I currently have the following code
select *
FROM list
WHERE name LIKE '___'
ORDER BY name;
I am trying to get it so that it only shows names with three or more words.
It only displays names with three characters
I cannot seem to work out the correct syntax for this.
Any help is appreciated
Thankyou
If you assume that there are no double spaces, you can do:
WHERE name like '% % %'
To get names with three or more words.
If you can have double spaces (or other punctuation), then you are likely to want a regular expression. Something like:
WHERE name REGEXP '^[^ ]+[ ]+[^ ]+.*$'
you can count number of words and then select those who are equal or greater then 3 words.
SELECT * FROM list
HAVING LENGTH(name) - LENGTH(REPLACE(name, ' ', ''))+1 >= 3
ORDER BY name
DEMO HERE
*Even if you have multi spaces it will not affect here check this
I do believe what you are after is
SELECT *
FROM list
WHERE name LIKE '% % %'
ORDER BY name;
Related
I have a field where I save strings, like:
"one two-tree"
"one-two-tree"
"one-two tree"
"one two tree"
When I do a SELECT, I want to retrieve strings that have either "-" or " " (space). Example:
When I do:
Select name from table where name="one two tree"
I want it to bring also results where there is either space or -, in this case returning all string exemplified above.
Is there a wildcard for this?
As far as standard SQL, you must use "or", or "like". depending on what exactly you want. EXAMPLE: Select name from table where name like "one?two?tree".
However, mySQL supports a REGEX extension that will give you what you want:
http://dev.mysql.com/doc/refman/5.7/en/pattern-matching.html
One option is to use replace:
select name
from yourtable
where replace(name, '-', ' ') = 'one two tree'
There is, but it is slow: you can use REGEXP:
SELECT name
FROM table
WHERE name REGEXP 'one[- ]two[- ]tree'
or you can use replacements:
SELECT name
FROM table
WHERE REPLACE(name, '-', ' ') = 'one two three'
but your best bet is to make an additional column where you will have a normalised name (with dashes always replaced with spaces, for example) so you can take advantage of indices.
You can use the LIKE condition with '%' (wildcard operator) for this.
e.g.
SELECT name from table_name WHERE name LIKE '%-%' OR name LIKE '% %'
-- will return all names that have `-` or ` `.
I have a column A contains of ID
ex: 1113 992 981 778 718
and my query
select product
from something
where userInvite like '% 1113 %'
It will be fine, but if it on only contains one string
ex: 1113
and query returns there is no results, but if I remove space, it will be fine
E.g.:
select product
from something
where userInvite like '%1113%'
How can I fix it, to search ID in case the column A is not contains a space and contains space.
The best way to do it, would be to use comma as separator, then you could just simply use mysql built in function: find_in_set.
In this case, we can do a little trick:
select product
from something
where find_in_set('123', replace(product, ' ', ',')) > 0
See demo here: http://sqlfiddle.com/#!9/bfcbc/2/0
select product from something where userInvite like '%1113%' should work
Heres a demo
How to return rows where a column has 2 words (that is, strings separated by a space) in it?
It must be purely using SQL.
SELECT * FROM table WHERE name (has 2 strings in it);
I dont know the names when querying. Its a big dataset. I only have to check if the name contains a spacebar basically (from a comment).
If you want to distinguish names that have two parts from one-part and three-plus-part names, you can use regular expression:
SELECT * FROM my_table WHERE name REGEXP '^[^ ]+[ ]+[^ ]+$'
This regular expression matches when the entire string consists of two non-empty parts containing no spaces, with one or more space separating them.
This perfectly works for me
You can use 'AND' condition and Like Operator with wildcards (%).
SELECT * FROM table_name WHERE name LIKE '%Word1%' AND name LIKE '%Word2%'
How about simply:
...
WHERE [name] LIKE '%Word1%'
AND [name] LIKE '%Word2%'
SELECT * FROM table WHERE concat(' ',name,' ') like '% str1 %'
AND concat(' ',name,' ') like '% str2 %'
The extra blanks are there to separate words.
You can use the following technique
mysql> select length('first name')-length(replace('first name',' ','')) as diff;
+------+
| diff |
+------+
| 1 |
+------+
i.e. get the difference of actual name and the name after replacing space, and if its 1 then you have the value as firstname lastname
So the query may look like
select * from table
where
length(col_name)-length(replace(col_name,' ','')) = 1
Use % between words.
Example:
SELECT * FROM Table WHERE Col LIKE '%word1%word2%'
SELECT * FROM table_name WHERE name LIKE '% %'
schema:
customers(name, mailid, city)
What to find:
Find all customer names consisting of three or more words (for example King George V).
What I tried:
select name from customers
where name like
'%[A-Za-z0-9][A-Za-z0-9]% %[A-Za-z0-9][A-Za-z0-9]% %[A-Za-z0-9][A-Za-z0-9]%'
what is surprising me:
If I am trying for two words (removing the last %[A-Za-z0-9]% from my query), its working fine but its not working for three words :(
MySQL Solution:
If a name has words separated by space character, then,
Try the following:
select name from customers
where ( length( name )
-
length( replace( name, ' ', '' ) ) + 1
) >= 3
In t-sql, the like clause can contain multiple wild card checks - eg:
SELECT * FROM Customers WHERE Name like '% % %'
will return those Names where two spaces are contained.
If you are consistent with the spacing between names, you could use this logic
SELECT LENGTH(name)-LENGTH(REPLACE(name,' ',''))
FROM customers
Or you can try this too if your sql dont have length function ( which is the situation I have when I'm doing an online exercies...) Inspired by answers above
SELECT name FROM customers WHERE (replace(name,' ','*')) LIKE '%*%*%'
I would like to sort a result from a MySQL query by the number of words in a specified column.
Something like this:
SELECT bar FROM foo ORDER BY WordCountFunction(bar)
Is it possible?
As far as I know, there is no word count function in MySQL, but you can count the number of spaces and add one if your data is formatted properly (space separator for words, no spaces at beginning/end of entry).
Here is the query listing longest words first:
SELECT bar FROM foo ORDER BY (LENGTH(bar) - LENGTH(REPLACE(bar, ' ', ''))+1) DESC
Using this method to count the number of words in a column, your query would look like this:
SELECT bar FROM foo ORDER BY (LENGTH(bar) - LENGTH(REPLACE(bar, ' ', ''))+1) DESC
Yes, sort of. It won't be 100% accurate though:
SELECT SUM(LENGTH(bar) - LENGTH(REPLACE(bar, ' ', ''))+1) as count
FROM table
ORDER BY count DESC
But this assumes the words are separated by a space ' ' and doesn't account for punctuation. You can always replace it with another char and it doesn't account for double spaces or other chars either.
For complete accuracy, you could always pull the result out and word-count in your language of choice - where accurate word-count function do exist!
Hope this helps.