How i can convert this scientific notation in synpase ? thanks
In the mapping section, you would have to explicitly map the column as decimal :
else it would consider it as string and copy it as E+11
Related
Below is the value of a string in a text column.
select col1 from tt_d_tab;
'A:10000000,B:50000000,C:1000000,D:10000000,E:10000000'
I'm trying to convert it into json of below format.
'{"A": 10000000,"B": 50000000,"C": 1000000,"D": 10000000,"E": 10000000}'
Can someone help on this?
If you know that neither the keys nor values will have : or , characters in them, you can write
select json_object(regexp_split_to_array(col1,'[:,]')) from tt_d_tab;
This splits the string on every colon and comma, then interprets the result as key/value pairs.
If the string manipulation gets any more complicated, SQL may not be the ideal tool for the job, but it's still doable, either by this method or by converting the string into the form you need directly and then casting it to json with ::json.
If your key is a single capital letter as in your example
select concat('{',regexp_replace('A:10000000,B:50000000,C:1000000,D:10000000,E:10000000','([A-Z])','"\1"','g'),'}')::json json_field;
A more general case with any number of letters caps or not
select concat('{',regexp_replace('Ac:10000000,BT:50000000,Cs:1000000,D:10000000,E:10000000','([a-zA-Z]+)','"\1"','g'),'}')::json json_field;
I have values like these in data input and I have not been able to convert those with parenthesis to a negative numeric. I am using TO_NUMBER(a.Total_Paid,'L999999D99')
Example ($123.45)
It should be -123.45
Does this work for you? (Note that it does not work for me if I include the L in the format string)
to_number(translate('($123.45)', '()', '<>'), '999999d99PR')
String Concatenation Instead of String Interpolation.
Why do we have to use string concatenation instead of string interpolation? I think in my own honest opinion. String interpolation is one of the magic of ruby on rails while string concatenation is just an ordinary one.
kindly please explain why is it use in this chapter and what is the advantage of over the other and vice versa.
Thanks
-Eric John Iglesia
String concatenation is faster (needs less processing power) and safer because you will not need to match parameters with format string and so. With interpolation, you may later remove a parameter (variable) without its matching placeholder in the format string. Concatenation accepts null values as well, this is not possible in a format string.
Suppose I have a value of type json, say y. One may obtain such a value through, for example, obj->'key', or any function that returns values of type json.
This value, when cast to text, includes quotation marks i.e. "y" instead of y. In cases where using json types is unavoidable, this poses a problem, especially when we wish to compare the value with literal strings e.g.
select foo(x)='bar';
The API Brainstorm page suggests a from_json function that will intelligently unwrap JSON strings, but I doubt that is available yet. In the meantime, how can one convert JSON strings to text without the quotation marks?
Text:
To extract a value as text, use #>>:
SELECT to_json('foo'::text) #>> '{}';
From: Postgres: How to convert a json string to text?
PostgreSQL doc page: https://www.postgresql.org/docs/11/functions-json.html
So it addresses your question specifically, but it doesn't work with any other types, like integer or float for example. The #> operator will not work for other types either.
Numbers:
Because JSON only has one numeric type, "number", and has no concept of int or float, there's no obvious way to cast a JSON type to a "correct" numeric type. It's best to know the schema of your JSON, extract the text and then cast to the correct type:
SELECT (('{"a":2.01}'::json)->'a'#>>'{}')::float
PostgreSQL does however have support for "arbitrary precision numbers" ("up to 131072 digits before the decimal point; up to 16383 digits after the decimal point") with its "numeric" type. JSON also supports 'e' notation for large numbers.
Try this to test them both out:
SELECT (('{"a":2e99999}'::json)->'a'#>>'{}')::numeric
The ->> operator unwraps quotation marks correctly. In order to take advantage of that operator, we wrap up our value inside an array, and then convert that to json.
CREATE OR REPLACE FUNCTION json2text(IN from_json JSON)
RETURNS TEXT AS $$
BEGIN
RETURN to_json(ARRAY[from_json])->>0;
END; $$
LANGUAGE plpgsql;
For completeness, we provide a CAST that makes use of the function above.
CREATE CAST (json AS text) WITH json2text(json) AS ASSIGNMENT;
As the title suggests I'm looking to detect where the numbers are in a string and then to just take the substring from the larger string. EG
If I have say zero89 or eight78, I would just like zero or nine returned. When using the strsplit function I have:
strsplit('zero89', but what do I put here?)
Interested in regexp that will provide you more options to explore with?
Extract numeric digits -
regexp('zero89','\d','match')
Extract anything other than digits -
regexp('zero89','\d+','Split')
strsplit('zero89', '\d', 'DelimiterType', 'RegularExpression')
Or just using regexp:
regexp('zero89','\D+','match')
I got the \D+ from here
Assuming you mean this strsplit?
strsplit('zero89', '8')