Please help me with this issue. I'm very bad at regex things. I need to remove country code and blank spaces at once from phone number. Something like:
'+12 345 678' to '345678'. Thanks for any help!
demo: db<>fiddle
Assuming the country code always is the first three characters:
SELECT replace(right('+12 345 678', -3), ' ', '')
right('xyz', -3) removes the first three characters
replace('xyz', ' ', '') removes the spaces.
More general:
SELECT
replace(right(numbers, -3), ' ', '')
FROM
phone
Related
How to return last n words using Postgres.
I have tried using LEFT method.
SELECT DISTINCT LEFT(name, -4) FROM my_table;
but it return last 4 characters ,i want to return last 3 words.
demo:db<>fiddle
You can do this using a the SUBSTRING() function and regular expressions:
SELECT
SUBSTRING(name FROM '((\S+\s+){0,3}\S+$)')
FROM my_table
This has been explained here: How can I match the last two words in a sentence in PostgreSQL?
\S+ is a string of non-whitespace characters
\s+ is a string of whitespace characters (e.g. one space)
(\S+\s+){0,3} Zero to three words separated by a space
\S+$ one word at the end of the text.
-> creates 4 words (or less if there are no more).
One way is to use regexp_split_to_array() to split the string into the words it contains and then put a string back together using the last 3 words in that array.
SELECT coalesce(w.words[array_length(w.words, 1) - 2] || ' ', '')
|| coalesce(w.words[array_length(w.words, 1) - 1] || ' ', '')
|| coalesce(w.words[array_length(w.words, 1)], '')
FROM mytable t
CROSS JOIN LATERAL (SELECT regexp_split_to_array(t."name", ' ') words) w;
db<>fiddle
RIGHT() should do
SELECT RIGHT('MYCOLUMN', 4); -- returns LUMN
UPD
You can convert to array and then back to string
SELECT array_to_string(sentence[(array_length(sentence,1)-3):(array_length(sentence,1))],' ','*')
FROM
(
SELECT regexp_split_to_array('this is the one of the way to get the last four words of the string', E'\\s+') AS sentence
) foo;
DEMO HERE
I am currently working on DB data which contains whitespaces and hyphens. I searched over the net and found this Remove/replace special characters in column values? . I tried to follow the answer but I am still getting hyphens. I tried playing around with it, I can only remove the whitespace
conn_p = p.connect("dbname='p_test' user='postgres' password='postgres' host='localhost'")
conn_t = p.connect("dbname='t_mig1' user='postgres' password='postgres' host='localhost'")
cur_p = conn_p.cursor()
cur_t = conn_t.cursor()
cur_t.execute("SELECT CAST(REGEXP_REPLACE(studentnumber, ' ', '') as integer), firstname, middlename, lastname FROM sprofile")
rows = cur_t.fetchall()
for row in rows:
print "Inserting ", row[0], row[1], row[2], row[3]
cur_p.execute(""" INSERT INTO "a_recipient" (id, first_name, middle_name, last_name) VALUES ('%s', '%s', '%s', '%s') """ % (row[0], row[1], row[2], row[3]))
cur_p.commit()
cur_pl.close()
cur_t.close()
What I would like to achieve is if I got a studentnumber of 001-2012-1456, it will be displayed as 000120121456.
To wipe out all characters in a set efficiently use translate. It takes a set of characters to translate into another set of characters. If the other set is empty it deletes them.
test=> select translate('001-2012-145 6', '- ', '');
translate
-------------
00120121456
While translate is simpler and faster for this particular job, it's important to know how to use regexes for others. To do it with regexp_replace there's two changes you need to make.
First, you have to match the set of - and as [- ].
Then, you have to specify to replace all occurrences, otherwise it will stop after the first one. That's done with the g flag.
test=> select regexp_replace('001-2012-145 6', '[- ]', '', 'g');
regexp_replace
----------------
00120121456
Here's a tutorial on POSIX regular expressions and character sets.
Its very simple to use inbuilt translate function.
Example:
select translate('001-2012-145 6', '- ', '');
Output of above command :
00120121456
I am wondering how I can remove all newline characters in Redshift from a field. I tried something like this:
replace(replace(body, '\n', ' '), '\r', ' ')
and
regexp_replace(body, '[\n\r]+', ' ')
But it didn't work. Please share if you know how to do this.
Use chr(10) instead of \n
example:
select replace(CONCAT('Text 1' , chr(10), 'Text 2'), chr(10), '-') as txt
This should help
regexp_replace(column, '\r|\n', '')
To remove line breaks:
SELECT REPLACE('This line has
a line break', CHR(10), '');
This gives output: This line hasa line break. You can see more ASCII or CHR() codes here: https://www.petefreitag.com/cheatsheets/ascii-codes/
To remove special characters like \r, \n, \t
Assuming col1 has text like This line has\r\n special characters.
Using replace()
SELECT REPLACE(REPLACE(col1, '\\r', ''), '\\n', '');
We need to escape \ because backslash is special character in SQL (used to escape double quotes, etc...)
Using regexp_replace()
SELECT REGEXP_REPLACE(col1, '(\\\\r|\\\\n)', '');
We need to escape \ because it is a special character in SQL and we need to escape the resulting backslashes again because backslash is a special character in regex as well.
Both replace() and regexp_replace() gives output: This line has special characters.
I'm trying to truncate leading zero from the address. example:
input
1 06TH ST
12 02ND AVE
123 001St CT
expected output
1 6TH ST
12 2ND AVE
123 1St CT
Here is what i have:
update table
set address = regexp_replace(address,'(0\d+(ST|ND|TH))','?????? need help here')
where address ~ '\s0\d+(ST|ND|TH)\s';
many thanks in advance
assuming that the address always has some number/letter address (1234, 1a, 33B) followed by a sequence of 1 or more spaces followed by the part you want to strip leading zeroes...
select substr(address, 1, strpos(address, ' ')) || ltrim(substr(address, strpos(address, ' ')), ' 0') from table;
or, to update the table:
update table set address = substr(address, 1, strpos(address, ' ')) || ltrim(substr(address, strpos(address, ' ')), ' 0');
-g
What you are looking for is the back references in the regular expressions:
UPDATE table
SET address = regexp_replace(address, '\m0+(\d+\w+)', '\1', 'g')
WHERE address ~ '\m0+(\d+\w+)'
Also:
\m used to match the beginning of a word (to avoid replacing inside words (f.ex. in 101Th)
0+ truncates all zeros (does not included in the capturing parenthesis)
\d+ used to capture the remaining numbers
\w+ used to capture the remaining word characters
a word caracter can be any alphanumeric character, and the underscore _.
I have a list of names and I want to separate the first and last words in a person's name.
I was trying to use the "trim" function without success.
Can someone explain how could I do it?
table:
Names
Mary Johnson Angel Smith
Dinah Robertson Donald
Paul Blank Power Silver
Then I want to have as a result:
Names
Mary Smith
Dinah Donald
Paul Silver
Thanks,
You can do it simply with regular expressions, like:
substring(trim(name) FROM '^([^ ]+)') || ' ' || substring(trim(name) FROM '([^ ]+)$')
Of course it would only work you are 100% there is always supplied at least a first and a last name. I'm not 100% sure it is the case for everybody in the World. For instance: would that work for names in Chinese? I'm not sure and I avoid doing any assumption about people names. The best is to simply ask the user two fields, one for the "name" and another for "How would you like to be called?".
Another approach, which takes advantage of Postgres string processing built-in functions:
SELECT split_part(name, ' ', 1) as first_token,
split_part(name, ' ', array_length(regexp_split_to_array(name, ' '), 1)) as last_token
FROM mytable
Here's how I extracted full names from emails with a dot in them, eg Jeremy.Thompson#abc.com
SELECT split_part(email, '.', 1) || ' ' || replace(split_part(email, '.', 2), '#abc','')
FROM people
Result:
Jeremy | Thompson
You can easily replace the dot with a space:
SELECT split_part(email, ' ', 1) || ' ' || replace(split_part(email, ' ', 2), '#abc','')
FROM people