mysql - SQL JOIN many-to-many -

-

sorry minimalistic title don't know how describe in short. have 3 tables:

the table of groups

id | genre ----------------- 1  | action 2  | adventure 3  | drama

many many table

groupid | elementid -----------------     3   |    1     1   |    2     2   |    2     2   |    3     3   |    3

and table of elements

id | element ----------------- 1  | pride , prejudice 2  | alice in wonderland 3  | curious incident of dog in night time

all fine , simple. select trying achieve following

id | element                                         |  genre ------------------------------------------------------------- 1  | pride , prejudice                             | drama 2  | alice in wonderland                             | null 3  | curious incident of dog in night time     | drama

i want select all elements table elements , set genre field drama or null.

i'm trying in mysql.

thank in advance

it's possible little trick (outer join on many-to-many table, constraint groupid has 3 (for drama)

http://sqlfiddle.com/#!2/b7c18/2

select elements.id, elements.element, groups.genre   elements left outer join group_elements   on elements.id = group_elements.elementid  , group_elements.groupid = 3 left outer join groups   on group_elements.groupid = groups.id

left outer join means : take lines tables preceded (the ones on left hand side of left outer join, if will), if there's no lines corresponding them in following tables. condition on elements.id = group_elements.elementid , group_elements.groupid = 3 says if find matches our elementid, must drama (groupid = 3). left outer join on groups table, enables display genre column, or null if element not drama.


Comments

Post a Comment

Popular posts from this blog

html5 - What is breaking my page when printing? -

-
Image
i have questioner , don't know why when printing it's creating break before first question. live demo (link) i able resolve issue removing fieldset container had wrapped in.. having fieldset element causing of fields grouped together, , since not fit on first page, being pushed second page. removing fieldset element, able firefox print first 2 questions on page 1 identically chrome. it seems not new problem firefox , fieldset not playing nice found old bugzilla ticket 2008 has persisted , still being commented on in 2014 (link) if retaining fieldset important, found solution else came here (link) <script type='text/javascript'> $(window).bind('beforeprint', function(){ $('fieldset').each( function(item) { $(this).replacewith($('<div class="fieldset">' + this.innerhtml + '</div>')); } ) }); $(window).bind(
Read more

c# - must be a non-abstract type with a public parameterless constructor in redis -

-
when save object, got below error must non-abstract type public parameterless constructor in order use parameter 't' in generic type or method 'servicestack.redis.redisclient.store<t>(t) redisclass.getinstnace().store(msg); // error here redisclass.getinstnace().save(); as third party's class, can not edit how save object? could create wrapper around third party object call constructor, store wrapper? e.g. public class mywrapper { public thirdpartyobject thirdpartyinstance { get; set; } public mywrapper() { thirdpartyinstance = new thirdpartyobject("constructors"); } }
Read more

ajax - PHP/JSON Login script (Twitter style) not setting sessions -

-
i'm having trouble getting twitter style login work properly, reason gets stuck at session_register("useremail"); session_register("userpass"); and doesn't redirect after it. if enter wrong account data, works , appropriate error message pops up. this full code: <?php if(empty($_post['email']) || empty($_post['password'])) { die(msg(0,"all fields required")); } if(!(preg_match("/^[\.a-z0-9_\-\+]+[@][a-z0-9_\-]+([.][a-z0-9_\-]+)+[a-z]{1,4}$/", $_post['email']))) die(msg(0,"you haven't provided valid e-mail")); $host="localhost"; $username="root"; $password="x"; $db_name="x"; $tbl_name="x"; mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select db"); $useremail=$_post['email']; $userpass=
Read more