I try get name of city's from string '{"travelzoo_hotel_name":"Graduate Minneapolis","travelzoo_hotel_id":"223","city":"Minneapolis","country":"USA","sales_manager":"Stephen Conti"}'
I try this regexp:
SELECT REGEXP_SUBSTR('{\"travelzoo_hotel_name\":\"Graduate Minneapolis\",\"travelzoo_hotel_id\":\"223\",\"city\":\"Minneapolis\",\"country\":\"USA\",\"sales_manager\":\"Stephen Conti\"}'
,'(?:.city...)([[:alnum:]]+)');
I have: '"city":"Minneapolis'
Me need only name of city:Minneapolis.
How to use groups in queries?
My example in regex101
Help me Please
I assume you are using MySQL 8.x that uses ICU regex expressions.
It looks like the string you want to process is JSON. You may use JSON_EXTRACT with JSON_UNQUOTE and a '$.city' as JSON path then:
JSON_UNQUOTE(JSON_EXTRACT('{"travelzoo_hotel_name":"Graduate Minneapolis","travelzoo_hotel_id":"223","city":"Minneapolis","country":"USA","sales_manager":"Stephen Conti"}', '$.city'))
will return Minneapolis.
In your regex, the non-capturing group pattern is still matched and appended to the match value. "Non-capturing" only means no separate memory buffer is alotted to the text captured with a grouping construct. So, you may fix it with '(?<="city":")[^"]+' pattern where (?<="city":") is a positive lookbehind that matches "city":" but does not put it into the match value. The only text you will have in the output is the one matched with [^"]+, 1+ chars other than ".
Related
I have a small mysql database with a column which has format of a field as following:
x_1_1,
x_1_2,
x_1_2,
x_2_1,
x_2_12,
x_3_1,
x_3_2,
x_3_11,
I want to extra the data where it matches last '_1'. So if I run a query on above sample dataset, it would return
x_1_1,
x_2_1,
x_3_1,
This should not return x_2_12 or x_3_11.
I tried like '%_1' but it returns x_2_12 and x_3_11 as well.
Thank you!
A simple method is the right() function:
select t.*
from t
where right(field, 2) = '_1';
You can use like but you need to escape the _:
where field like '%$_1' escape '$'
Or use regular expressions:
where field regexp '_1$'
The underscore character has special significance in a LIKE clause. It acts as a wildcard and represent one single character. So you would have to escape it with a backslash:
LIKE '%\_1'
RIGHT does the job too, but it requires that you provide the proper length for the string being sought and is thus less flexible.
Duh, I found the answer.
Use RIGHT (col_name, 2) = '_1'
Thank you!
If I try to use Regex inside locate it fails
Select Locate(FieldA regexp '[a-z][A-Z][a-z]',Binary FieldA) from PatternTester
as per http://sqlfiddle.com/#!9/403c36/2.
If I search for the explicit letter pattern it locates it correctly:
Select Locate('lC',Binary FieldA) from PatternTester
as per http://sqlfiddle.com/#!9/403c36/6
Is there something I need to do to make locate 'obey' Regex or will it simply not?
As mysql document says, LOCATE() will returns the position of the first occurrence of substring substr in string st, so it will not take any regex as input argument.
Also from checking your fiddle reference, you don't have REGEX_INSTR in this version!
I'm attempting to query our MSSQL database but I'm getting no data when there clearly is data there.
First I query
SELECT id, instruction_link FROM work_instructions WHERE instruction_link LIKE "%\\\\cots-sbs%";
Which returns 100+ lines.
http://tinypic.com/r/ief8td/8
(sorry couldn't post as actual picture, don't have enough rep :(
However if I query
SELECT id, instruction_link FROM work_instructions WHERE instruction_link LIKE "%\\\\cots-sbs\\%";
http://tinypic.com/r/33ksw3q/8
I get no results with the 2nd query. I have no idea what I'm doing wrong here. Seems pretty simple but I can't make any sense of it..
Thanks in advance.
As documented under LIKE:
Note
Because MySQL uses C escape syntax in strings (for example, “\n” to represent a newline character), you must double any “\” that you use in LIKE strings. For example, to search for “\n”, specify it as “\\n”. To search for “\”, specify it as “\\\\”; this is because the backslashes are stripped once by the parser and again when the pattern match is made, leaving a single backslash to be matched against.
\\% is parsed as a string containing a literal backslash followed by a percentage character, which is then interpreted as a pattern containing only a literal percentage sign.
I need to search for a value like 1234-abc. The database doesn't have this particular value, but has another value 1234. Now the problem is when I write my query like
SELECT * FROM words WHERE tval='1234-abc'
instead of fetching an empty recordset, it fetches the 1234 value, it seems to ignore anything after the -, any idea what's going on?
http://sqlfiddle.com/#!2/9de62/3
You can use the BINARY keyword for the exact match
SELECT tval FROM words WHERE BINARY tval='1223-abc';
Binary is a built-in keyword that after your WHERE clause that forces a comparison for an exact case-sensitive match
Fiddle
The existing expression is implicitly converting the string expression to a number - you need to explicitly convert the number to a character strng, like so:
SELECT tval FROM words WHERE convert(tval,char(20))='1223-1ABCDE';
SQLFiddle here.
One column returns such values:
Something";s:5:"value";s:3:"900";s:11:"print_
I want to extract all numbers that are at least 3 digits long, in the above case thats 900. How can I do that in MySQL? Maybe using a regex? I cant use any index, the length of the string and the number in the string can be different.
Thanks!
Try unserialize() it if you are using PHP! And then var_dump it to see the strings and arrays
You can't extract them using MySQL, use any other language for that.
What you can do is include a Where Clause, that will make the work easier for your script.
Assuming your column is called "serialized" in the table "example"
SELECT serialized FROM example WHERE serialized REGEXP '[0-9]{3,}'
Please note that REGEXP is just outputting 1 or 0
After you did the query, use the regex functions of your language do extract the numbers like so:
([0-9]{3,})*