Let's say I have the following entries in my database:
Id
Name
12
John Doe
13
Mary anne
13
little joe
14
John doe
In my program I have a string variable that is always capitalized, for example:
myCapString = "JOHN DOE"
Is there a way to retrieve the rows in the table by using a WHERE on the name column with the values capitalized and then matching myCapString?
In this case the query would return two entries, one with id=12, and one with id=14
A solution is NOT to change the actual values in the table.
A general solution in Postgres would be to capitalize the Name column and then do a comparison against an all-caps string literal, e.g.
SELECT *
FROM yourTable
WHERE UPPER(Name) = 'JOHN DOE';
If you need to implement this is Knex, you will need to figure out how to uppercase a column. This might require using a raw query.
Related
I have a column that can have the following possible values -
ITO26218361281- JANE
SBC28791827135 VATS
SOT21092832917 JOHN DOE
TIM INQ12109283291
JANE DOE 12/15
I only want to extract the 14 characters
alphanumeric phrase from the strings that can look like above. If the record is like (5), I still want that record to exist to be able to call it out as an error. I don't need the exact text to be the same, I just need it to be flagged for error.
Result expected -
ITO26218361281
SBC28791827135
SOT21092832917
INQ12109283291
JANE DOE 12/15 (or flagged as error)
You can use a regular expression to match the pattern you need, 3 letters and 11 numbers.
Using this in WHERE clause, you can match all "valid" values:
SELECT *
FROM TableName
WHERE ColumnName like '%[A-Z][A-Z][A-Z][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]%'
I'm not a master of patterns nor regular expression, so used a "simple to understand" pattern here.
With this query, you extract the data to other table, or UPDATE a table column with the flag you want.
You can see this query working here in sqlfiddle.com
I'm writing a SQL query where I am taking the substring of 2 names (First name/last name) to create an initials column, the data is unstructured to a certain extent (Can't show for GDPR reasons) but where there is a company name it is just in the surname column.
I'm trying to use Regex to say when the already present initials column is 1 letter (I.e not an initial) and if it is not an initial run a command that I wrote which successfully works.
CAST(CASE
WHEN [DATA_TABLE].[INITIALS] = '\d' THEN (CONCAT(substring([DATA_TABLE].[FIRSTNAMES],1,1),substring([DATA_TABLE].[SURNAME],1,1)) AS char) AS INITIALS
ELSE [DATA_TABLE].[INITIALS]
end as char) as INITIALS,
An example of the data format:
First name last name initials
John smith JS
Electrical company E
Sam Craig SC
I want the names that are just in the surname (Company names) to just remain as they are with no change (I.e The \d regex). Ones which don't will become the substring of their first name as (1,1) and a substring of their last name to also be (1,1).
Name Correction
As the wedding season is on, John has been given the work of printing guest names on wedding cards. John has written code to print only those names that start with upper-case alphabets and reject those that start with lower-case alphabets or special characters.
Your job is to do the following:
1.Correct the rejected names (names which start with lower case or with a special character). You have to change the first alphabet of the rejected name to Upper case and in the case of special character there will be no change.
2.Output the newly corrected names in ascending order.
Table format
Table: person
Field Type
name varchar(20)
Sample
Sample person table
name
mohit
Kunal
manoj
Raj
tanya
#man
Sample output table
name
#man
Manoj
Mohit
Tanya
Solution Attempted: IN SQL SERVER 2014
select name
from person as per
where (left(per.name,1) like '%[^A-Z]%' or left(per.name,1) like '% %')
union
select Upper(left(per.name,1))+right(per.name,len(per.name)-1)
from person as per
where left(per.name,1)<>left(Upper(per.name),1)
collate Latin1_General_CS_AI
order by per.name
Sample Test Cases Passes,
Still getting wrong answer in some competitor exam.
Please suggest what test case i have not handled.
Since you are only interested in correcting lower case and reporting special characters in the first character position I would use ascii comparision rather than regex.
select name, ascii(left(name,1)),
case
when ascii(left(name,1)) between 97 and 122 then
concat(char(ascii(left(name,1)) - 32),substring(name,2,len(name) -1))
else name
end name
from t
where ascii(left(name,1)) <= 64 or
ascii(left(name,1)) >= 91
I have a table called Students.
This table has two fields (ID, Name)
i need to Select all the students whose last name have 5 characters.
For example if i have in this table two records.
Student 1: ID - 1
Name - Roman Jatt Pearce
Student 2: ID:2
Name: Matt Crazy
The query i wanted should only return Matt Crazy since his last name has 5 characters and roman pearce's doesnt.
someone told me to use charindex but i dont really know how to implement it
any suggestion?
Assuming the format of Name is always "First Middle Last", no names contain spaces, and there are no other things like generation listed (Jr., Sr., et al).
SELECT *
FROM Students
WHERE CHARINDEX(' ', REVERSE(Name)) = 6
How about
select * from Students where Name like '% _____'
with dash symbol coming five times
I have a column called NAME, I have 2000 rows in that column that are filled with people's full names, e.g. ANN SMITH. How do I do a query that will list all the people whose first name is ANN? There are about 20 different names whose first name is ANN but the surname is different.
I tried
and (NAME = 'ANN')
but it returned zero results.
I have to enter the FULL name and (NAME = 'ANN SMITH') ANN SMITH to even get a result .
I just want to list all the people with there first name as ANN
Try in your where clause:
Where Name like 'ANN %'
Should work mate.
ANN% will find all results where ANN is first then anything after.
%ANN% will find the 3 letters ANN in any part of that rows field.
Hope it helps
Also usually Name is separated into First names and second name columns.
this will save Having to use wild cards in your SQL and provide A bit more normalized data.
SELECT NAME
FROM NAMES
WHERE NAME LIKE 'ANN %'
This should wildcard select anything that begins with 'ANN' followed by a space.