I am trying to select a record using the LIKE function. However, it does not show results if there are characters in between. I want it to select data that contains those strings. For example:
$value = "Mark Anthony";
$qry ="SELECT * FROM students WHERE name LIKE '%$value%'";
returns these results:
John Mark Anthony
Mark Anthony Philipps
but I also want to have results like these
Mark James Anthony
Mark Gabriel Anthony Fernandez
Marko Julian Anthonyo
Any ideas?
UPDATE: 'Mark' must be before 'Anthony'
I think Full TEXT works
But Full-text searches are supported for MyISAM tables only
SELECT * FROM students
WHERE MATCH (name)
AGAINST ('Mark Anthony' IN BOOLEAN MODE)
UPDATE:- As per question update OP need to search Marko as well, then you can get like this:-
SELECT * FROM students
WHERE
(
name LIKE '%Mark%'
OR name LIKE '%Anthony%'
)
You could split the value string into two parts. Like so:
WHERE name LIKE '%Mark%' AND name LIKE '%Anthony%'
last try:
$value = "Mark Anthony";
$value = str_replace(" ","%",$value);
$qry ="SELECT * FROM students WHERE name LIKE '%$value%'";
not tested!
Try this solution -
SELECT * FROM table WHERE column REGEXP '(Mark).+(Anthony)';
The proplem is, you say "bring all results which contains Mark Anthony" so it does. You should set it like
$value = "Mark%Anthony";
Related
Currently I have a query that searchs for sentences/words, it works almost as expected,
I have a regex expresion that searches for names in a table, expample:
function getNames($str){
$stmt = "SELECT * FROM users WHERE name = :name
OR name REGEXP :reg1
OR name REGEXP :reg2
OR name LIKE :lik1";
$query = self::$connection->prepare($stmt);
$query->execute(array(":name"=>$str,
":reg1"=>"^$str" . "[a-zA-Z\s]*$",
":reg2"=>"^[a-zA-Z]*[\s]*[$str][a-zA-Z]"
":lik1"=>"%" . $str . "%"
));
return $query->fetchAll(PDO::FETCH_ASSOC);
}
Let's suppose my table contains the following values
Bob
Peter
Mark
David
John
If I run my query with Bob as $name value it gets it but I would like to be able to find Bob when I run the query using BobsomeLettersExtra or Bob something as $name value
Is there a way to do this using REGEXP or LIKE ?
SELECT * FROM users WHERE name LIKE '%".$name."%'
above query should be enough to get the result. You should validate data before you enter data to the table if not please use the regex as well
"SELECT * FROM users WHERE name LIKE '%".$name."%' AND REGEXP ^".$name."[a-zA-Z]*$"
UPDATE
sorry if i have misunderstand the question please try this
"Select * from users WHERE '".$name."' LIKE CONCAT(name , '%')"
You may try below Query with where Clause for LIKE :
"SELECT * FROM users WHERE name = ".$name." OR name LIKE '%".$name."%' OR name REGEXP ^".$name."[a-zA-Z]*$"
My database structure is like this:
Movie Name GenreID**
Movie1 1,2,3
Movie2 2,4
Movie3 4,5,16
I need to select a Movie name based on the Genre ID the user selected which I put inside the genreIDArray[]
Let's say for example the genreIDArray has values: $genreIDArray = ['1','2','3'];
My current query method is the ff:
Here I prepared each ID into parts so the result won't become genreID LIKE (%1,2,3%) because I checked this doesn't work.
So I did this separation loop:
$queryParts = array();
foreach($genresIDArray as $genreID) {
$queryParts[] = "'%".$genreID."%'";
}
After the separation loop I put together the final query:
$genreString = implode(" OR genreID LIKE ",$queryParts);
$genreQuery = " SELECT * FROM movies WHERE (genreID LIKE {$genreString}) ";
gave me this final query output:
SELECT * FROM movies WHERE (genreID LIKE '%1%' OR genreID LIKE '%2%' OR genreID LIKE '%3%')
This actually works, but apparently not that efficient because genreID 11,12,13 and so on that start with 1 is also selected. I think I'm missing the MYSQL LIKE logic here. I've tried '%$genreID' which means to select the starting or first number/letter of a table data, but that's still the same thing, $genreID% doesn't and would not work because this only means genreid ENDING letters/number will be selected.
I hope I spelled that out clear enough. I'm in a bind here. Please help.
Thank you so much.
There is a very cool function for that. You can use FIND_IN_SET.
SELECT * FROM test WHERE FIND_IN_SET(1,colors)
But if its possible you should avoid such structures in your database and normalize your database.
I want to select record from table with key.My code is working but it's not select all record regarding Php and Mysql it select only PHP,Mysql record.
My code is
$skill=PHP,Mysql;
mysql_query("select * from tablename(skill) where fieldname(key) like '%$skill%'");
Try to do this
mysql_query("select * from skill where (key like '%PHP%' OR key like '%Mysql%')");
As you have comma separated values in $skill variable like operator will not work correctly, try using following query
select * from skill where FIND_IN_SET(keyn,'$skill')
Demo sql fiddle at http://sqlfiddle.com/#!2/181c74/5
$skill="PHP,Mysql";
mysql_query("select * from skill where key like '%$skill%'");
I am assuming you want to search for records that contain "PHP", "MySQL", or both. What you're doing is searching for the exact string of "PHP,MySQL".
A quick and dirty solution would be to turn $skill into an array that is split on the commas and then perform the search for each term using the OR keyword.
For example:
$skill = array("PHP", "MySQL");
$query = "select * from tablename(skill) where ";
foreach ($skil in $skill)
$query . "fieldname(key) like '%$skil%' OR ";
//code to remove the OR at the very end
Use in into your Query
select * from `skills` WHERE skillslist in ('Php','Javascript')
Run demo Fiddle
I currently use a mysql statement like the one below to search post titles.
select * from table where title like %search_term%
But problem is, if the title were like: Acme launches 5 pound burger and a user searched for Acme, it'll return a result. But if a user searched for Acme burger or Acme 5 pound, it'll return nothing.
Is there a way to get it to return results when a users searches for more than one word? Is LIKE the correct thing to use here or is there something else that can be used?
You could use a REGEXP to match any of the words in your search string:
select *
from tbl
where
title REGEXP CONCAT('[[:<:]](', REPLACE('Acme burger', ' ', '|'), ')[[:>:]]')
Please notice that this will not be very efficient. See fiddle here.
If you need to match every word in your string, you could use a query like this:
select *
from tbl
where
title REGEXP CONCAT('[[:<:]]', REPLACE('Acme burger', ' ', '[[:>:]].*[[:<:]]'), '[[:>:]]')
Fiddle here. But words have to be in the correct order (es. 'Acme burger' will match, 'burger Acme' won't). There's a REGEXP to match every word in any order, but it is not supported by MySql, unless you install an UDF that supports Perl regexp.
To search for a string against a text collection use MATCH() and AGAINST()
SELECT * FROM table WHERE MATCH(title) AGAINST('+Acme burger*')
or why not RLIKE
SELECT * FROM table WHERE TITLE RLIKE 'Acme|burger'
or LIKE searching an array, to have a compilation of $keys
$keys=array('Acme','burger','pound');
$mysql = array('0');
foreach($keys as $key){
$mysql[] = 'title LIKE %'.$key.'%'
}
SELECT * FROM table WHERE '.implode(" OR ", $mysql)
What you need to do is construct a SQL such that, for example:
select * from table where title like "%Acme%" and title like "%burger%"
In short: split the string and create one like for each part.
It might also work with replacing spaces with %, but I'm not sure about that.
The best thing is thing use perform union operation by splitting your search string based on whitespaces,
FOR Acme 5 pound,
SELECT * FROM TABLE WHERE TITLE LIKE '%ACME 5 POUND%'
UNION
SELECT * FROM TABLE WHERE TITLE LIKE '%ACME%'
UNION
SELECT * FROM TABLE WHERE TITLE LIKE '%5%'
UNION
SELECT * FROM TABLE WHERE TITLE LIKE '%POUND%'
Find out a way to give the first query a priority. Or pass the above one as four separate queries with some priority. I think you are using front end tp pass query to data bases, so it should be easy for you.
<?php
$search_term = 'test1 test2 test3';
$keywords = explode(" ", preg_replace("/\s+/", " ", $search_term));
foreach($keywords as $keyword){
$wherelike[] = "title LIKE '%$keyword%' ";
}
$where = implode(" and ", $wherelike);
$query = "select * from table where $where";
echo $query;
//select * from table where title LIKE '%test1%' and title LIKE '%test2%' and title LIKE '%test3%'
I have the followings examples in be_user_profiles.subject. These are subject ids which each teacher teaches.
1// English
1,2 // English and Math etc
1,2,14
2,4,114
12,24,34
15, 23
I want to select where be_user_profiles.subject has 1. When I use the following, it outputs all which has 1 in it. So it will outputs all. I tried HAVING but it picks up only exact matches. So it shows only the first one. How can I pick up data which has the be_user_profiles.subject?
$this->db->select('*');
$this->db->from('be_user_profiles');
$this->db->join('be_users', 'be_users.id = be_user_profiles.user_id');
$this->db->where('be_users.group', $teachergrp);
$this->db->like('be_user_profiles.subject', $subjectid);
//$this->db->having("be_user_profiles.subject = $subjectid");// this picks up only exact match
$query = $this->db->get();
Thank you in advance.
be_user_profiles table
row1: 1,2,14
row2: 2,4,114
row3: 12,24,34
row4: 15, 23
To get data with exact match use this query
$this->db->query("
SELECT * FROM `be_user_profiles`
WHERE subject LIKE '1'
UNION
SELECT * FROM `be_user_profiles`
WHERE subject LIKE '1,%'
UNION
SELECT * FROM `be_user_profiles`
WHERE subject LIKE '%,1,%'
UNION
SELECT * FROM `be_user_profiles`
WHERE subject LIKE '%,1'
");
The both clause that you put into the like query means to add % widcard in front and after of the string to search, so it returns 1 as long as 12, 21, 121 etc. If you remove it it will search only for exact match.
You could add this like clause and add commas to it and i think that it will work. Try to add this instead of the like you have now:
$this->db->like("," . "be_users.group" . "," , "," . $subjectid. "," , both);
I think you can use a regex pattern here.
$pattern = "(^|.*,)1(,.*|$)";
...
...
$this->db->select('*');
....etc
$this->db->where("be_user_profiles.subject REGEXP $pattern");
This regex pattern assumes that there are no spaces in the comma string.
However, as #halfer said in the comments you really, really should split this out into a "teachersubject" table with teacherid and subjectid columns otherwise it will bite you in the backside very, very soon. [Been there, done that :-) ]
eg Imagine trying to expand the above into searching for a teacher that teaches ((maths or physics) and English). Nightmare!