"
)
There is no FIELDS clause.
The ID is initially up to 4000 characters from the line (just a large value to hopefully capture any data you have). A regex replace is then applied to that; the pattern defines a first group as any characters (non-greedy), optionally followed by a second group comprising a pipe and a third inner group of zero or more characters after that pipe. The original value is replaced by group 1.
The TEXT is defined as an EXPRESSION, meaning it isn't obtained directly from the file; instead the same regex pattern is applied to the original ID value, but now that is replaced by the third group, which is everything after the first pipe (if there is one).
An equivalent in plain SQL as a demo would be:
with data (id) as (
select '123|test 1' from dual
union all
select '234|test 2|with pipe' from dual
union all
select '345|test 3|with|multiple|pipes|' from dual
union all
select null from dual
union all
select '678' from dual
union all
select '789|' from dual
)
select id as original,
regexp_replace(ID, '^(.*?)(\|(.*))?$', '\1') as id,
regexp_replace(ID, '^(.*?)(\|(.*))?$', '\3') as text
from data;
which gives:
ORIGINAL ID TEXT
------------------------------- ---- ------------------------------
123|test 1 123 test 1
234|test 2|with pipe 234 test 2|with pipe
345|test 3|with|multiple|pipes| 345 test 3|with|multiple|pipes|
567 567
678| 678
If you don't need to worry about records without that first pipe, or with that first pipe but followed by nothing, then the regex could be simpler:
(
ID CHAR(4000) "regexp_replace(:ID, '^(.*?)\\|(.*)$', '\\1')",
TEXT EXPRESSION "regexp_replace(:ID, '^(.*?)\\|(.*)$', '\\2')"
)
Question: For every part description that begins with the letter “b”, list the part description, and then pad each part description with a “+”on the left side so that all these part descriptions are 15 characters in length.
And I wrote like
SELECT
LENGTH(PART_PART_DESCRIPTION), LPAD(PART_PART_DESCRIPTION,15,'+'),
PART_PART_DESCRIPTION, CONCAT('+', PART_PART_DESCRIPTION) FROM PART
WHERE SUBSTR(PART_PART_DESCRIPTION,1,1)='B'
but the output doesn't show 15 of '+' on left side.
Here is the output table
Your column PART_PART_DESCRIPTION is of CHAR data type with 285 data length. so BLENDER in your column has a total 285 (7 + 278 trailing spaces) length. that is why you are facing the problem.
See this:
SQL> select LPAD(CAST('BLENDER' AS CHAR(285)),15,'+') FROM DUAL;
LPAD(CAST('BLENDER'ASCHAR(285)),15,'+')
------------------------------------------------------------
BLENDER
SQL> select LPAD('BLENDER',15,'+') FROM DUAL;
LPAD('BLENDER',
---------------
++++++++BLENDER
SQL>
You need to use TRIM to properly use the LPAD on CHAR datatype column Something like the following:
LPAD(trim(PART_PART_DESCRIPTION),15,'+')
Most probably your data is padded with spaces. Try this
SELECT
LENGTH(PART_PART_DESCRIPTION), LPAD(TRIM(PART_PART_DESCRIPTION),15,'+'),
PART_PART_DESCRIPTION, CONCAT('+', PART_PART_DESCRIPTION) FROM PART
WHERE SUBSTR(PART_PART_DESCRIPTION,1,1)='B'
I'm trying to all rows where a column contains any control charters with the exception of the line feed character (hex value of A). I've tried the following, but this only returns results that have a control character and don't have a line feed. I really want the set of characters that are control characters, LESS the line feed character. Is there a 'minus' operation for character sets, where you can exclude particular ones from it?
SELECT *
FROM MyTable
WHERE REGEXP_LIKE(MyColumn, '[:cntrl: &&[^' || UTL_RAW.CAST_TO_VARCHAR2(HEXTORAW('A')) || ']]{1,}');
Any thoughts?
Thanks!
Well here's a first try that will work but I'm sure this can be made more elegant and efficient:
SELECT *
FROM MyTable
WHERE regexp_like(MyColumn, '[[:cntrl:]]')
AND MyColumn NOT like '%' || chr(10) || '%';
I am trying to select a column from a table that contains newline (NL) characters (and possibly others \n, \r, \t). I would like to use the REGEXP to select the data and replace (only these three) characters with a space, " ".
No need for regex. This can be done easily with the ASCII codes and boring old TRANSLATE()
select translate(your_column, chr(10)||chr(11)||chr(13), ' ')
from your_table;
This replaces newline, tab and carriage return with space.
TRANSLATE() is much more efficient than its regex equivalent. However, if your heart is set on that approach, you should know that we can reference ASCII codes in regex. So this statement is the regex version of the above.
select regexp_replace(your_column, '([\x0A|\x0B|`\x0D])', ' ')
from your_table;
The tweak is to reference the ASCII code in hexadecimal rather than base 10.
select translate(your_column, chr(10)||chr(11)||chr(13), ' ') from your_table;
to clean it is essential to serve non-null value as params ...
(oracle function basically will return null once 1 param is null, there are few excpetions like replace-functions)
select translate(your_column, ' '||chr(10)||chr(11)||chr(13), ' ') from your_table;
this examples uses ' '->' ' translation as dummy-value to prevent Null-Value in parameter 3
I am loading some data to Oracle via SQLLDR. The source file is "pipe delimited".
FIELDS TERMINATED BY '|'
But some records contain pipe character in data, and not as separator. So it breaks correct loading of records as it understands indata pipe characters as field terminator.
Can you point me a direction to solve this issue?
Data file is about 9 GB, so it is hard to edit manually.
For example,
Loaded row:
ABC|1234567|STR 9 R 25|98734959,32|28.12.2011
Rejected Row:
DE4|2346543|WE| 454|956584,84|28.11.2011
Error:
Rejected - Error on table HSX, column DATE_N.
ORA-01847: day of month must be between 1 and last day of month
DATE_N column is the last one.
You could not use any separator, and do something like:
field FILLER,
col1 EXPRESSION "REGEXP_REPLACE(:field,'^([^|]*)\\|([^|]*)\\|(.*)\\|([^|]*)\\|([^|]*)\\|([^|]*)$', '\\1')",
col2 EXPRESSION "REGEXP_REPLACE(:field,'^([^|]*)\\|([^|]*)\\|(.*)\\|([^|]*)\\|([^|]*)\\|([^|]*)$', '\\2')",
col3 EXPRESSION "REGEXP_REPLACE(:field,'^([^|]*)\\|([^|]*)\\|(.*)\\|([^|]*)\\|([^|]*)\\|([^|]*)$', '\\3')",
col4 EXPRESSION "REGEXP_REPLACE(:field,'^([^|]*)\\|([^|]*)\\|(.*)\\|([^|]*)\\|([^|]*)\\|([^|]*)$', '\\4')",
col5 EXPRESSION "REGEXP_REPLACE(:field,'^([^|]*)\\|([^|]*)\\|(.*)\\|([^|]*)\\|([^|]*)\\|([^|]*)$', '\\5')",
col6 EXPRESSION "REGEXP_REPLACE(:field,'^([^|]*)\\|([^|]*)\\|(.*)\\|([^|]*)\\|([^|]*)\\|([^|]*)$', '\\6')"
This regexp takes six capture groups (inside parentheses) separated by a vertical bar (I had to escape it because otherwise it means OR in regexp). All groups except the third cannot contain a vertical bar ([^|]*), the third group may contain anything (.*), and the regexp must span from beginning to end of the line (^ and $).
This way we are sure that the third group will eat all superfluous separators. This only works because you've only one field that may contain separators. If you want to proofcheck you can for example specify that the fourth group starts with a digit (include \d at the beginning of the fourth parenthesized block).
I have doubled all backslashes because we are inside a double-quoted expression, but I am not really sure that I ought to.
It looks to me that it's not really possible for SQL*Loader to handle your file because of the third field which: can contain the delimiter, is not surrounded by quotes and is of a variable length. Instead, if the data you provide is an accurate example then I can provide a sample workaround. First, create a table with one column of VARCHAR2 with length the same as the maximum length of any one line in your file. Then just load the entire file into this table. From there you can extract each column with a query such as:
with CTE as
(select 'ABC|1234567|STR 9 R 25|98734959,32|28.12.2011' as CTETXT
from dual
union all
select 'DE4|2346543|WE| 454|956584,84|28.11.2011' from dual)
select substr(CTETXT, 1, instr(CTETXT, '|') - 1) as COL1
,substr(CTETXT
,instr(CTETXT, '|', 1, 1) + 1
,instr(CTETXT, '|', 1, 2) - instr(CTETXT, '|', 1, 1) - 1)
as COL2
,substr(CTETXT
,instr(CTETXT, '|', 1, 2) + 1
,instr(CTETXT, '|', -1, 1) - instr(CTETXT, '|', 1, 2) - 1)
as COL3
,substr(CTETXT, instr(CTETXT, '|', -1, 1) + 1) as COL4
from CTE
It's not perfect (though it may be adaptable to SQL*Loader) but would need a bit of work if you have more columns or if your third field is not what I think it is. But, it's a start.
OK, I recomend you to parse the file and replace the delimiter.
In command line in Unix/linux you should do:
cat current_file | awk -F'|' '{printf( "%s,%s,", $1, $2); for(k=3;k<NF-2;k++) printf("%s|", $k); printf("%s,%s,%s", $(NF-2),$(NF-1),$NF);print "";}' > new_file
This command will not change your current file.
Will create a new file, comma delimited, with five fields.
It splits the input file on "|" and take first, second, anything to antelast, antelast, and last chunk.
You can try to sqlldr the new_file with "," delimiter.
UPDATE:
The command can be put in a script like (and named parse.awk)
#!/usr/bin/awk
# parse.awk
BEGIN {FS="|"}
{
printf("%s,%s,", $1, $2);
for(k=3;k<NF-2;k++)
printf("%s|", $k);
printf("%s,%s,%s\n", $(NF-2),$(NF-1),$NF);
}
and you can run in this way:
cat current_file | awk -f parse.awk > new_file