I have a TEXT column where each text is formatted as such:
/customers/{customer_id}/views/{id1}~{id2}/
I am trying to fetch the id2 only.
My idea is how to split the string by the / character first, where I will have:
customers, {customer_id}, views, {id1}~{id2}.
And then, get the last position:
{id1}~{id2}. And then split it again by the ~ character, and finally get the last position.
The issue is that I am new to SQL and I have no idea if this is even possible. How can I do that and end up with only one column?
SELECT
split_part(thetext, '/', 4) as temp
// how do I proceed from here?
FROM mytable
EDIT:
Some examples:
/customers/1231341/views/1312391293~3432491/
/customers/2213441/views/424131~231321341/
The IDs are of different lengths as well.
Use regexp_replace() to capture the part you want while matching the whole input, and replacing (the whole input) with the capture:
select regexp_replace(thetext, '.*~(.*)/', '\1') as temp
from mytable
See live demo.
Related
I need some help with the next. I have a field text in SQL, this record a list of times sepparates with '|'. For example
'14613|15474|3832|148|5236|5348|1055|524' Each value is a time in milliseconds. This field could any length, for example is perfect correct '3215|2654' or '4565' (only 1 value). I need get this field and replace all number with -1000 value.
So '14613|15474|3832|148|5236|5348|1055|524' will be '-1000|-1000|-1000|-1000|-1000|-1000|-1000|-1000'
Or '3215|2654' => '-1000|-1000' Or '4565' => '-1000'.
I try use regexp_replace(times_field,'[[:digit:]]','-1000','g') but it replace each digit, not the complete number, so in this example:
'3215|2654' than must be '-1000|-1000', i get:
'-1000-1000-1000-1000|-1000-1000-1000-1000', I try with other combinations and more options of regexp but i'm done.
Please need your help, thanks!!!.
We can try using REGEXP_REPLACE here:
UPDATE yourTable
SET times_field = REGEXP_REPLACE(times_field, '\y[0-9]+\y', '-1000', 'g');
If instead you don't really want to alter your data but rather just view your data this way, then use a select:
SELECT
times_field,
REGEXP_REPLACE(times_field, '\y[0-9]+\y', '-1000', 'g') AS times_field_replace
FROM yourTable;
Note that in either case we pass g as the fourtb parameter to REGEXP_REPLACE to do a global replacement of all pipe separated numbers.
[[:digit:]] - matches a digit [0-9]
+ Quantifier - matches between one and unlimited times, as many times as possible
your regexp must look like
regexp_replace(times_field,'[[:digit:]]+','-1000','g')
I need to return the fields that have more than one . in a specific column.
Now I have this query:
select *
from table
where column ~ '\.{2,}?';
But for some reason it returns nothing. If I use something like 'A{2,}?' it works. Apparently the problem is the dot.
It returns null since the dots are not next two each other. You have to consider the occurrences of the characters in the order of your regex meta characters. You could try this instead:
select *
from table
where column ~ '\.\d{3}\.';
Or instead of just focusing on the dot characters start parsing the string as a whole and consider the numbers as well:
where column ~ '^\d{3}\.\d{3}\.';
Why not just use like?
where column like '%.%.%'
What Oracle SQL query could return the second, third and fourth positions of characters contained within rows of a specific column using the REGEXP_SUBSTR method instead of using SUBSTR method like my example provided below?
SELECT SUBSTR(city,2,3) AS "2nd, 3rd, 4th"
FROM student.zipcode;`
One way that works for me (with test data) is:
SELECT REGEXP_SUBSTR(city, '\S{3}', 2) AS partial FROM student.zipcode;
Note that this is set to find three non-whitespace characters beginning at the second position of the string.
You could also use:
SELECT REGEXP_SUBSTR(city, '.{3}', 2) AS partial FROM student.zipcode;
which will instead match any three characters in the 2nd to 4th position.
However, I'm not sure what advantage this has over simply:
SELECT SUBSTR(city,2,3) AS partial FROM student.zipcode;
The REGEXP_INSTR function is not what you want, as it returns an index (position number) for the search item in the searched string. You can read about it here: http://www.techonthenet.com/oracle/functions/regexp_instr.php
I know I'm close to figuring this out but need a little help. What I'm trying to do is all grab a column from a particular table, but chop off the first 4 characters. For example if in a column the value is "KPIT08L", the result I was is 08L. Here is what I have so far but not getting the desired results.
SELECT LEFT(FIELD_NAME, 4)
FROM TABLE_NAME
First up, left will give you the leftmost characters. If you want the characters starting at a specific location, you need to look into mid:
select mid (field_name,5) ...
Secondly, if you value performance,portability and scalability at all, this sort of "sub-column" manipulation should generally be avoided. It's usually far easier (and faster) to patch columns together than to split them apart.
In other words, keep the first four characters in their own column and the rest in a separate column, and do your selects on the relevant one. If you're using anything less than a full column, then it's technically not one attribute of the row.
Try with
SELECT MID(FIELD_NAME, 5) FROM TABLE_NAME
Mid is very powerfull, it let you select the starting point and all the remainder, or,
if specified, the length desidered as in
SELECT MID(FIELD_NAME, 5, 2) FROM TABLE_NAME ' gives 08 in your example text
SELECT RIGHT(FIELD_NAME,LEN(FIELD_NAME)-4)
FROM TABLE_NAME;
If it is for a generic string then the above one will work...
Don't have Access at my current location, but please try this.
SELECT RIGHT(FIELD_NAME, LEN(FIELD_NAME)-4)
FROM TABLE_NAME
The LEFT(FIELD_NAME, 4) will return the first 4 caracters of FIELD_NAME.
What you need to do is :
SELECT MID(FIELD_NAME, 5)
FROM TABLE_NAME
If you have a FIELD_NAME of 10 caracters, the function will return the 6 last caracters (chopping the first 4)!
If I have a query to return all matching entries in a DB that have "news" in the searchable column (i.e. SELECT * FROM table WHERE column LIKE %news%), and one particular row has an entry starting with "In recent World news, Somalia was invaded by ...", can I return a specific "chunk" of an SQL entry? Kind of like a teaser, if you will.
select substring(column,
CHARINDEX ('news',lower(column))-10,
20)
FROM table
WHERE column LIKE %news%
basically substring the column starting 10 characters before where the word 'news' is and continuing for 20.
Edit: You'll need to make sure that 'news' isn't in the first 10 characters and adjust the start position accordingly.
You can use substring function in a SELECT part. Something like:
SELECT SUBSTRING(column, 1,20) FROM table WHERE column LIKE %news%
This will return the first 20 characters from column column
I had the same problem, I ended up loading the whole field into C#, then re-searched the text for the search string, then selected x characters either side.
This will work fine for LIKE, but not full text queries which use FORMS OF INFLECTION because that may match "women" when you search for "woman".
If you are using MSSQL you can perform all kinds VB-like of substring functions as part of your query.