I search my table with query contains LIKE clause %str%.
Is here a way to know where string 'str' was finded in sentence?
I would like to print out 'str' as markup (bold).
For this I need information where exact 'str' begins in any row which contain 'str'.
you can get the string position using the POSITION function ( http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_position ) however thats probably not the best way to do it since if they use the markup more than once it will only return the first position. It would be easier just to replace the string with the string wrapped with whatever markup you want.
If you want an all MySQL solution this would probably work:
SELECT REPLACE(exampleTable.field, 'search_string', '<b>search_string</b>')
FROM exampleTable
WHERE exampleTable.field LIKE '%search_string%';
However i would recommend doing any replacement like this on the PHP / ASP side... using string replacement tools from the respective language.
Sure, you want INSTR() .
You could also use it in your where clause, though you'd want to compare performance between that and LIKE
SELECT INSTR(`field`, 'str') FROM `table` WHERE 0 < INSTR(`field`, 'str')
Remember that INSTR() returns a 1-based index, that is, the first character is postion 1, not position 0; saving 0 for "not found".
Related
I am working on a search function, where the matches are weighted based on certain conditions. One of the conditions I want to add weight to is matches where the character length of the query string in a LIKE match is longer than 4.
This is what I want to the query to look like, roughly. %s is meant to represent the actual match found by LIKE, but I don't think it does. I'm wondering if there is a special variable in MySQL that does represent the precise character match found by LIKE.
SELECT help.*,
IF(CHAR_LENGTH(%s) > 4, 2, 0) w
FROM help
WHERE (
(title LIKE '%this%' OR title LIKE '%testy%' OR title LIKE '%test%') OR
(content LIKE '%this%' OR content LIKE '%testy%' OR content LIKE '%test%')
) LIMIT 1000
edit: I could in the PHP split the search string array into two arrays based on the character length of the elements, with two separate queries that return different values for 'w', then combine the results, but I'd rather not do that, as it seems to me that would be awkward, messy, and slow.
Check out FULLTEXT as another way to discover rows. It will be faster, but won't address your question.
This probably has the effect you want.
SELECT ....
IF ( (title LIKE '%testy%' OR
content LIKE '%testy%'), 2, 0)
....
Note that the "match" in your LIKEs includes the %, so it is the entire length of the string. I don't think that is what you wanted.
REGEXP "(this|testy|that)" will match either 4 or 5 characters (in this example). It may be possible to do something with REGEXP_REPLACE to replace that with the empty string, then see how much it shrank.
I think the answer to my question is that what I wanted to do isn't possible. There is no special variable in MySQL representing the core character match in a WHERE condtional where LIKE is the operator. The match is the contents of the returned data row.
What I did to reach my objective was took the original dynamic list of search tokens, iterated through that list, and performed a search on each token, with the SQL tailored to the conditions that matched each token.
As I did this I built an array of the search results, using the id for the database row as the index for the array. This allowed me to perform calculations with the array elements, while avoiding duplicates.
I'm not posting the PHP code because the original question was about the SQL.
I have the following strings:
SDZ420-1241242,
AS42-9639263,
SPF3-2352353
I want to "escape" the SDZ420- part while searching and only search using the last digits, so far I've tried RLIKE '^[a-zA-Z\d-]' which works but I am confused on how to add the next digits (user input, say 1241242) to it. I cannot use LIKE '%$input' since that would return a row even if I just input '242' as the search string.
In simple words, a user input of '1241242' should return the row with 'SDZ420-1241242'. Is there any other approach other than creating a separate table with the numbers only?
Note that without jumping through some crazy hoops, this search needs to hit every row in the table; if you have an index on this, it's not going to use that (an index is generally used, assuming it's of the proper kind, which they tend to be, when you search on start, and generally only when using LIKE 'needle%' and not RLIKE. If that's a problem, storing the digits separately, and then putting an index on that, is probably the simplest way to solve your problem here.
To query for the final few digits, why not:
SELECT * FROM foo WHERE colName LIKE ?
with the string made in your programming language via:
String searchTerm = "%-" + digits;
You can also pass in the number as a string and use:
where substring_index(colname, '-', -1) = ?
This does not require changing the value in the application code.
I have one column(varchar) containing only json string within one table. I want replace all keys with "" on that column. How can I do that using sql? My database is MySQL.
For example:
|--------------------------------------------------------------------|
| t_column |
|--------------------------------------------------------------------|
| {"name":"mike","email":"xxx#example.com","isManage":false,"age":22}|
|--------------------------------------------------------------------|
SELECT replace(t_column, regexp, "") FROM t_table
I expect:
mikexxx#example.comfalse22
How to write that regexp?
Start from
select t_column->'$.*' from test
This will return a JSON array of attribute values:
[22, "mike", "xxx#example.com", false]
This might be already all you need, and you can try something like
select *
from test
where t_column->'$.*' like '%mike%';
Unfortunately there seems to be no native way to join array values to a single string like JSON_ARRAY_CONCAT(). In MySQL 8.0 you can try REGEXP_REPLACE() and strip all JSON characters:
select regexp_replace(t_column->'$.*', '[" ,\\[\\]]', '') from test
which will return '22mikexxx#example.comfalse'.
If the values can contain one of those characters, they will also be removed.
Note: That isn't very reliable. But it's all I can do in a "simple" way.
See demo on db-fiddle.
I could be making it too simplistic, but this is just a mockup based on your comment. I can formalize it into a query if it fits your requirement.
Let's say you get your JSON string to this format where you replace all the double quotes and curly brackets and then add a comma at the end. After playing with replace and concat_ws, you are now left with:
name:mike,email:xxx#example.com,isManage:false,age:22,
With this format, every value is now preceded by a semicolon and followed by a comma, which is not true for the key. Let's say you now want to see if this JSON string has the value "mike" in it. This, you could achieve using
select * from your_table where json_col like '%:mike,%';
If you really want to solve the problem with your approach then the question becomes
What is the regex that selects all the undesired text from the string {"name":"mike","email":"xxx#example.com","isManage":false,"age":22} ?
Then the answer would be: {\"name\":\"|\"email\":\"|\",\"isManage\":|,\"age\":|}
But as others let you notice I would actually approach the problem parsing JSONs. Look up for functions json_value and json_query
Hope I helped
PS: Keep close attention on how I structured the bolded sentence. Any difference changes the problem.
EDIT:
If you want a more generic expression, something like select all the text that is not a value on a json-formatted string, you can use this one:
{|",|"\w+\":|"|,|}
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,})*
I've been trying to get a table row with this query:
SELECT * FROM `table` WHERE `field` LIKE "%\u0435\u0442\u043e\u0442%"
Field itself:
Field
--------------------------------------------------------------------
\u0435\u0442\u043e\u0442 \u0442\u0435\u043a\u0441\u0442 \u043d\u0430
Although I can't seem to get it working properly.
I've already tried experimenting with the backslash character:
LIKE "%\\u0435\\u0442\\u043e\\u0442%"
LIKE "%\\\\u0435\\\\u0442\\\\u043e\\\\u0442%"
But none of them seems to work, as well.
I'd appreciate if someone could give a hint as to what I'm doing wrong.
Thanks in advance!
EDIT
Problem solved.
Solution: even after correcting the syntax of the query, it didn't return any results. After making the field BINARY the query started working.
As documented under String Comparison Functions:
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.
Therefore:
SELECT * FROM `table` WHERE `field` LIKE '%\\\\u0435\\\\u0442\\\\u043e\\\\u0442%'
See it on sqlfiddle.
it can be useful for those who use PHP, and it works for me
$where[] = 'organizer_info LIKE(CONCAT("%", :organizer, "%"))';
$bind['organizer'] = str_replace('"', '', quotemeta(json_encode($orgNameString)));