I want to select executed query from log file. Specifically an example would look something like this:
2019-01-10 10:33:21 +07 dvdrentalLOG: statement: SELECT last_update
From public.actor
2019-03-06 14:07:06 +07 dvdrentalLOG: statement: SELECT film_id, title
FROM public.film
WHERE film_id = 1
I want to get the queries using looping. desired output:
query1 : SELECT last_update From public.actor
query2 : SELECT film_id, title FROM public.film WHERE film_id = 1
This I have tried:
import re
def parseFile(filepath):
line=[]
with open(filepath,'r') as log:
regex = re.compile(r'(\d{4}-\d{2}-\d{2})(.*)',re.MULTILINE|re.DOTALL)
for line in log:
date = regex.findall(line)
if date == []:
print()
else:
print(date)
filepath = 'text.txt'
parseFile(filepath)
output:
[('2019-01-10', ' 10:33:21 +07 dvdrentalLOG: statement: SELECT last_update \n')]
[('2019-03-06', ' 14:07:06 +07 dvdrentalLOG: statement: SELECT film_id, title\n')]
the output don't select all the queries. what should I do?
You can adapt your code like this (you need to read the whole file before parsing it, if you read line by line as you did in your code, your regex will only parse a line after another and will never be able to select the whole SQL queries split on several lines) :
import re
def parseFile(filepath):
line=[]
with open(filepath,'r') as log:
regex = re.compile(r'(\d{4}-\d{2}-\d{2})(.*?)(?=\d{4}-\d{2}-\d{2}|$)',re.MULTILINE|re.DOTALL)
lines = re.sub('\n|\s{2,}',' ',log.read())#.replace('\n', '')
date = regex.findall(lines)
if date == []:
print()
else:
print(date)
filepath = 'query.log'
parseFile(filepath)
output:
[('2019-01-10', ' 10:33:21 +07 dvdrentalLOG: statement: SELECT last_update From public.actor '), ('2019-03-06', ' 14:07:06 +07 dvdrentalLOG: statement: SELECT film_id, title FROM public.film WHERE film_id = 1 ')]
Where the regex (using positive lookahead to limit the number of characters matched by .*?) used is detailed here: https://regex101.com/r/nE0omm/1/
(\d{4}-\d{2}-\d{2})(.*?)(?=\d{4}-\d{2}-\d{2}|$)
You're only processing a single line at a time (via the for line in log: loop), so your regex only applies to a single line at a time. It couldn't match across lines because you're not giving it multiple lines at a time to match across.
You could instead read the entire file via log.read() and then call .findall on that.
Related
I have a regular expression like the following: (Running on Oracle's regexp_like(), despite the question isn't Oracle-specific)
abc|bcd|def|xyz
This basically matches a tags field on database to see if tags field contains abc OR bcd OR def OR xyz when user has input for the search query "abc bcd def xyz".
The tags field on the database holds keywords separated by spaces, e.g. "cdefg abcd xyz"
On Oracle, this would be something like:
select ... from ... where
regexp_like(tags, 'abc|bcd|def|xyz');
It works fine as it is, but I want to add an extra option for users to search for results that match all keywords. How should I change the regular expression so that it matches abc AND bcd AND def AND xyz ?
Note: Because I won't know what exact keywords the user will enter, I can't pre-structure the query in the PL/SQL like this:
select ... from ... where
tags like '%abc%' AND
tags like '%bcd%' AND
tags like '%def%' AND
tags like '%xyz%';
You can split the input pattern and check that all the parts of the pattern match:
SELECT t.*
FROM table_name t
CROSS APPLY(
WITH input (match) AS (
SELECT 'abc bcd def xyz' FROM DUAL
)
SELECT 1
FROM input
CONNECT BY LEVEL <= REGEXP_COUNT(match, '\S+')
HAVING COUNT(
REGEXP_SUBSTR(
t.tags,
REGEXP_SUBSTR(match, '\S+', 1, LEVEL)
)
) = REGEXP_COUNT(match, '\S+')
)
Or, if you have Java enabled in the database then you can create a Java function to match regular expressions:
CREATE AND COMPILE JAVA SOURCE NAMED RegexParser AS
import java.util.regex.Pattern;
public class RegexpMatch {
public static int match(
final String value,
final String regex
){
final Pattern pattern = Pattern.compile(regex);
return pattern.matcher(value).matches() ? 1 : 0;
}
}
/
Then wrap it in an SQL function:
CREATE FUNCTION regexp_java_match(value IN VARCHAR2, regex IN VARCHAR2) RETURN NUMBER
AS LANGUAGE JAVA NAME 'RegexpMatch.match( java.lang.String, java.lang.String ) return int';
/
Then use it in SQL:
SELECT *
FROM table_name
WHERE regexp_java_match(tags, '(?=.*abc)(?=.*bcd)(?=.*def)(?=.*xyz)') = 1;
Try this, the idea being counting that the number of matches is == to the number of patterns:
with data(val) AS (
select 'cdefg abcd xyz' from dual union all
select 'cba lmnop xyz' from dual
),
targets(s) as (
select regexp_substr('abc bcd def xyz', '[^ ]+', 1, LEVEL) from dual
connect by regexp_substr('abc bcd def xyz', '[^ ]+', 1, LEVEL) is not null
)
select val from data d
join targets t on
regexp_like(val,s)
group by val having(count(*) = (select count(*) from targets))
;
Result:
cdefg abcd xyz
I think dynamic SQL will be needed for this. The match all option will require individual matching with logic to ensure every individual match is found.
An easy way would be to build a join condition for each keyword. Concatenate the join statements in a string. Use dynamic SQL to execute the string as a query.
The example below uses the customer table from the sample schemas provided by Oracle.
DECLARE
-- match string should be just the values to match with spaces in between
p_match_string VARCHAR2(200) := 'abc bcd def xyz';
-- need logic to determine match one (OR) versus match all (AND)
p_match_type VARCHAR2(3) := 'OR';
l_sql_statement VARCHAR2(4000);
-- create type if bulk collect is needed
TYPE t_email_address_tab IS TABLE OF customers.EMAIL_ADDRESS%TYPE INDEX BY PLS_INTEGER;
l_email_address_tab t_email_address_tab;
BEGIN
WITH sql_clauses(row_idx,sql_text) AS
(SELECT 0 row_idx -- build select plus beginning of where clause
,'SELECT email_address '
|| 'FROM customers '
|| 'WHERE 1 = '
|| DECODE(p_match_type, 'AND', '1', '0') sql_text
FROM DUAL
UNION
SELECT LEVEL row_idx -- build joins for each keyword
,DECODE(p_match_type, 'AND', ' AND ', ' OR ')
|| 'email_address'
|| ' LIKE ''%'
|| REGEXP_SUBSTR( p_match_string,'[^ ]+',1,level)
|| '%''' sql_text
FROM DUAL
CONNECT BY LEVEL <= LENGTH(p_match_string) - LENGTH(REPLACE( p_match_string, ' ' )) + 1
)
-- put it all together by row_idx
SELECT LISTAGG(sql_text, '') WITHIN GROUP (ORDER BY row_idx)
INTO l_sql_statement
FROM sql_clauses;
dbms_output.put_line(l_sql_statement);
-- can use execute immediate (or ref cursor) for dynamic sql
EXECUTE IMMEDIATE l_sql_statement
BULK COLLECT
INTO l_email_address_tab;
END;
Variable
Value
p_match_string
abc bcd def xyz
p_match_type
AND
l_sql_statement
SELECT email_address FROM customers WHERE 1 = 1 AND email_address LIKE '%abc%' AND email_address LIKE '%bcd%' AND email_address LIKE '%def%' AND email_address LIKE '%xyz%'
Variable
Value
p_match_string
abc bcd def xyz
p_match_type
OR
l_sql_statement
SELECT email_address FROM customers WHERE 1 = 0 OR email_address LIKE '%abc%' OR email_address LIKE '%bcd%' OR email_address LIKE '%def%' OR email_address LIKE '%xyz%'
I have the following dataframe:
Column1 Column2
0 .com<br><br>Finance<br><br><br><br><br><br><br><br><br><br><br><br> .comFinance
1 .com<br><br>Finance<br><br><br><br><br>DO<br><br><br><br><br><br><br> .comFinanceDO
2 <br><br>Finance<br><br><br>ISV<br><br>DO<br>DO Prem<br><br><br><br><br><br> FinanceISVDODO Prem
3 <br><br>Finance<br><br><br><br><br><br><br><br><br><br><br><br> Finance
4 <br><br>Finance<br><br><br>TTY<br><br><br><br><br><br><br><br><br> ConsultingTTY
I used to following line of code to get Column2:
df['Column2'] = df['Column1'].str.replace('<br>', '', regex=True)
I want to remove all instances of "< b >" and so I want the column to look like this:
Column2
.com, Finance
.com, Finance, DO
Finance, ISV, DO, DO Prem
Finance
Consulting, TTY
Given the following dataframe:
Column1
.com<br><br>Finance<br><br><br><br><br><br><br><br><br><br><br><br>
.com<br><br>Finance<br><br><br><br><br>DO<br><br><br><br><br><br><br>
<br><br>Finance<br><br><br>ISV<br><br>DO<br>DO Prem<br><br><br><br><br><br>
<br><br>Finance<br><br><br><br><br><br><br><br><br><br><br><br>
<br><br>Finance<br><br><br>TTY<br><br><br><br><br><br><br><br><br>
df['Column2'] = df['Column1'].str.replace('<br>', ' ', regex=True).str.strip().replace('\\s+', ', ', regex=True) doesn't work because of sections like <br>DO Prem<br>, which will end of like DO, Prem, not DO Prem.
Split on <br> to make a list, then use a list comprehension to remove the '' spaces.
This will preserve spaces where they're supposed to be.
Join the list values back into a string with (', ').join([...])
import pandas as pd
df['Column2'] = df['Column1'].str.split('<br>').apply(lambda x: (', ').join([y for y in x if y != '']))
# output
Column1 Column2
.com<br><br>Finance<br><br><br><br><br><br><br><br><br><br><br><br> .com, Finance
.com<br><br>Finance<br><br><br><br><br>DO<br><br><br><br><br><br><br> .com, Finance, DO
<br><br>Finance<br><br><br>ISV<br><br>DO<br>DO Prem<br><br><br><br><br><br> Finance, ISV, DO, DO Prem
<br><br>Finance<br><br><br><br><br><br><br><br><br><br><br><br> Finance
<br><br>Finance<br><br><br>TTY<br><br><br><br><br><br><br><br><br> Finance, TTY
### Replace br with space
df['Column 2'] = df['column 1'].str.replace('<br>', ' ')
### Get rid of spaces before and after the string
df['Column 2'] = df['Column 2'].strip()
### Replace the space with ,
df['Column 2'] = df['Column 2'].str.replace('\\s+', ',', regex=True)
As pointed out by TrentonMcKinney, his solution is better. This one doesn't solve the issue when there is a space between the string values in Column 1
My question: how do I create a dictionary from a list by assigning dictionary keys based on a regex pattern match ('^--L-[0-9]{8}'), and assigning the values by using all lines between each key.
Example excerpt from the raw file:
SQL> --L-93752133
SQL> --SELECT table_name, tablespace_name from dba_tables where upper(table_name) like &tablename_from_developer;
SQL>
SQL> --L-52852243
SQL>
SQL> SELECT log_mode FROM v$database;
LOG_MODE
------------
NOARCHIVELOG
SQL>
SQL> archive log list
Database log mode No Archive Mode
Automatic archival Disabled
Archive destination USE_DB_RECOVERY_FILE_DEST
Oldest online log sequence 3
Current log sequence 5
SQL>
SQL> --L-42127143
SQL>
SQL> SELECT t.name "TSName", e.encryptionalg "Algorithm", d.file_name "File Name"
2 FROM v$tablespace t
3 , v$encrypted_tablespaces e
4 , dba_data_files d
5 WHERE t.ts# = e.ts#
6 AND t.name = d.tablespace_name;
no rows selected
Some additional detail: The raw file can be large (at least 80K+ lines, but often much larger) and I need to preserve the original spacing so the output is still easy to read. Here's how I'm reading the file in and removing "SQL>" from the beginning of each line:
with open(rawFile, 'r') as inFile:
content = inFile.read()
rawList = content.splitlines()
for line in rawList:
cleanLine = re.sub('^SQL> ', '', line)
Finding the dictionary keys I'm looking for is easy:
pattern = re.compile(r'^--L-[0-9]{8}')
if pattern.search(cleanLine) is not None:
itemID = pattern.search(cleanLine)
print(itemID.group(0))
But how do I assign all lines between each key as the value belonging to the most recent key preceding them? I've been playing around with new lists, tuples, and dictionaries but everything I do is returning garbage. The goal is to have the data and keys linked to each other so that I can return them as needed later in my script.
I spent a while searching for a similar question, but in most other cases the source file was already in a dictionary-like format so creating the new dictionary was a less complicated problem. Maybe a dictionary or tuple isn't the right answer, but any help would be appreciated! Thanks!
In general, you should question why you would read the entire file, split the lines into a list, and then iterate over the list. This is a Python anti-pattern.
For line oriented text files, just do:
with open(fn) as f:
for line in f:
# process a line
It sounds, however, that you have multi-line block oriented patterns. If so, with smaller files, read the entire file into a single string and use a regex on that. Then you would use group 1 and group 2 as the key, value in your dict:
pat=re.compile(pattern, flags)
with open(file_name) as f:
di={m.group(1):m.group(2) for m in pat.finditer(f.read())}
With a larger file, use a mmap:
import re, mmap
pat=re.compile(pattern, flags)
with open(file_name, 'r+') as f:
mm = mmap.mmap(f.fileno(), 0)
for i, m in enumerate(pat.finditer(mm)):
# process each block accordingly...
As far as the regex, I am a little unclear on what you are trying to capture or not. I think this regex is what I am understanding you want:
^SQL> (--L-[0-9]{8})(.*?)(?=SQL> --L-[0-9]{8}|\Z)
Demo
In either case, running that regex with the example string yields:
>>> pat=re.compile(r'^SQL> (--L-[0-9]{8})\s*(.*?)\s*(?=SQL> --L-[0-9]{8}|\Z)', re.S | re.M)
>>> with open(file_name) as f:
... di={m.group(1):m.group(2) for m in pat.finditer(f.read())}
...
>>> di
{'--L-52852243': 'SQL> \nSQL> SELECT log_mode FROM v;\n\n LOG_MODE\n ------------\n NOARCHIVELOG\n\nSQL> \nSQL> archive log list\n Database log mode No Archive Mode\n Automatic archival Disabled\n Archive destination USE_DB_RECOVERY_FILE_DEST\n Oldest online log sequence 3\n Current log sequence 5\nSQL>',
'--L-93752133': 'SQL> --SELECT table_name, tablespace_name from dba_tables where upper(table_name) like &tablename_from_developer;\nSQL>',
'--L-42127143': 'SQL> \nSQL> SELECT t.name TSName, e.encryptionalg Algorithm, d.file_name File Name\n 2 FROM v t\n 3 , v e\n 4 , dba_data_files d\n 5 WHERE t.ts# = e.ts#\n 6 AND t.name = d.tablespace_name;\n\n no rows selected'}
Something like this?
with open(rawFile, 'r') as inFile:
content = inFile.read()
rawList = content.splitlines()
keyed_dict = {}
in_between_lines = ""
last_key = 0
for line in rawList:
cleanLine = re.sub('^SQL> ', '', line)
pattern = re.compile(r'^--L-[0-9]{8}')
if pattern.search(cleanLine) is not None:
itemID = pattern.search(cleanLine)
if last_key: keyed_dict[last_key] = in_between_lines
last_key = itemID.group(0)
in_between_lines = ""
else:
in_between_lines += cleanLine
from googlefinance import getQuotes
import json
import time as t
import re
List = ["A","AA","AAB"]
Time=t.localtime() # Sets variable Time to retrieve date/time info
Date2= ('%d-%d-%d %dh:%dm:%dsec'%(Time[0],Time[1],Time[2],Time[3],Time[4],Time[5])) #formats time stamp
while True:
for i in List:
try: #allows elements to be called and if an error does the next step
Data = json.dumps(getQuotes(i.lower()),indent=1) #retrieves Data from google finance
regex = ('"LastTradePrice": "(.+?)",') #sets parse
pattern = re.compile(regex) #compiles parse
price = re.findall(pattern,Data) #retrieves parse
print(i)
print(price)
except: #sets Error coding
Error = (i + ' Failed to load on: ' + Date2)
print (Error)
It will display the quote as: ['(number)'].
I would like it to only display the number, which means removing the brackets and quotes.
Any help would be great.
Changing:
print(price)
into:
print(price[0])
prints this:
A
42.14
AA
10.13
AAB
0.110
Try to use type() function to know the datatype, in your case type(price)
it the data type is list use print(price[0])
you will get the output (number), for brecess you need to check google data and regex.
When i use * i receive the error
raise error, v # invalid expression
error: nothing to repeat
other wildcard characters such as ^ work fine.
the line of code:
df.columns = df.columns.str.replace('*agriculture', 'agri')
am using pandas and python
edit:
when I try using / to escape, the wildcard does not work as i intend
In[44]df = pd.DataFrame(columns=['agriculture', 'dfad agriculture df'])
In[45]df
Out[45]:
Empty DataFrame
Columns: [agriculture, dfad agriculture df]
Index: []
in[46]df.columns.str.replace('/*agriculture*','agri')
Out[46]: Index([u'agri', u'dfad agri df'], dtype='object')
I thought the wildcard should output Index([u'agri', u'agri'], dtype='object)
edit:
I am currently using hierarchical columns and would like to only replace agri for that specific level (level = 2).
original:
df.columns[0] = ('grand total', '2005', 'agriculture')
df.columns[1] = ('grand total', '2005', 'other')
desired:
df.columns[0] = ('grand total', '2005', 'agri')
df.columns[1] = ('grand total', '2005', 'other')
I'm looking at this link right now: Changing columns names in Pandas with hierarchical columns
and that author says it will get easier at 0.15.0 so I am hoping there are more recent updated solutions
You need to the asterisk * at the end in order to match the string 0 or more times, see the docs:
In [287]:
df = pd.DataFrame(columns=['agriculture'])
df
Out[287]:
Empty DataFrame
Columns: [agriculture]
Index: []
In [289]:
df.columns.str.replace('agriculture*', 'agri')
Out[289]:
Index(['agri'], dtype='object')
EDIT
Based on your new and actual requirements, you can use str.contains to find matches and then use this to build a dict to map the old against new names and then call rename:
In [307]:
matching_cols = df.columns[df.columns.str.contains('agriculture')]
df.rename(columns = dict(zip(matching_cols, ['agri'] * len(matching_cols))))
Out[307]:
Empty DataFrame
Columns: [agri, agri]
Index: []