I want to get the data from mariadb on a specific column to match the specific word in that column and give me all the rows that math the word. For example I want to match the word "was". I am trying too much but not
found any regular expression that find the exact word.
Help me...
When I write query I give me all rows in which the word "was" is alone or appended in some string. But I
want only those rows in word "was" is a complete word not appended with some string
As described in the MariaDB Documentation you have to use word boundaries
[[:<:]] and [[:>:]]
SELECT column1, column2, .. FROM yourtable
WHERE was_column REGEXP "[[:<:]]was[[:>:]]"
Related
I'd like to replace the content of a column in a data frame with only a specific word in that column.
The column always looks like this:
Place(fullName='Würzburg, Germany', name='Würzburg', type='city', country='Germany', countryCode='DE')
Place(fullName='Iphofen, Deutschland', name='Iphofen', type='city', country='Germany', countryCode='DE')
I'd like to extract the city name (in this case Würzburg or Iphofen) into a new column, or replace the entire row with the name of the town. There are many different towns so having a gsub-command for every city name will be tough.
Is there a way to maybe just use a gsub and tell Rstudio to replace whatever it finds inside the first two ' '?
Might it be possible to tell it, "give me the word after "name=' until the next '?
I'm very new to using R so I'm kind of out of ideas.
Thanks a lot for any help!
I know of the gsub command, but I don't think it will be the most appropriate in this case.
Yes, with a regular expression you can do exactly that:
string <- "Place(fullName='Würzburg, Germany', name='Würzburg', type='city', country='Germany', countryCode='DE')"
city <- gsub(".*name='(.*?)'.*", "\\1", string)
The regular expression says "match any characters followed by name=', then capture any characters until the next ' and then match any additional characters". Then you replace all of that with just the captured characters ("\\1").
The parentheses mean "capture this part", and the value becomes "\\1". (You can do multiple captures, with subsequent captures being \\2, \\3, etc.
Note the question mark in (.*?). This means "match as little as possible while still satisfying the rest of the regex". If you don't include the question mark, the regular expression will match "greedily" and you will capture the entire rest of the line instead of just the city since that would also satisfy the regular expression.
More about regular expression (specific to R) can be found here
A particular column in my table is marked as a String type, however, more than likely, it will have an integer value mixed in with the string.
Here are some example rows:
"hello1"
"keys"
"Thom27"
"3for5"
I would like to be able to select the rows that have numbers in the string. Thank you.
You can use GLOB for that:
SELECT *
FROM tblNames
WHERE Name GLOB '*[0-9]*'
GLOB is the best answer (as mentioned above). GLOB supports REGEX like matching, which is more efficient than individual matches with LIKE.
I do a full text search using LIKE clause and the text can contain a '%'.
What is a good way to search for a % sign in an sqlite database?
I did try
SELECT * FROM table WHERE text_string LIKE '%[%]%'
but that doesn't work in sqlite.
From the SQLite documentation
If the optional ESCAPE clause is present, then the expression following the ESCAPE keyword must evaluate to a string consisting of a single character. This character may be used in the LIKE pattern to include literal percent or underscore characters. The escape character followed by a percent symbol (%), underscore (_), or a second instance of the escape character itself matches a literal percent symbol, underscore, or a single escape character, respectively.
We can achieve same thing with the below query
SELECT * FROM table WHERE instr(text_string, ?)>0
Here :
? => your search word
Example :
You can give text directly like
SELECT * FROM table WHERE instr(text_string, '%')>0
SELECT * FROM table WHERE instr(text_string, '98.9%')>0 etc.
Hope this helps better.
I need to find any special character. If I put it in the middle of a word, SQLite FTS match can ignore it as if it does not exist, e.g.:
Text Body: book's
If my match string is 'books' I need to get result of "book's"..
No problem using porter or simple tokenizer.
I tried many characters for that like: book!s, book?s, book|s, book,s, book:s…, but when searching by match for 'books' no results of these returned.
I don't understand, why?
I am using: Contentless FTS4 Tables, and External Content FTS4 Tables, my text body has many characters in each word, should be changed to ignore it when searching..
I cannot change match query because I do not know where the special character in the word is. Also, I need to leave the original word length equal to the length of FTS Index word to use match info or snippet(); as such, I cannot remove these characters from text body.
The default tokenizers do not ignore punctuation characters but treat them as word separators.
So the text body or match string book's will end up as two words, book and s.
These will never match a single work like books.
To ignore characters like ', you have to install your own custom tokenizer.
I would like to know if it is possible to use [] in SQLite query as we used to in Access and other DB.
e.g. SELECT * FROM mytable WHERE fwords like '%b[e,i,a]d%'
this will retrieve all rows have fwords containing bad, bed, bid
Thanks a lot
From http://www.sqlite.org/lang_expr.html:
The LIKE operator does a pattern matching comparison. The operand to the right of the LIKE operator contains the pattern and the left hand operand contains the string to match against the pattern. A percent symbol ("%") in the LIKE pattern matches any sequence of zero or more characters in the string. An underscore ("_") in the LIKE pattern matches any single character in the string. Any other character matches itself or its lower/upper case equivalent (i.e. case-insensitive matching).
Does that help?
You can have a look at the regex section here.