I have a column with numbers - 1,2,3...100. 5 of them looks like 1_, 3_, 5_
So, actually I have 1, 1_, 2, 3, 3_, 4, 5, 5_, 6, 7, 8 ...
I did the query:
SELECT code
FROM table.column1
WHERE column1 LIKE '%_'
to get the list of this numbers with _. Instesd of it I got the list of all values, including values with _. What's your opinion - what's the reason, where is a mistake?
The character _ is, similar to %, a wildcard in LIKE-operations. While % matches multiple characters, _ only matches a single character. To get your expected result you need to escape the _ character:
SELECT code
FROM table.column1
WHERE column1 LIKE '%\_'
Here is a dbfiddle showing it.
You use the "E" before the string to escape the symbol.
SELECT col1
FROM table1
WHERE col1 LIKE E'%\_';
Related
I want to select the column which has 4th character is number.
example:
column
----------
Game1234
nothing12
demo:db<>fiddle
SELECT
*
FROM
mytable
WHERE substring(col, 4, 1) ~ '[0-9]'
substring() extracts a certain substring (in this case from the 4th position a substring with length 1)
~ '[0-9]' checks if the extracted character is a digit (using regular expressions)
More efficient would be:
substr(col, 4, 1) BETWEEN '0' and '9'
I need to remove non-numeric characters in a column (character varying) and keep numeric values in postgresql 9.3.5.
Examples:
1) "ggg" => ""
2) "3,0 kg" => "3,0"
3) "15 kg." => "15"
4) ...
There are a few problems, some values are like:
1) "2x3,25"
2) "96+109"
3) ...
These need to remain as is (i.e when containing non-numeric characters between numeric characters - do nothing).
Using regexp_replace is more simple:
# select regexp_replace('test1234test45abc', '[^0-9]+', '', 'g');
regexp_replace
----------------
123445
(1 row)
The ^ means not, so any character that is not in the range 0-9 will be replaced with an empty string, ''.
The 'g' is a flag that means all matches will be replaced, not just the first match.
For modifying strings in PostgreSQL take a look at The String functions and operators section of the documentation. Function substring(string from pattern) uses POSIX regular expressions for pattern matching and works well for removing different characters from your string.
(Note that the VALUES clause inside the parentheses is just to provide the example material and you can replace it any SELECT statement or table that provides the data):
SELECT substring(column1 from '(([0-9]+.*)*[0-9]+)'), column1 FROM
(VALUES
('ggg'),
('3,0 kg'),
('15 kg.'),
('2x3,25'),
('96+109')
) strings
The regular expression explained in parts:
[0-9]+ - string has at least one number, example: '789'
[0-9]+.* - string has at least one number followed by something, example: '12smth'
([0-9]+.\*)* - the string similar to the previous line zero or more times, example: '12smth22smth'
(([0-9]+.\*)*[0-9]+) - the string from the previous line zero or more times and at least one number at the end, example: '12smth22smth345'
there is a table's fields on MSSQL Serrver 2005 as VARCHAR. It contains alphanumeric values like "A,B,C,D ... 1,2,3,...,10,11,12" etc.
When i use below codes;
....
ORDER BY TableFiledName
Ordering result is as follow 11,12,1,2,3 etc.
When i use codes as below,
....
ORDER BY
CASE WHEN ISNUMERIC(TableFiledName) = 0 THEN CAST(TableFiledNameAS INT) ELSE TableFiledName END
I get error message as below;
Msg 8114, Level 16, State 5, Line 1 Error converting data type varchar
to float.
How can get like this ordering result: 1,2,3,4,5,6,7,8,9,10,11,12 etc..
Thanks in advance.
ISNUMERIC returns 1 when the field is numeric.
So your first problem is that it should be ...
CASE WHEN ISNUMERIC(TableFiledName) = 1 THEN
But this alone won't work.
You need to prefix the values with zeroes and take the rightmost
order by
case when ISNUMERIC(FieldName) =1
then right('000000000'+FieldName, 5)
else FieldName
end
Using 5 allows for numbers up to 99999 - if your numbers are higher, increase that number.
This will put the numbers before the letters. If you want the letters before the numbers, then you can add an isnumeric to the sort order - ie:
order by
isnumeric(FieldName),
case...
This won't cope with decimals, but you haven't mentioned them
maybe someone can help me out with a postgres query.
the table structure looks like this
nummer nachname vorname cash
+-------+----------+----------+------+
2 Bert Brecht 0,758
2 Harry Belafonte 1,568
3 Elvis Presley 0,357
4 Mark Twain 1,555
4 Ella Fitz 0,333
…
How can I coalesce the fields where "nummer" are the same and sum the cash values?
My output should look like this:
2 Bert, Brecht 2,326
Harry, Belafonte
3 Elvis, Presley 0,357
4 Mark, Twain 1,888
Ella, Fitz
I think the part to coalesce should work something like this:
array_to_string(array_agg(nachname|| ', ' ||coalesce(vorname, '')), '<br />') as name,
Thanks for any help,
tony
SELECT
nummer,
string_agg(nachname||CASE WHEN vorname IS NULL THEN '' ELSE ', '||vorname END, E'\n') AS name,
sum(cash) AS total_cash
FROM Table1
GROUP BY nummer;
See this SQLFiddle; note that it doesn't display the newline characters between names, but they're still there.
The CASE statement is used instead of coalesce so you don't have a trailing comma on entries with a last name but no first name. If you want a trailing comma, use format('%s, %s',vorname,nachname) instead and avoid all that ugly string concatenation business:
SELECT
nummer, string_agg(format('%s, %s', nachname, vorname), E'\n'),
sum(cash) AS total_cash
FROM Table1
GROUP BY nummer;
If string_agg doesn't work, get a newer PostgreSQL, or mention the version in your questions so it's clear you're using an obsolete version. The query is trivially rewritten to use array_to_string and array_agg anyway.
If you're asking how to sum numbers that're actually represented as text strings like 1,2345 in the database: don't do that. Fix your schema. Format numbers on input and output instead, store them as numeric, float8, integer, ... whatever the appropriate numeric type for the job is.
I want to search a column and get values where value containts \ .
I tried select * from "Values" where "ValueName" like '\'. But returns no value.
Also tried like "\" and like'\''%' etc. But no results.
See the DB2 Documentation on the LIKE predicate, in particular the parts about escape expressions.
What you want is
select * from Values where ValueName like '\\%' escape '\'
To give an example of usage:
create table backslash_escape_test
(
backslash_escape_test_column varchar(20)
);
insert into backslash_escape_test(backslash_escape_test_column)
values ('foo\');
insert into backslash_escape_test(backslash_escape_test_column)
values ('no slashes here');
insert into backslash_escape_test(backslash_escape_test_column)
values ('foo\bar');
insert into backslash_escape_test(backslash_escape_test_column)
values ('\bar');
select count(*) from backslash_escape_test where
backslash_escape_test_column like '%\\%' escape '\';
returns 3 (all 3 rows with \ in them).
select count(*) from backslash_escape_test where
backslash_escape_test_column like '\\%' escape '\';
returns 1 (the \bar row).
select * from Values where ValueName like '%\\%'
values is a not so good name because it may be confused with the values keyword
Don't escape it. You just need wildcards around it like this:
select count(*)
from escape_test
where test_column like '%\%'
But, suppose you really do need to escape the slash. Here's a simpler, more straightforward answer:
The escape-expression allows you to specify whatever character for escaping that you wish. So why use a character that you're looking for, thus requiring you to escape it? Use any other character instead. I'll use a plus sign as an example, but it could be a backslash, pound-sign, question-mark, anything other than a character you are looking for or one of the wildcard characters (% or _).
select count(*)
from escape_test
where test_column like '%\%' escape '+';
Now you don't have to add anything into your like-pattern.
To hold myself to the same standard of proof that #Michael demonstrated --
create table escape_test
( test_column varchar(20) );
insert into escape_test
(test_column)
values ('foo\'),
('no slashes here'),
('foo\bar'),
('\bar');
select 'test1' trial, count(*) result
from escape_test
where test_column like '%\%'
UNION
select 'test2', count(*)
from escape_test
where test_column like '%\\%' escape '\'
UNION
select 'test3', count(*)
from escape_test
where test_column like '%\%' escape '+'
;
Which returns the same number of rows for each method:
TRIAL RESULT
----- ------
test1 3
test2 3
test3 3