I want to use regexp_replace to replace all blank with '_'.
I use this statment:
select regexp_replace('"<div_class="CCL-temp-border"><div_class="input-group_moveDivEnd"_style="margin-bottom:_5px;_top:_auto;_left:_auto;_width:_100%;_position:_relative;_opacity:_1;_filter:_none;"_data-id="moveDivEnd_1545116285310">_; <span_class="input-group-addon_CCL-te (...)"', '\s', '_', 'g')
But the result is this:
"<div_class="CCL-temp-border"><div_class="input-group_moveDivEnd"_style="margin-bottom:_5px;_top:_auto;_left:_auto;_width:_100%;_position:_relative;_opacity:_1;_filter:_none;"_data-id="moveDivEnd_1545116285310">_;_______<span_class="input-group-addon_CCL-t (...)"
My statment is this:
select case when length(topiccontent)=0 THEN '_' else coalesce(regexp_replace(replace(replace(replace(topiccontent,chr(13), '_'),chr(10),'_'),' ','_'),'\s', '_', 'g'),'_') end as topiccontent
from ccl_topics
You can see the blank still exists, why?
I know why it can't be replaced.
There are some character restrictions when the data is pasted out from database.
The omitted part is converted to (...).
So (...) is not real characters, but ellipses.
For example, more than 600 characters exist in a column of a table, and then paste it,the result with ellipsis marks.
Related
I have a query like this which is not retrieving the values from DB table even if the required value exist there.
Here's the query, which return zero rows:
Select * from SitePanel_FieldValue WHere SiteFieldIdfk =111
And SiteFieldvalue like '%!##$%&*()-_=+{}|:"<>?[]\;'',./%'
Following is the value in the table:
'!##$%&*()-_=+{}|:"<>?[]\;'',./'
When I run the query without ";" it is returning the value.
Can any one help me in figuring this out?
Thanks
Ritu
You are using multiple characters which are reserved when using LIKE statement.
i.e. %, _, []
Use the escape character clause (where I have used backtick to treat special characters as regular) such as
Select * from SitePanel_FieldValue WHere SiteFieldIdfk =111
And SiteFieldvalue like '%!##$`%&*()-`_=+{}|:"<>?`[`]\;'',./%' escape '`'
The value in your table is:
!##$%&*()-_=+{};; :"<>?[]\;'',./
And the one in the like is:
(!##$%&*()-_=+{};;
Starting with ( it will never match, also you should scape the percent (%) in the middle of the string like this:
Select *
FROM SitePanel_FieldValue
WHERE SiteFieldIdfk =111
AND SiteFieldvalue like '%!##$\%&*()-_=+{};;%' ESCAPE '\'
The problem is your brackets ([]), it has nothing to do with semicolons. If we remove the brackets, the above works:
SELECT CASE WHEN '!##$%&*()-_=+{}|:"<>?\;'',./' LIKE '%!##$%&*()-_=+{}|:"<>?\;'',./%' THEN 1 END AS WithoutBrackets,
CASE WHEN '!##$%&*()-_=+{}|:"<>?[]\;'',./' LIKE '%!##$%&*()-_=+{}|:"<>?[]\;'',./%' THEN 1 END AS WithBrackets
Notice that WithoutBrackets returns 1, where as WithBrackets returns NULL.
Brackets in a LIKE are to denote a pattern. For example SomeExpress LIKE '[ABC]' would match the characters, A, B, and C. If you are going to include special characters, you need to ESCAPE them. You have both brackets, a percent sign (%) and an underscore (_) you need to escape. You don't need to escape the hyphen (-), as it doesn't appear in a pattern (for example [A-Z]). I choose to use a backtick as the ESCAPE character, as it doesn't appear in your string, and demonstrate with a CASE expression again:
SELECT CASE WHEN '!##$%&*()-_=+{}|:"<>?[]\;'',./' LIKE '%!##$`%&*()-`_=+{}|:"<>?`[`]\;'',./%' ESCAPE '`' THEN 1 END;
If you wanted to use a backslash (\ ), which many do, you would need to also escape the backslash in your string:
SELECT CASE WHEN '!##$\%&*()-_=+{}|:"<>?[]\;'',./' LIKE '%!##$%&*()-\_=+{}|:"<>?\[\]\\;'',./%' ESCAPE '\' THEN 1 END;
db<>fiddle
I think the issue is actually with the backslash. This is an escape character and so if you want it to be included, you have to put it in twice.
Select * from SitePanel_FieldValue WHere SiteFieldIdfk =111
And SiteFieldvalue like '%!##$%&*()-_=+{}|:"<>?[]\\;'',./%'
Using Oracle db,
Select name from name_table where name like 'abc%';
returns one row with value "abc, cd" but when I do a select query with a comma before % in my like query, it fails to return any value.
Select name from name_table where name like 'abc,%';
returns no row. How can I handle a comma before % in the like query?
Example:
Database has "Sam, Smith" in the name column when the like has "Sam%" it returns one row, when i do "Sam,%" it doesn't return any row
NOT AN ANSWER but posting it as one since I can't format in a comment.
Look at this and use DUMP() on your own machine... see if this helps.
SQL> select dump('Smith, Stan') from dual;
DUMP('SMITH,STAN')
-----------------------------------------------------
Typ=96 Len=11: 83,109,105,116,104,44,32,83,116,97,110
If you count, the string is 11 characters (including the comma and the space). The comma is character 44, and the space is character 32. If you look at YOUR string and you don't see 44 where the comma should be, you will know that's the problem. You could then let us know what you see there (just for that character, I understand posting "Leno, Jay" would be a violation of privacy).
Also, make sure you don't have any extra characters (perhaps non-printable ones!) right before the comma. Just compare the two strings you are using as inputs and see where the differences may be.
This one matches column_name like 'CharEndsHere%'
and
This one doesn't column_name like 'CharEndsHere'
I know that like operator will consider even the trailing spaces, so I just copied the exact column value (with trailing spaces) and pasted it.
Something like column_name like 'CharEndsHere ' yet it doesn't match -- why?.
I haven't used '=' operator since the columns type is ntext
Is there something I am missing here or shouldn't I use like operator in this way?
Edited : column_name like 'CharEndsHere__' (__ denoted the spaces) 'CharEndsHere ' is the exact value in that cell, using like in this way valid or no?
Edit :
This is the code I tried,
SELECT *
FROM [DBName].[dbo].[TableName]
WHERE [DBName].[dbo].[TableName].Address1 LIKE rtrim('4379 Susquehanna Trail S ')
I have also tried without using rtrim, yet the same result
Edit: According to Blindy's answer,
If a comparison in a query is to return all rows with the string LIKE 'abc' (abc
without a space), all rows that start with abc and have zero or more trailing
blanks are returned.
But in my case, I have queried, Like 'abc' and there is a cell containing 'abc '(with trailing spaces) which is not returned. That's my actual problem
This is a case of reading the documentation, it's very explicitly stated here: http://msdn.microsoft.com/en-us/library/ms179859.aspx
When you perform string comparisons by using LIKE, all characters in the pattern string are significant. This includes leading or trailing spaces. If a comparison in a query is to return all rows with a string LIKE 'abc ' (abc followed by a single space), a row in which the value of that column is abc (abc without a space) is not returned. However, trailing blanks, in the expression to which the pattern is matched, are ignored. If a comparison in a query is to return all rows with the string LIKE 'abc' (abc without a space), all rows that start with abc and have zero or more trailing blanks are returned.
Edit: According to your comments, you seem to be looking for a way to use like while ignoring trailing spaces. Use something like this: field like rtrim('abc '). It will still use indexes because rtrim() is a scalar operand and it's evaluated before the lookup phase.
I'm trying to select some rows from an Oracle database like so:
select * from water_level where bore_id in ('85570', '112205','6011','SP068253');
This used to work fine but a recent update has meant that bore_id in water_level has had a bunch of whitespace added to the end for each row. So instead of '6011' it is now '6011 '. The number of space characters added to the end varies from 5 to 11.
Is there a way to edit my query to capture the bore_id in my list, taking account that trialling whitespace should be ignored?
I tried:
select * from water_level where bore_id in ('85570%', '112205%','6011%','SP068253%');
which returns more rows than I want, and
select * from water_level where bore_id in ('85570\s*', '112205\s*','6011\s*', 'SP068253\s*');
which didn't return anything?
Thanks
JP
You should RTRIM the WHERE clause
select * from water_level where RTRIM(bore_id) in ('85570', '112205','6011');
To add to that, RTRIM has an overload which you can pass a second parameter of what to trim, so if the trailing characters weren't spaces, you could remove them. For example if the data looked like 85570xxx, you could use:
select * from water_level where RTRIM(bore_id, 'x') IN ('85570','112205', '6011');
You could use the replace function to remove the spaces
select * from water_level where replace(bore_id, ' ', '') in ('85570', '112205', '6011', 'SP068253');
Although, a better option would be to remove the spaces from the data if they are not supposed to be there or create a view.
I'm guessing bore_id is VARCHAR or VARCHAR2. If it were CHAR, Oracle would use (SQL-standard) blank-padded comparison semantics, which regards 'foo' and 'foo ' as equivalent.
So, another approach is to force comparison as CHARs:
SELECT *
FROM water_level
WHERE CAST(bore_id AS CHAR(16)) IN ('85570', '112205', '6011', 'SP068253');
I am using this query to replace one character in a cell
select replace(id,',','')id from table
But I want to replace two characters in a cell.
If the cell is having this data (1,3.1), and I want it to look like this (131).
How can I replace two different characters in one cell?
Use TRANSLATE instead of REPLACE(). It replaces each occurrence of a character in the first pattern with its matched character in the second. To remove characters, simply leave cut short the replacement string:
select translate(id, '1,.', '1') id from table
Note that the second string cannot be null. Hence the need to include 1 (or some other character) in both strings.
Find out more.
Obviously the more characters you need to convert/remove the more attractive TRANSLATE() becomes. The main use for REPLACE is changing patterns (such as words) rather than individual characters.
Can use
select replace(translate(id,',.',' '),' ','') from table;
or
select regexp_replace('1,3.1','[,.]','') from dual;
or
select replace(replace(id,',',''),'.','') from table;
Call the replace again.
select replace(replace(id,',',''), '.','') id from table
Do this:
select REPLACE(REPLACE(id,',',''),'.','')
Or use a regular expression:
select regexp_replace(id, '[.,]', '') id from table
Find out more