need to see if a taxid field contains letters or any special characters:. *,!,?,#,#,$,&,+,(,),/
How can I write this SQL?
Oracle has regexp_like:
select * from tablename where regexp_like(columnname, '[*!?##$&+()/]');
Here is the best way to display all the taxid's that are NOT entirely composed of digits:
select taxid
from your_table
where translate(taxid, '.0123456789', '.') is not null
TRANSLATE will "translate" (replace) each period in the input with a period in the output. Since the other characters in the second argument do not have a corresponding character in the third argument, they will simply be deleted. If the result of this operation is not null, then the taxid contains at least one character that is not a digit. If taxid is all digits, the result of the operation is null. Note that the period character used here is needed, due to an oddity in Oracle's definition of TRANSLATE: if any of its arguments is null, then so is its return value. That doesn't make a lot of sense, but we must work with the functions as Oracle defined them.
Related
I am using SQL Developer over a backend Oracle DB. I have a record where the buyer name is Pete Hansen.
Why I try
select * from data1 where buyer = 'Pete Hansen';
I get the result, no problem. However, when I use the following, I do not get any results:
select * from data1 where buyer like 'Pete Hansen';
Also, when I try the following, I do not get any results
select * from data1 where buyer like 'Pete Hans_n';
but the following works well:
select * from data1 where buyer like 'Pete Hans_n%';
Could you please help me understand? Thanks in advance.
Your buyer column seems to be defined as char; you can see the issue reproduced in this db<>fiddle, but not when the column is varchar2.
The documentation for character comparison explains the difference between blank-padded or nonpadded comparison semantics. When you compare them with =, because both the column and the string literal are `char, blank-padded semantics are used:
With blank-padded semantics, if the two values have different lengths, then Oracle first adds blanks to the end of the shorter one so their lengths are equal. ... If two values have no differing characters, then they are considered equal. This rule means that two values are equal if they differ only in the number of trailing blanks. Oracle uses blank-padded comparison semantics only when both values in the comparison are either expressions of data type CHAR, NCHAR, text literals, or values returned by the USER function.
When the column is `varchar2 then nonpadded semantics are used:
With nonpadded semantics, ... If two values of equal length have no differing characters, then the values are considered equal. Oracle uses nonpadded comparison semantics whenever one or both values in the comparison have the data type VARCHAR2 or NVARCHAR2.
LIKE works differently. Only your final pattern with % matches, because that is allowing for the trailing spaces in the char value, while the other two patterns do not. With the varchar2 version there aren't any trailing spaces to the other two patterns also match.
It's unusual to need or want to user char columns; varchar2 is more usual. Tom Kyte opined on this many years ago.
I suspect it may have to do with trailing white spaces, which like operator is not forgiving about but = is. To test that, try
select * from data1 where trim(buyer) like 'Pete Hansen';
I have a questions regarding below data.
You clearly can see each EMP_IDENTIFIER has connected with EMP_ID.
So I need to pull only identifier which is 10 characters that will insert another column.
How would I do that?
I did some traditional way, using INSTR, SUBSTR.
I just want to know is there any other way to do it but not using INSTR, SUBSTR.
EMP_ID(VARCHAR2)EMP_IDENTIFIER(VARCHAR2)
62049 62049-2162400111
6394 6394-1368000222
64473 64473-1814702333
61598 61598-0876000444
57452 57452-0336503555
5842 5842-0000070666
75778 75778-0955501777
76021 76021-0546004888
76274 76274-0000454999
73910 73910-0574500122
I am using Oracle 11g.
If you want the second part of the identifier and it is always 10 characters:
select t.*, substr(emp_identifier, -10) as secondpart
from t;
Here is one way:
REGEXP_SUBSTR (EMP_IDENTIFIER, '-(.{10})',1,1,null,1)
That will give the 1st 10 character string that follows a dash ("-") in your string. Thanks to mathguy for the improvement.
Beyond that, you'll have to provide more details on the exact logic for picking out the identifier you want.
Since apparently this is for learning purposes... let's say the assignment was more complicated. Let's say you had a longer input string, and it had several groups separated by -, and the groups could include letters and digits. You know there are at least two groups that are "digits only" and you need to grab the second such "purely numeric" group. Then something like this will work (and there will not be an instr/substr solution):
select regexp_substr(input_str, '(-|^)(\d+)(-|$)', 1, 2, null, 2) from ....
This searches the input string for one or more digits ( \d means any digit, + means one or more occurrences) between a - or the beginning of the string (^ means beginning of the string; (a|b) means match a OR b) and a - or the end of the string ($ means end of the string). It starts searching at the first character (the second argument of the function is 1); it looks for the second occurrence (the argument 2); it doesn't do any special matching such as ignore case (the argument "null" to the function), and when the match is found, return the fragment of the match pattern included in the second set of parentheses (the last argument, 2, to the regexp function). The second fragment is the \d+ - the sequence of digits, without the leading and/or trailing dash -.
This solution will work in your example too, it's just overkill. It will find the right "digits-only" group in something like AS23302-ATX-20032-33900293-CWV20-3499-RA; it will return the second numeric group, 33900293.
I have a column error_desc with values like:
Failure occurred in (Class::Method) xxxxCalcModule::endCustomer. Fan id 111232 is not Effective or not present in BL9_XXXXX for date 20160XXX.
What SQL query can I use to display only the number 111232 from that column? The number is placed at 66th position in VARCHAR column and ends 71st.
SELECT substr(ERROR_DESC,66,6) as ABC FROM bl1_cycle_errors where error_desc like '%FAN%'
This solution uses regular expressions.
The challenge I faced was on pulling out alphanumerics. We have to retain only numbers and filter out string,alphanumerics or punctuations in this case, to detect the standalone number.
Pure strings and words not containing numbers can be easily filtered out using
[^[:digit:]]
Possible combinations of alphanumerics are :
1.Begins with a character, contains numbers, may end with characters or punctuations :
[a-zA-Z]+[0-9]+[[:punct:]]*[a-zA-Z]*[[:punct:]]*
2.Begins with numbers and then contains alphabets,may contain punctuations :
[0-9]+[[:punct:]]*[a-zA-Z]+[[:punct:]]*
Begins with numbers then contains punctuations,may contain alphabets :
-- [0-9]+[a-zA-Z][[:punct:]]+[a-zA-Z] --Not able to highlight as code, refer solution's last regex combination
Combining these regular expressions using | operator we get:
select trim(REGEXP_REPLACE(error_desc,'[^[:digit:]]|[a-zA-Z]+[0-9]+[[:punct:]]*[a-zA-Z]*[[:punct:]]*|[0-9]+[[:punct:]]*[a-zA-Z]+[[:punct:]]*|[0-9]+[a-zA-Z]*[[:punct:]]+[a-zA-Z]*',' '))
from error_table;
Will work in most cases.
Column xy of type 'nvarchar2(40)' in table ABC.
Column consists mainly of numerical Strings
how can I make a
select to_number(trim(xy)) from ABC
query, that ignores non-numerical strings?
In general in relational databases, the order of evaluation is not defined, so it is possible that the select functions are called before the where clause filters the data. I know this is the case in SQL Server. Here is a post that suggests that the same can happen in Oracle.
The case statement, however, does cascade, so it is evaluated in order. For that reason, I prefer:
select (case when NOT regexp_like(xy,'[^[:digit:]]') then to_number(xy)
end)
from ABC;
This will return NULL for values that are not numbers.
You could use regexp_like to find out if it is a number (with/without plus/minus sign, decimal separator followed by at least one digit, thousand separators in the correct places if any) and use it like this:
SELECT TO_NUMBER( CASE WHEN regexp_like(xy,'.....') THEN xy ELSE NULL END )
FROM ABC;
However, as the built-in function TO_NUMBER is not able to deal with all numbers (it fails at least when a number contains thousand separators), I would suggest to write a PL/SQL function TO_NUMBER_OR_DEFAULT(numberstring, defaultnumber) to do what you want.
EDIT: You may want to read my answer on using regexp_like to determine if a string contains a number here: https://stackoverflow.com/a/21235443/2270762.
You can add WHERE
SELECT TO_NUMBER(TRIM(xy)) FROM ABC WHERE REGEXP_INSTR(email, '[A-Za-z]') = 0
The WHERE is ignoring columns with letters. See the documentation
I have a table with a field that denotes whether the data in that row is valid or not. This field contains a string of undetermined length. I need a query that will only pull out rows where all the characters in this field are N. Some possible examples of this field.
NNNNNNNNNNNNNNNNNNN
NNNNNNNNNNNNNNNNNNNNNNN
NNNNNEEEENNNNNNNNNNNN
NNNNNOOOOOEEEENNNNNNNNNNNN
Any suggestions on a postcard please.
Many thanks
This should do the trick:
SELECT Field
FROM YourTable
WHERE Field NOT LIKE '%[^N]%' AND Field <> ''
What it's doing is a wildcard search, broken down:
The LIKE will find records where the field contains characters other than N in the field. So, we apply a NOT to that as we're only interested in records that do not contain characters other than N. Plus a condition to filter out blank values.
SELECT *
FROM mytable
WHERE field NOT LIKE '%[^N]%'
I don't know which SQL dialect you are using. For example Oracle has several functions you may use. With oracle you could use condition like :
WHERE LTRIM(field, 'N') = ''
The idea is to trim out all N's and see if the result is empty string. If you don't have LTRIM, check if you have some kind of TRANSLATE or REPLACE function to do the same thing.
Another way to do it could be to pick length of your field and then construct comparator value by padding empty string with N. Perhaps something like:
WHERE field = RPAD('', field, 'N)
Oracle pads that empty string with N's and picks number of pad characters from length of the second argument. Perhaps this works too:
WHERE field = RPAD('', LENGTH(field), 'N)
I haven't tested those, but hopefully that give you some ideas how to solve your problem. I guess that many of these solutions have bad performance if you have lot of rows and you don't have other WHERE conditions to select proper index.