Bug Tracker

Opened 14 years ago

Closed 13 years ago

Last modified 11 years ago

#3279 closed bug (invalid)

[autocomplete] trimWords splits on multipleSeparator regardless of options.multiple setting

Reported by: SDraconis Owned by:
Priority: minor Milestone: 1.3
Component: plugin Version: 1.2.6
Keywords: Cc:
Blocked by: Blocking:


This is for Jörn Zaefferer's AutoComplete plugin (http://bassistance.de/jquery-plugins/jquery-plugin-autocomplete/).

The trimWords function will always split the value on options.multipleSeparator, even if options.multiple == false. The fix is to change the line

var words = value.split( options.multipleSeparator );


var words = options.multiple ? value.split( options.multipleSeparator ) : value;

Change History (4)

comment:1 Changed 14 years ago by SDraconis

Sorry, the line should be

var words = options.multiple ? value.split( options.multipleSeparator ) : [value];

comment:2 Changed 13 years ago by dmethvin

Resolution: invalid
Status: newclosed

This is not a jQuery core bug. Please report plugin bugs to the plugin's author, or ask on the jQuery forums. jQuery UI bugs should be reported on the UI bug tracker, http://dev.jqueryui.com .

comment:3 Changed 11 years ago by anonymous

comment:4 Changed 11 years ago by anjumasim

function selectCurrent() {

var selected = select.selected(); if (!selected)

return false;

var v = selected.result;

v = v.split(":")[1]; previousValue = v;

if (options.multiple) {

var words = trimWords($input.val());

if (words.length > 1) {

v = words.slice(0, words.length - 1).join(options.multipleSeparator) + options.multipleSeparator + v;

v = $input.val().slice(0, $input.val().length - words[words.length - 1].length) + v;


v += options.multipleSeparator;


$input.val(v); hideResultsNow(); $input.trigger("result", [selected.data, selected.value]); return true;


Version 0, edited 11 years ago by anjumasim (next)
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