mysql - SQL Stored Procedure Syntax error missing 'end' -


i've been working on sql stored procedure , new process. i'm trying write procedure retrieves value 1 table , inserts part of new row.

i'm using mysql workbench, , it's giving me error @ line

where blockid = blk;

as part of following code:

create definer=`root`@`localhost` procedure `add_userblock`(in user int(11), in blk int(11)) begin     -- determine how many blocks can add     declare blocknum int default -1;     select addcount blocknum         block         blockid = blk;      -- determine if block exists user     declare entryexists int default 0;     select count(*) entryexists         userblock         blockid = blk , userid = user;      if (entryexists = 0)         -- new entry         begin         insert userblock (userid, blockid, num) values (user, blk, blocknum);         select last_insert_id() 'id';         end     else         -- existing entry         begin         update userblock             set num = (num + blocknum)             userid = user , blockid = blk;         end end

the error "syntax error: missing 'end'". i'm not entirely sure means, , i'm having trouble solving issue.

thanks, appreciate time!

add end if after else-block

create  definer=`root`@`localhost`  procedure `add_userblock`(   in user int(11),    in blk int(11) ) begin -- determine how many blocks can add declare blocknum int default -1; select addcount blocknum   block  blockid = blk;  -- determine if block exists user declare entryexists int default 0; select count(*) entryexists   userblock  blockid = blk    , userid = user;  if (entryexists = 0)     -- new entry     begin     insert userblock (userid, blockid, num) values (user, blk, blocknum);     select last_insert_id() 'id';     end else     -- existing entry     update userblock        set num = (num + blocknum)      userid = user , blockid = blk; end if end

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