Postgresql: Extract text before number starts
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I have a table os which contains below data
id name
-- ----
1 windows server 2012 R2
2 windows 2016 SQL
3 Oracle linux 7.5
I need to update os table to something like below
id name
-- ----
1 windows server
2 windows
3 Oracle linux
Basically I need to extract the text from name starting from to before the number.
I am not getting an idea about how to do this. How can I do this in a Postgresql query?
postgresql
edited Nov 23 '18 at 17:26
halfer
14.8k759117
asked Nov 22 '18 at 8:32
Hemadri DasariHemadri Dasari
10.4k32358
add a comment |
I have a table os which contains below data
id name
-- ----
1 windows server 2012 R2
2 windows 2016 SQL
3 Oracle linux 7.5
I need to update os table to something like below
id name
-- ----
1 windows server
2 windows
3 Oracle linux
Basically I need to extract the text from name starting from to before the number.
I am not getting an idea about how to do this. How can I do this in a Postgresql query?
postgresql
edited Nov 23 '18 at 17:26
halfer
14.8k759117
asked Nov 22 '18 at 8:32
Hemadri DasariHemadri Dasari
10.4k32358
add a comment |
I have a table os which contains below data
id name
-- ----
1 windows server 2012 R2
2 windows 2016 SQL
3 Oracle linux 7.5
I need to update os table to something like below
id name
-- ----
1 windows server
2 windows
3 Oracle linux
Basically I need to extract the text from name starting from to before the number.
I am not getting an idea about how to do this. How can I do this in a Postgresql query?
postgresql
edited Nov 23 '18 at 17:26
halfer
14.8k759117
asked Nov 22 '18 at 8:32
Hemadri DasariHemadri Dasari
10.4k32358
I have a table os which contains below data
id name
-- ----
1 windows server 2012 R2
2 windows 2016 SQL
3 Oracle linux 7.5
I need to update os table to something like below
id name
-- ----
1 windows server
2 windows
3 Oracle linux
Basically I need to extract the text from name starting from to before the number.
I am not getting an idea about how to do this. How can I do this in a Postgresql query?
postgresql
postgresql
edited Nov 23 '18 at 17:26
halfer
14.8k759117
asked Nov 22 '18 at 8:32
Hemadri DasariHemadri Dasari
10.4k32358
edited Nov 23 '18 at 17:26
halfer
14.8k759117
asked Nov 22 '18 at 8:32
Hemadri DasariHemadri Dasari
10.4k32358
edited Nov 23 '18 at 17:26
halfer
14.8k759117
edited Nov 23 '18 at 17:26
halfer
14.8k759117
edited Nov 23 '18 at 17:26
halfer
14.8k759117
14.8k759117
asked Nov 22 '18 at 8:32
Hemadri DasariHemadri Dasari
10.4k32358
asked Nov 22 '18 at 8:32
Hemadri DasariHemadri Dasari
10.4k32358
asked Nov 22 '18 at 8:32
Hemadri DasariHemadri Dasari
10.4k32358
10.4k32358
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
You may try this. It starts from beginning and finds anything that's not a digit before a digit and replaces it with the matched string.
SELECT s,RTRIM(regexp_replace (s, '^([^d]+)d(.*)$', '1')) as m
FROM ( VALUES ('windows server 2012 R2'),
('windows 2016 SQL'),
('Oracle linux 7.5' ) ) AS t(s);
s | m
------------------------+----------------
windows server 2012 R2 | windows server
windows 2016 SQL | windows
Oracle linux 7.5 | Oracle linux
(3 rows)
answered Nov 22 '18 at 9:51
Kaushik NayakKaushik Nayak
21.9k41332
add a comment |
The function substring allows to specify regular expression:
So some solution can looks like:
postgres=# select name, trim(substring(name from '[^d]+')) from the_table;
+----------------------------+----------------+
| name | btrim |
+----------------------------+----------------+
| windows server 2012 R2 foo | windows server |
| windows 2016 SQL | windows |
| Oracle linux 7.5 | Oracle linux |
+----------------------------+----------------+
(3 rows)
More in documentation
answered Nov 22 '18 at 10:07
Pavel StehulePavel Stehule
23.6k34859
add a comment |
This is a case where the absence of a function that finds a string based on a regex is a bit cumbersome.
You can use regexp_matches() to find the first pattern of the first number in the string, then combine that with substr() and strpos()` to extract everything before that:
select id, substr(name, 1, strpos(name, (regexp_matches(name, '[0-9.]+'))[1]) - 1)
from the_table;
regexp_matches() returns an array of all matches. So wee need to extract the first first match from that array. That's what (regexp_matches(name, '[0-9.]+'))[1] does.
Online example: https://rextester.com/BIOV86746
answered Nov 22 '18 at 9:50
a_horse_with_no_namea_horse_with_no_name
308k46471573
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You may try this. It starts from beginning and finds anything that's not a digit before a digit and replaces it with the matched string.
SELECT s,RTRIM(regexp_replace (s, '^([^d]+)d(.*)$', '1')) as m
FROM ( VALUES ('windows server 2012 R2'),
('windows 2016 SQL'),
('Oracle linux 7.5' ) ) AS t(s);
s | m
------------------------+----------------
windows server 2012 R2 | windows server
windows 2016 SQL | windows
Oracle linux 7.5 | Oracle linux
(3 rows)
answered Nov 22 '18 at 9:51
Kaushik NayakKaushik Nayak
21.9k41332
add a comment |
You may try this. It starts from beginning and finds anything that's not a digit before a digit and replaces it with the matched string.
SELECT s,RTRIM(regexp_replace (s, '^([^d]+)d(.*)$', '1')) as m
FROM ( VALUES ('windows server 2012 R2'),
('windows 2016 SQL'),
('Oracle linux 7.5' ) ) AS t(s);
s | m
------------------------+----------------
windows server 2012 R2 | windows server
windows 2016 SQL | windows
Oracle linux 7.5 | Oracle linux
(3 rows)
answered Nov 22 '18 at 9:51
Kaushik NayakKaushik Nayak
21.9k41332
add a comment |
You may try this. It starts from beginning and finds anything that's not a digit before a digit and replaces it with the matched string.
SELECT s,RTRIM(regexp_replace (s, '^([^d]+)d(.*)$', '1')) as m
FROM ( VALUES ('windows server 2012 R2'),
('windows 2016 SQL'),
('Oracle linux 7.5' ) ) AS t(s);
s | m
------------------------+----------------
windows server 2012 R2 | windows server
windows 2016 SQL | windows
Oracle linux 7.5 | Oracle linux
(3 rows)
answered Nov 22 '18 at 9:51
Kaushik NayakKaushik Nayak
21.9k41332
You may try this. It starts from beginning and finds anything that's not a digit before a digit and replaces it with the matched string.
SELECT s,RTRIM(regexp_replace (s, '^([^d]+)d(.*)$', '1')) as m
FROM ( VALUES ('windows server 2012 R2'),
('windows 2016 SQL'),
('Oracle linux 7.5' ) ) AS t(s);
s | m
------------------------+----------------
windows server 2012 R2 | windows server
windows 2016 SQL | windows
Oracle linux 7.5 | Oracle linux
(3 rows)
answered Nov 22 '18 at 9:51
Kaushik NayakKaushik Nayak
21.9k41332
answered Nov 22 '18 at 9:51
Kaushik NayakKaushik Nayak
21.9k41332
answered Nov 22 '18 at 9:51
Kaushik NayakKaushik Nayak
21.9k41332
answered Nov 22 '18 at 9:51
Kaushik NayakKaushik Nayak
21.9k41332
21.9k41332
add a comment |
add a comment |
The function substring allows to specify regular expression:
So some solution can looks like:
postgres=# select name, trim(substring(name from '[^d]+')) from the_table;
+----------------------------+----------------+
| name | btrim |
+----------------------------+----------------+
| windows server 2012 R2 foo | windows server |
| windows 2016 SQL | windows |
| Oracle linux 7.5 | Oracle linux |
+----------------------------+----------------+
(3 rows)
More in documentation
answered Nov 22 '18 at 10:07
Pavel StehulePavel Stehule
23.6k34859
add a comment |
The function substring allows to specify regular expression:
So some solution can looks like:
postgres=# select name, trim(substring(name from '[^d]+')) from the_table;
+----------------------------+----------------+
| name | btrim |
+----------------------------+----------------+
| windows server 2012 R2 foo | windows server |
| windows 2016 SQL | windows |
| Oracle linux 7.5 | Oracle linux |
+----------------------------+----------------+
(3 rows)
More in documentation
answered Nov 22 '18 at 10:07
Pavel StehulePavel Stehule
23.6k34859
add a comment |
The function substring allows to specify regular expression:
So some solution can looks like:
postgres=# select name, trim(substring(name from '[^d]+')) from the_table;
+----------------------------+----------------+
| name | btrim |
+----------------------------+----------------+
| windows server 2012 R2 foo | windows server |
| windows 2016 SQL | windows |
| Oracle linux 7.5 | Oracle linux |
+----------------------------+----------------+
(3 rows)
More in documentation
answered Nov 22 '18 at 10:07
Pavel StehulePavel Stehule
23.6k34859
The function substring allows to specify regular expression:
So some solution can looks like:
postgres=# select name, trim(substring(name from '[^d]+')) from the_table;
+----------------------------+----------------+
| name | btrim |
+----------------------------+----------------+
| windows server 2012 R2 foo | windows server |
| windows 2016 SQL | windows |
| Oracle linux 7.5 | Oracle linux |
+----------------------------+----------------+
(3 rows)
More in documentation
answered Nov 22 '18 at 10:07
Pavel StehulePavel Stehule
23.6k34859
answered Nov 22 '18 at 10:07
Pavel StehulePavel Stehule
23.6k34859
answered Nov 22 '18 at 10:07
Pavel StehulePavel Stehule
23.6k34859
answered Nov 22 '18 at 10:07
Pavel StehulePavel Stehule
23.6k34859
23.6k34859
add a comment |
add a comment |
This is a case where the absence of a function that finds a string based on a regex is a bit cumbersome.
You can use regexp_matches() to find the first pattern of the first number in the string, then combine that with substr() and strpos()` to extract everything before that:
select id, substr(name, 1, strpos(name, (regexp_matches(name, '[0-9.]+'))[1]) - 1)
from the_table;
regexp_matches() returns an array of all matches. So wee need to extract the first first match from that array. That's what (regexp_matches(name, '[0-9.]+'))[1] does.
Online example: https://rextester.com/BIOV86746
answered Nov 22 '18 at 9:50
a_horse_with_no_namea_horse_with_no_name
308k46471573
add a comment |
This is a case where the absence of a function that finds a string based on a regex is a bit cumbersome.
You can use regexp_matches() to find the first pattern of the first number in the string, then combine that with substr() and strpos()` to extract everything before that:
select id, substr(name, 1, strpos(name, (regexp_matches(name, '[0-9.]+'))[1]) - 1)
from the_table;
regexp_matches() returns an array of all matches. So wee need to extract the first first match from that array. That's what (regexp_matches(name, '[0-9.]+'))[1] does.
Online example: https://rextester.com/BIOV86746
answered Nov 22 '18 at 9:50
a_horse_with_no_namea_horse_with_no_name
308k46471573
add a comment |
This is a case where the absence of a function that finds a string based on a regex is a bit cumbersome.
You can use regexp_matches() to find the first pattern of the first number in the string, then combine that with substr() and strpos()` to extract everything before that:
select id, substr(name, 1, strpos(name, (regexp_matches(name, '[0-9.]+'))[1]) - 1)
from the_table;
regexp_matches() returns an array of all matches. So wee need to extract the first first match from that array. That's what (regexp_matches(name, '[0-9.]+'))[1] does.
Online example: https://rextester.com/BIOV86746
answered Nov 22 '18 at 9:50
a_horse_with_no_namea_horse_with_no_name
308k46471573
This is a case where the absence of a function that finds a string based on a regex is a bit cumbersome.
You can use regexp_matches() to find the first pattern of the first number in the string, then combine that with substr() and strpos()` to extract everything before that:
select id, substr(name, 1, strpos(name, (regexp_matches(name, '[0-9.]+'))[1]) - 1)
from the_table;
regexp_matches() returns an array of all matches. So wee need to extract the first first match from that array. That's what (regexp_matches(name, '[0-9.]+'))[1] does.
Online example: https://rextester.com/BIOV86746
answered Nov 22 '18 at 9:50
a_horse_with_no_namea_horse_with_no_name
308k46471573
answered Nov 22 '18 at 9:50
a_horse_with_no_namea_horse_with_no_name
308k46471573
answered Nov 22 '18 at 9:50
a_horse_with_no_namea_horse_with_no_name
308k46471573
answered Nov 22 '18 at 9:50
a_horse_with_no_namea_horse_with_no_name
308k46471573
308k46471573
add a comment |
add a comment |
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