在某个字符之后从文本中提取每一行，并使用 postgres 将结果提取到 table

Question

我有这样一段文字：

>Sequenz: Test 1
MTEITAAMVKELRESTGAGMMDCKNALSETNGDFDKAVQLLREKGLGKAAKKADRLAAEG

>Sequenz 2 1234 Organism: Treponema
MTEITAAMVKELRESTGAGMMDCKNALSETNGDFDKAVQLLREKGLGKAAKKADRLAAEG

>Sequenz 3
MTEITAAMVKELRESTGAGMMDCKNALSETNGDFDKAVQLLREKGLGKAAKKADRLAAEG

文本块之间不一定有空行，它可能是 'MTEITAAMVKELRESTGAGM' 的各种行数。唯一可以确定的是每行前的 >。

我希望得到这样的 table：

HEADER 
----------
Sequenz: Test 1 
----------
Sequenz 2 1234 Organism: Treponema
----------
Sequenz 3

我试过了：

SELECT regexp_matches(regexp_split_to_table( 'text from above', '\n>'),'([A-Z,a-z,0-9]+\s)');

导致

HEADER
----------
Sequenz
----------
Sequenz
----------
Sequenz

和

Select regexp_split_to_table('text from bove', '[\\n>+(.)\\n]+')

导致

HEADER
----------
----------
Sequenz: Test 1 
----------
MTEITAAMVKELRESTGAGMMDCKNALSETNGDFDKAVQLLREKGLGKAAKKADRLAAEG
----------
----------
Sequenz 2 1234 Organism: Treponema 
----------
MTEITAAMVKELRESTGAGMMDCKNALSETNGDFDKAVQLLREKGLGKAAKKADRLAAEG
----------
----------
Sequenz 3 
----------
MTEITAAMVKELRESTGAGMMDCKNALSETNGDFDKAVQLLREKGLGKAAKKADRLAAEG

Answer 1

试试这个：

SELECT split_part(regexp_split_to_table(trim(leading '>' from '>Sequenz: Test 1
MTEITAAMVKELRESTGAGMMDCKNALSETNGDFDKAVQLLREKGLGKAAKKADRLAAEG

>Sequenz 2 1234 Organism: Treponema
MTEITAAMVKELRESTGAGMMDCKNALSETNGDFDKAVQLLREKGLGKAAKKADRLAAEG

>Sequenz 3
MTEITAAMVKELRESTGAGMMDCKNALSETNGDFDKAVQLLREKGLGKAAKKADRLAAEG'), E'>'),E'\n', 1) AS res

如果要保留第一个空行，请删除 trim() 函数。

演示：http://rextester.com/LQXY98290

在某个字符之后从文本中提取每一行，并使用 postgres 将结果提取到 table

Extracting each line from a text after a certain character and extract the result to a table using postgres

postgresql

fasta