Link Break in SQL - line

What's the symbol or expression for a line break in oracle sql?
Ive searched everywhere without success, so that an expression 'line1line2' is written :
line
line2 (One beneath the other)
This is the code:
SELECT NVL2(TRIM(NAME),TRIM(NAME),NULL)||CHR(13)|| NVL2(TRIM(STATE),TRIM(STATE),NULL)

Related

Cannot query my container using the GridDB Shell?

I have a device container with the name 1cbfce15ec4d which houses some my data. I know for a fact there's data in there, but when I try a simple query in the griddb shell, I got the following error:
gs[public]> sql select * from 1cbfce15ec4d;
D20332: An unexpected error occurred while executing a SQL. : msg=[[240001:SQL_COMPILE_SYNTAX_ERROR] Parse SQL failed, reason = Syntax error: 1cbfce15ec4d; on executing query (sql="select * from 1cbfce15ec4d") (db='public') (user='admin') (appName='gs_sh') (clientId='a6d92f48-e558-440-86dd-a05e949fa726:1') (clientNd='{clientId=3, address=127.0.0.1:55744}') (address=127.0.0.1:20001, partitionId=983)]
I am not exactly sure what is going on here -- at first I assumed my data must be corrupt or empty, but that is not the case. It seems to be a case of the shell dying trying to process something about that container name.
Any ideas?

According to the manual :
"If the name of a table or a column contains characters other than ASCII alphanumeric characters and underscore, or if the first character of the name is a number in a SQL statement, enclose the name with double quotation marks."
Try select * from "1cbfce15ec4d"

Replace string output from query in results tab

I have a DB with a large number of CLOB columns. Each of these columns contains a repeated set of characters, which are fillers designed to designate a paragraph break (I didn't design the tables).
Is there a way to write the script so that each time the script finds these characters it enters a paragraph or line break into the results that are returned, while still saying in the same row of the results?
The data would look like
"Hello all,XYZ!£$I can't get ... to work.XYZ!£$The error meassge says ..."
As an example:
SELECT *
FROM ALERTS
REPLACE(Alert_Text, 'XYZ!£$', CH(13))
(Obviously the above returns errors)
The ideal output of the query would be:
"Hello all,
I can't get ... to work.
The error meassge says ..."
I am using SqlDbx to connect to an Oracle DB.

The obvious error is ORA-00904: "CH": invalid identifier
The reason is, that the function name is CHR
select REPLACE(txt, 'XYZ!£$', CHR(13)) from tab;
REPLACE(TXT,'XYZ!£$',CHR(13))
--------------------------------------------------------------------------------
Hello all,
I can't get ... to work.
The error meassge says ...

sql code not working in oraclesqldeveloper: Invalid Character error

I wrote a code to find few items by a particular number but it keeps saying that is something is an "invalid character"."ORA-00911: invalid character
00000 - "invalid character"
*Cause: identifiers may not start with any ASCII character other than
letters and numbers. $#_ are also allowed after the first
character. Identifiers enclosed by doublequotes may contain
any character other than a doublequote. Alternative quotes
(q'#...#') cannot use spaces, tabs, or carriage returns as
delimiters. For all other contexts, consult the SQL Language
Reference Manual.
*Action: Error at Line: 14 Column: 23"
My brain is blown,I am not sure where the problem exists.
Need Help. Thanks.
select
a.app_num,
a.crs_pol_num,
kl.score,
kl.risk_level_desc
from
application a,
kyc_new_risk_level kl
where
a.app_num = kl.app_num
and a.app_num = (select max(to_number(a1.app_num)) from
application a1,
kyc_new_risk_level kl1 where kl1.app_num = a1.app_num and a1.crs_pol_num = a.crs_pol_num)
and a.crs_pol_num in (‘CG0147511’,
‘CG0133662’,
‘CG0138107’,
‘493186’,
‘CG0142230’,
‘CS0138382’,
‘CG0147509’,
‘CG0147545’,
‘921141048001’,
‘CG0347239’,
‘CG0142212’,
‘CG0147518’,
‘CG0134057’,
‘CG0143158’,
‘CG0147536’,
‘CG0244124’,
‘CG0134102’,
‘CG0241709’,
‘CG0147197’,
‘CG0137204’,
‘CG0347496’,
‘CG0147490’,
‘CG0143467’,
‘CG0135689’,
‘CG0146904’,
‘CS0132517’,
‘CG0145455’,
‘CG0147554’,
‘CG0133626’,
‘CG0147560’,
‘CG0135359’,
‘CG0133303’,
‘921165287001’,
‘CG0147546’,
‘CG0114581,
‘CG0122266’,
‘CG0236577’,
‘CG0345349’,
‘CG0132670’,
‘CG0147413’,
‘CG0241646’,
‘CG0143783’,
‘CG0245160’,
‘CG0124066’,
‘CG0124830’,
‘CG0145956’,
‘CG0232953’,
‘CG0144479’,
‘CG0147569’,
‘CG0147555’,
‘CG0244857’,
‘CG0147562’,
‘CG0347578’,
‘CG0346461’,
‘CS0133352’,
‘512097’,
‘CS0127026’,
‘CG0147583’,
‘CG0233314’,
‘CG0247096’,
‘CG0131282’,
‘CG0123462’,
‘CS0124502’,
‘CG0146034’,
‘CG0140236’,
‘CS0126420’,
‘CG0147557’,
‘CG0123182’,
‘CG0233300’,
‘CG0132782’,
‘CG0147501’,
‘CS0141693’,
‘CG0145237’,
‘CG0141763’,
‘CG0147591’,
‘CG0144107’,
‘CG0125208’,
‘CG0132306’);

SQL Developer is showing you the problem before you even run your query
That's not a quote, it's a smart quote. Which is nice for browsers and word processors - but not so much for the database engine.
If you click on that hint, it will replace it with a real single quote.
But with so many to fix, would be much faster, better to use search and replace.
But your cursor on the bad character, and hit Ctrl+R
And then hit 'replace all', then repeat for the closing smart quote.
But in case you missed all that, the error also back from the DB tells you where the issue is

Hive -character '<EOF>' not supported here

Select * from mytable where field=
'ce7bd3d4-dbdd-407e-a3c3-ce093a65abc9;cdb597073;7cf6cda5fc'
Getting Below Error while running above query in Hive
FAILED: ParseException line 1:92 character '' not supported here

<EOF> here means End Of File. When you get an "unexpected End Of File" error it means the parser reached the end of the query unexpectedly. This typically happens when the parser is expecting to find a closing character, such as when you have started a string with ' or " but have not closed the string (with the closing ' or ").
When you come across these types of errors it is good to check that your query can be parsed correctly. In addition, the error gives you the location where the parser failed: line 1:92 in this case. You can usually look at this location (character 92 of the query) and work backwards to find the problem character.

Try adding the database name to the "from" statement as below.
Select * from my_db_name.mytable where field= 'ce7bd3d4-dbdd-407e-a3c3-
ce093a65abc9;cdb597073;7cf6cda5fc';
Hive uses the default database when no database was previously specified.

Parsing an error string in Lua

Lets say I have the following error string:
err = "/mnt/cd4/autorun.lua:43: 'end' expected (to close 'while' at line 1)
near '-eof-'"
How would I parse the file path, line number, and the error message separately from the string?
I have no prior experience in parsing Lua strings, so I thought asking here would be useful. I also tried finding a topic solving the same matter but I could not find one.

Something like this should work:
err = "/mnt/cd4/autorun.lua:43: 'end' expected (to close 'while' at line 1) near '-eof-'"
local file, line, errmsg = err:match('^(.-):(%d+):(.+)')
print(file, line, errmsg)
The pattern says: capture starting at the end of the line (^) a shortest group of zero or more (-) of any symbol (.), followed by :, then a group of one or more digits (%d+), followed by :, and then a group of one of more symbols (.+). You can read about patterns here.