MySQL left join excludes non-matching records? -


when execute mysql query, correct results expected:

select h.id productid, h.producttitle, u.username, ht.producttype, hd.difficulty-- , ifnull(sum(pv.upvote),0) - ifnull(sum(pv.downvote),0) netvotes products h  join users u on h.userid = u.id  join producttypes ht on h.producttypeid = ht.id  join productdifficulties hd  on h.productdifficultyid = hd.id  -- left join productvotes pv -- on h.id = pv.productid

notice commented out last column in select (aggregate) , left join. when include in sql, records have matching productvotes record included:

select h.id productid, h.producttitle, u.username, ht.producttype, hd.difficulty, ifnull(sum(pv.upvote),0) - ifnull(sum(pv.downvote),0) netvotes products h  join users u on h.userid = u.id  join producttypes ht on h.producttypeid = ht.id  join productdifficulties hd  on h.productdifficultyid = hd.id  left join productvotes pv on h.id = pv.productid

i expect behavior inner join, not left join. why left join excluding records without matching productvotes record? , how can have included, in first query above?

edit: had limit 0, 10 @ end of query, , folks thought causing problem. removed limit clause , problem still remains, had nothing it. have removed limit clause original post because it's irrelevant. still awaiting answer. :-)

here full results of select * productvotes:

id productid userid upvote downvote  2         1      2      1 null  3         1      3      1 null  4         1      4      1 null  5         1      5      1 null  6         1      6      1 null  7         1      7      1 null  8         1      8      1 null  9         1      9   null    1 10         1     10   null    1 22         1      1      1 null

(notice 10 records in table.)

important: please notice aggregations in select statements: ifnull(sum(pv.upvote),0) - ifnull(sum(pv.downvote),0) netvotes

select h.id productid, h.producttitle, u.username, ht.producttype, hd.difficulty,  (select ifnull(sum(pv.upvote),0) - ifnull(sum(pv.downvote),0) productvotes pv h.id = pv.productid ) netvotes products h  join users u on h.userid = u.id  join producttypes ht on h.producttypeid = ht.id  join productdifficulties hd  on h.productdifficultyid = hd.id

Comments

Post a Comment

Popular posts from this blog

php - failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request -

this code i'm using find latitude , longitude of place: function coords($address){ echo $url="http://maps.googleapis.com/maps/api/geocode/json?address=".$address; $json = file_get_contents(urlencode($url)); $data = json_decode($json, true); print_r($data['results'][0]['geometry']['location']['lat']); print_r($data['results'][0]['geometry']['location']['lng']); } however returns warning: failed open stream: http request failed! http/1.0 400 bad request if use above code in simple procedure works fine not in function. note: have tried curl_init()... method produced same result? instead of encoding whole url encode address part. following code work fine $add='jamshoro phase 2'; $url="http://maps.googleapis.com/maps/api/geocode/json?address=".urlencode($add); $json = file_get_contents($url); $data = json_decode($json, true); print_r($data['r
Read more

java - How to filter a backspace keyboard input -

i have code gives message when key number pressed: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochararray().length - 1]; if (!(ch >= '0' && ch <= '9')) { system.out.println("the char entered not number"); t.consume(); } } }); now if press on wrong button , press backspace remove it, error message. how can add backspace input if statement? i have found answer: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochar
Read more

java - Show Soft Keyboard when EditText Appears -

i have code in android app, alert dialog , keyboard, , when dialog shown want keyboard appear. not , not sure why? here code: final alertdialog.builder alert = new alertdialog.builder(this); alert.settitle("title"); alert.setmessage("message"); final edittext input = new edittext(this); input.setinputtype(inputtype.type_class_number); alert.setview(input); input.setonfocuschangelistener(new view.onfocuschangelistener() { @override public void onfocuschange(view v, boolean hasfocus) { if (hasfocus) { log.i(tag, "in focus"); inputmethodmanager inputmethodmanager = (inputmethodmanager) getsystemservice(context.input_method_service); inputmethodmanager.showsoftinput(input, inputmethodmanager.show_implicit); } } }); but keyboard not appear though edit text in focus, , tell appear why, doesn't come up? thanks in advance.
Read more