I'm very new to access. I have a data in my column that looks similar to this:
JONES/KEN
SMITH/TAMMY
MILLER FRED
PICARD.JOHN
Am I able to grab the letters before the first non-alphanumeric?
So my result would be:
JONES
SMITH
MILER
PICARD
How about a derived table:
SELECT Left([Surname],InStr([Surname],[NonAlpha])-1) AS LeftName,
MainTable.Surname
FROM MainTable,
(SELECT " " As NonAlpha From Table1
UNION
SELECT "." As NonAlpha From Table1
UNION
SELECT "," As NonAlpha From Table1
UNION
SELECT "/" As NonAlpha From Table1) AS n
WHERE (((MainTable.Surname) Like "*" & [nonalpha] & "*"));
Table1 is a scratch table, it does contain records but the query will only
return the four assigned rows (,./ )
Maintable is the table with a field Surname, which is the field to be split.
Unfortunately, I don't know of a "Word" function that's available in some languages. I would do it with brute force checking using Instr and then Mid to extract the code. The construct would be very convoluted to get every kind of character.
I've used the iif function and nested it - this the basic format here:
iif (instr (fieldname,"{the character}") > 0,
mid(fieldname,1, instr(fieldname,"{the character}")-1,
fieldname{or go further into ifs})
Using your sample data with Client Name as the field and 3 conditions - space, / and period. IT does work, but it's ugly - you will have to scroll pretty far to the right to get everything:
ShortName: IIf(InStr(1,[client_name]," ")>0,
mid(client_name,1,InStr(1,[client_name]," ")-1),
IIf(InStr(1,[client_name],"/")>0,
mid(client_name,1,InStr(1,[client_name],"/")-1),
IIf(InStr(1,[client_name],".")>0,
mid(client_name,1,InStr(1,client_name],".")-1),
Client_Name)))
Put this in a query based on your table.
Related
I am trying to pull a product code from a long set of string formatted like a URL address. The pattern is always 3 letters followed by 3 or 4 numbers (ex. ???### or ???####). I have tried using REGEXP and LIKE syntax, but my results are off for both/I am not sure which operators to use.
The first select statement is close to trimming the URL to show just the code, but oftentimes will show a random string of numbers it may find in the URL string.
The second select statement is more rudimentary, but I am unsure which operators to use.
Which would be the quickest solution?
SELECT columnName, SUBSTR(columnName, LOCATE(columnName REGEXP "[^=\-][a-zA-Z]{3}[\d]{3,4}", columnName), LENGTH(columnName) - LOCATE(columnName REGEXP "[^=\-][a-zA-Z]{3}[\d]{3,4}", REVERSE(columnName))) AS extractedData FROM tableName
SELECT columnName FROM tableName WHERE columnName LIKE '%___###%' OR columnName LIKE '%___####%'
-- Will take a substring of this result as well
Example Data:
randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz123&hello_world=us&etc_etc
In this case, the desired string is "xyz123" and the location of said pattern is variable based on each entry.
EDIT
SELECT column, LOCATE(column REGEXP "([a-zA-Z]{3}[0-9]{3,4}$)", column), SUBSTR(column, LOCATE(column REGEXP "([a-zA-Z]{3}[0-9]{3,4}$)", column), LENGTH(column) - LOCATE(column REGEXP "^.*[a-zA-Z]{3}[0-9]{3,4}", REVERSE(column))) AS extractData From mainTable
This expression is still not grabbing the right data, but I feel like it may get me closer.
I suggest using
REGEXP_SUBSTR(column, '(?<=[&?]random_code=[^&#]{0,256}-)[a-zA-Z]{3}[0-9]{3,4}(?![^&#])')
Details:
(?<=[&?]random_code=[^&#]{0,256}-) - immediately on the left, there must be & or &, random_code=, and then zero to 256 chars other than & and # followed with a - char
[a-zA-Z]{3} - three ASCII letters
[0-9]{3,4} - three to four ASCII digits
(?![^&#]) - that are followed either with &, # or end of string.
See the online demo:
WITH cte AS ( SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz123&hello_world=us&etc_etc' val
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz4567&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz89&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz00000&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-aaaaa11111&hello_world=us&etc_etc')
SELECT REGEXP_SUBSTR(val,'(?<=[&?]random_code=[^&#]{0,256}-)[a-zA-Z]{3}[0-9]{3,4}(?![^&#])') output
FROM cte
Output:
I'd make use of capture groups:
(?<=[=\-\\])([a-zA-Z]{3}[\d]{3,4})(?=[&])
I assume with [^=\-] you wanted to capture string with "-","\" or "=" in front but not include those chars in the result. To do that use "positive lookbehind" (?<=.
I also added a lookahead (?= for "&".
If you'd like to fidget more with regex I recommend RegExr
I have a table with part numbers. sometimes people put xxx at the end of the part number when they want to include all of the different possible endings (a lot like the wildcard ### in Access). How do I write a query that will give me all of the part numbers and replace anything ending in xxx with ###?
So if my table has:
1234
1235-xxx
1236
How do I write a query that will give me:
1234
1235-###
1236
In Access, you can use the replace function to change some text into some other text: http://office.microsoft.com/en-ca/access-help/replace-function-HA001228898.aspx
Together with a select statement and the iif function to choose the exact rows that need the replacement, it's a fairly simple operation:
select
iif(
part_num like '*-xxx'
, replace(part_num, '-xxx', '-###')
, part_num
)
from my_table
Im optimizing my SQL statement to make it faster.
I have a comma seperated list with zipcodes like
1111, 1112,1115,1112 etc etc
Now in my query I want to match if the iput matches 1 of those zipcodes. If so.. then it will return a ID of the object that has all those zipcodes.
But what is the best way to do this now im doing
AND ( loc.loc_zip LIKE '%".$_REQUEST['zip']."%'
Validation of the input will be added ofcourse.. but this is just for testing. But I have tested this and it seems a bit slow.
Is this the best way to do this ?
you should use 'in'
select * from Users where userid in (1,2,3,4,45,6,656)
Edit:
if the ZipCodes are Chars, you can only use in if you separate them by ''
select * from loc where loc.loc_zip in ('1111','1112','1115','1112')
if .$_REQUEST['zip']. has ' ' then->
select * from loc where loc.loc_zip in (.$_REQUEST['zip'].)
if the ZipCodes are int you can use the first statement
Bare in mid you must intersect your list with '' or it wont work
I have a field of comma-separated lists in MySQL:
id field1
1 aa,bb,cc
2 aa
I would like to count the total number of elements, with overlap. In this case that would be 4: aa appears twice and so should be double-counted.
It would suffice to count the number of commas in each field and add 1, since my lists do not have quotes or escaping.
Let's try that:
select count(field1) + sum(CHAR_LENGTH(field1) - CHAR_LENGTH(REPLACE(field1,',','')))
from Table1;
Why do you use comma separated string?
Databases aren't made for that.
You have 2 possibilities :
- make a loop in a stored procedure,
- count it into the code which call this MySQL query...
I have this table under user_name='high'
function_description :
akram is in a date
test
akram is studying
test4
kheith is male
test3
I want a query that returns results of field that have at least an 'akram'
SELECT *
FROM functions
WHERE 'isEnabled'=1
AND 'isPrivate'=1
AND user_name='high'
AND function_description LIKE '%akram%'
and this returns absolutely nothing!
Why?
You are listing the column names as if they are strings. This is why it returns nothing.
Try this:
SELECT *
FROM functions
WHERE user_name='high'
AND function_description LIKE '%akram%'
edit: After trying to re-read your question... are isEnabled and isPrivate columns in this table?
edit2: updated.. remove those unknown columns.
You are comparing strings 'isEnabled' with integer 1, which likely leads to the integer being converted to a string, and the comparison then fails. (The alternative is that the string is converted to an integer 0 and the comparison still fails.)
In MySQL, you use back-quotes, not single quotes, to quote column and table names:
SELECT *
FROM `functions`
WHERE `isEnabled` = 1
AND `isPrivate` = 1
AND `user_name` = 'high'
AND `function_description` LIKE '%akram%'
In standard SQL, you use double quotes to create a 'delimited identifier'; in Microsoft SQL Server, you use square brackets around the names.
Please show the schema more carefully (column names, sample values, types if need be) next time.