I want to fetch the names of employees from a table upto the character ':' but couldn't as substr and ltrim is not working as expected. Below given are given some examples:
ABINERI:REBECCA C
CARRINGTON:JAMES M
But I want them in the way given below:
REBECCA C ABINERI
JAMES M CARRINGTON
I just used the query below in Toad for Oracle:
<pre>
<b>select name from employees</b>
</pre>
Please try below query:
select SUBSTR(name,(INSTR(name,':')+1)) || ' ' || SUBSTR(name,1,(INSTR(name,':'))-1) from employees;
hope above query will resolve your issue.
Related
I have a column with Names, and I am trying to split the column into First and Last Name using Text functions such as LEFT/SUBSTRING/CHARINDEX.
Data in the column:
Name
Yang, Jon
Huang, Eugene
Torres, Ruben
Zhu, Christy
Johnson, Elizabeth
Everything works fine as long as I use this code:
SELECT
[Name]
--,LEFT([Name], CHARINDEX(' ', [Name])) AS FirstName
,SUBSTRING([Name], 1, CHARINDEX(' ', [Name] )) AS FirstName
FROM
DataModeling.Customer
But the problem arises when I try to subtract 1 from CHARINDEX to exclude the Comma from the result and it throws this error:
I have done this operation many times in Excel so trying to replicate it with TSQL. Any suggestion on what I am doing wrong is helpful.
You get that error when CHARINDEX(' ', [Name] ) return 0. So minus 1 will make it negative and it is invalid value for substring()
You can use CASE expression to check the return value from CHARINDEX() and return the correct value to substring()
Or, you can "cheat" by using
CHARINDEX( ' ', [Name] + ' ' )
So CHARINDEX() will always return a value that is more than 0
I am working on Data standardization rules, and one of the rules says, "If the last name is part of first name, then remove the last name from first name".
my Query- how do i check if first name column has the last name in it using oracle sql developer?
I tried using :
select fst_nm, lst_nm from emp where fst_nm = fst_nm || lst_nm ;
but this query returns '0' results.
Also, I tried another query:
select fst_nm, lst_nm, regexp_substr(fst_nm, '[^ ]+',1,1) from emp ;
I tried using the below query
select fst_nm, lst_nm from emp where fst_nm = fst_nm || lst_nm ;
but this query returns nothing, I mean '0' results.
Also, I tried another query:
select fst_nm, lst_nm, regexp_substr(fst_nm, '[^ ]+',1,1) from emp ;
expected result is:
fst_nm = john smith ;
lst_nm = smith
Actual result showing up is :
fst_nm = john ;
lst_nm = smith
Please help
You should be able to just do a blanket replace on the entire table:
UPDATE emp
SET fst_nm = REPLACE(fst_nm, lst_nm, '');
The reason this should work is that for those records where the last name does not appear as part of the first name, the replace would have no effect. Otherwise, the last name would be stripped from the first name.
You can use below logic
select length('saurabh rai'),instr('saurabh rai',' '),case when length('saurabh rai') > instr('saurabh rai',' ') then substr('saurabh',1,instr('saurabh rai',' ')-1) else 'saurabh rai' end as a from dual;
Update emp set fst_nm=(Case when length(fst_nm) > instr(fst_nm, ' ') then substr(fst_nm,1,instr(fst_nm,' ')-1) else fst_nm end);
I have a SQL query and I would like to insert a hashtag between one column and another to be able to reference in Excel, using an import option in fields delimited by #. Anyone have an idea how to do it? A query is as follows:
SELECT FC.folha, folha->folhames,folha->folhaano, folha->folhaseq, folha->folhadesc, folha->TipoCod as Tipo_Folha,
folha->FolhaFechFormatado as Folha_Fechada, folha->DataPagamentoFormatada as Data_Pgto,
Servidor->matricula, Servidor->nome, FC.rubrica,
FC.Rubrica->Codigo, FC.Rubrica->Descricao, FC.fator, FC.TipoRubricaFormatado as TipoRubrica,
FC.ValorFormatado,FC.ParcelaAtual, FC.ParcelaTotal
FROM RHFolCalculo FC WHERE folha -> FolhaFech = 1
AND folha->folhaano = 2018
and folha->folhames = 06
and folha->TipoCod->codigo in (1,2,3,4,6,9)
You are generating delimited output from the query, so the first row should be a header row, with all following rows the data rows. You will really only have one column due to concat. So remove the alias from the columns, output the first row like so (using the alias here) . . .
SELECT 'folha#folhames#folhaano#folhaseq#folhadesc#Tipo_Folha#
Folha_Fechada#Data_Pgto#
matricula#nome#rubrica#
Codigo#Descricao#fator#TipoRubrica#
ValorFormatado#ParcelaAtual#ParcelaTotal'
UNION
SELECT FC.folha || '#' || folha->folhames || '#' || folha->folhaano . . .
The UNION will give the remaining rows. Note some conversion may be necessary on the columns data if not all strings.
I'm trying to concatenate 2 columns using delimiter as "."
code :
PCollection<BeamRecord> first = apps.apply(BeamSql.query(
"SELECT *,('CatLib' || 'ProdKey') AS CatLibKey from PCOLLECTION"));
How shall I specify delimiter between 2 columns ?
I'd say go for
SELECT
COALESCE(CatLib, '') || '.' || COALESCE(ProdKey, '') AS CatLibKey,
(any other columns here...)
FROM
PCOLLECTION;
but in SQL there is no "Select everything but column X" or "Select everything else" so you'd have to write down every name of the column you want to select.
Thanks #Impulse The Fox.
I have modified my query to :
PCollection<BeamRecord> first = apps.apply(BeamSql.query(
"SELECT Outlet, CatLib, ProdKey, Week, SalesComponent, DuetoValue, PrimaryCausalKey, CausalValue, ModelIteration, Published, (CatLib || '.' || ProdKey) AS CatLibKey from PCOLLECTION"));
and this worked perfectly.
We have a musicians table containing records with multiple string fields, say:
"Jimi", "Hendrix", "Guitar"
"Phil", "Collins", "Drums"
"Sting", "", "Bass"
"Ringo", "Starr", "Drums"
"Paul", "McCartney", "Bass"
I want to pass postgres a long string, say:
"It is known that Jimi liked to set light to his guitar and smash up
all the drums while on stage."
and i want to get returned the fields that have any matches - preferably in order of the most matches first:
"Jimi", "Hendrix", "Guitar"
"Phil", "Collins", "Drums"
"Ringo", "Starr", "Drums"
because i need the search to be case insensitive, i'm constructing a query like this...
select * from musicians where lowercase_string like '%'||firstname||'%' or lowercase_string like '%'||lastname||'%' or lowercase_string like '%'||instrument||'%'
and then looping through (in ruby in my case) to capture the result with the most matches.
this is however very slow in the sql stage (1 minute+).
i've tried adding lower-case GIN index using pg_trgm as suggested here - but it's not helping - presumably because the like query is back to front?
Thanks!
With my testing, it seems that no trigram index could help your query at all. And no other index type could possibly speed up an (I)LIKE / FTS based search.
I should mention that all of the queries below use the trigram indexes, when they are queried "reversed": when the table contains the document (which is indexed), and your parameter is the query. The (I)LIKE variant variant f.ex. 2-3 times faster with it.
These the queries I've tested:
select *
from musicians
where :input_string ilike '%' || firstname || '%'
or :input_string ilike '%' || lastname || '%'
or :input_string ilike '%' || instrument || '%'
At first, FTS seemed a great idea, but my testing shows that even without ranking, it is 60-100 times slower than the (I)LIKE variant. (So even, when you don't have to post-process results with these methods, these are not worth it).
select *
from musicians
where to_tsvector(:input_string) ## (plainto_tsquery(firstname) || plainto_tsquery(lastname) || plainto_tsquery(lastname))
However, ORDER BY rank doesn't slow down that much further: it is 70-120 times slower than the (I)LIKE variant.
select *
from musicians
where to_tsvector(:input_string) ## (plainto_tsquery(firstname) || plainto_tsquery(lastname) || plainto_tsquery(lastname))
order by ts_rank(to_tsvector(:input_string), plainto_tsquery(firstname) || plainto_tsquery(lastname) || plainto_tsquery(lastname))
Then, for a last effort, I tried the (fairly new) "word similarity" operators of the trigram module: <% and %> (available from PostgreSQL 9.6).
select *
from musicians
where :input_string %> firstname
or :input_string %> lastname
or :input_string %> instrument
select *
from musicians
where firstname <% :input_string
or lastname <% :input_string
or instrument <% :input_string
These were somewhat faster then FTS: around 50-70 times slower than the (I)LIKE variant.
(Partially working) rextester: it is run against PostgreSQL 9.5, so the 9.6 operators obviously won't run here.
Update: IF full word match is enough for you, you can actually reverse your query, to be able to use indexes. You'll need to "parse" your query (aka. "long string") though:
with long_string(ls) as (
values (:input_string)
),
words(word) as (
select s
from long_string, regexp_split_to_table(ls, '[^[:alnum:]]+') s
where s <> ''
)
select musicians.*
from musicians, words
where firstname ilike word
or lastname ilike word
or instrument ilike word
group by musicians.id
Note: I parsed the query for every complete word. You can have some other logic there, or it can even be parsed in client side.
The default, btree index shines here, as it is much faster than the trigram index with (I)LIKE (we won't need them anyway, as we are looking for complete word match here):
with long_string(ls) as (
values (:input_string)
),
words(word) as (
select s
from long_string, regexp_split_to_table(lower(ls), '[^[:alnum:]]+') s
where s <> ''
)
select musicians.*
from musicians, words
where lower(firstname) = word
or lower(lastname) = word
or lower(instrument) = word
group by musicians.id
http://rextester.com/PSABJ6745
You could even get the match count with something like
sum((lower(firstname) = word)::int
+ (lower(lastname) = word)::int
+ (lower(instrument) = word)::int)
The ilike option with match ordering:
with long_string (ls) as (values
('It is known that Jimi liked to set light to his guitar and smash up all the drums while on stage.')
)
select musicians.*, matches
from
musicians
cross join
long_string
cross join lateral
(select
(ls ilike format ('%%%s%%', first_name) and first_name != '')::int +
(ls ilike format ('%%%s%%', last_name) and last_name != '')::int +
(ls ilike format ('%%%s%%', instrument) and instrument != '')::int
as matches
) m
where matches > 0
order by matches desc
;
first_name | last_name | instrument | matches
------------+-----------+------------+---------
Jimi | Hendrix | Guitar | 2
Phil | Collins | Drums | 1
Ringo | Starr | Drums | 1