Javascript: If you add a selection, don't allow that selection to be added again (no duplicates)

Question

I currently have many languages for an Airports project I'm working on.

-Currently you can select whatever language you would like from the drop-down menu and it will appear in the list. Working fine. Can delete them too.

-The goal: once a language has been added to the list, you should not be able to add it again. For example, you can hit the 'Add Languages' for French and add it as many times as you want. This goes for any language.

The current js code for adding a language:

function addLanguage()
{
        var languages = $("#languages_dd").val();
        language_display = languages.split("-");
        alert(languages);

        var units = $("#units_dd").val();
        var unit_display = $("#units_dd :selected").text();

        $(".none_class").hide();
        $("#error_msg").html("");
        $("#summary").append("<li><input type='radio' name='language_item'> <span class='route_summary_field_big'>"+language_display[0]+"</span>"+unit_display+"<input type='hidden' name='languages[]' value='"+languages+"'><input type='hidden' name='units[]' value='"+units+"'></li>");
}

I'm not too familiar with javascript and have been searching around online. I know it's going to be a conditional, something along the lines of:

if($("#languages_dd :selected")
{
//do something;
}
else if
//do something else;
}

Any input is appreciated to steer me in the right direction.

http://stackoverflow.com/questions/784012/javascript-equivalent-of-phps-in-array - Store your languages in an array, and check if such a language is already in the array. — Ariane, Aug 22 '13 at 19:24
If you don't want duplicates, remove it from the selection once its been added. This way your users don't have the option to add it again. If you delete it from your list just add the option back into the choices list. — AJak, Aug 22 '13 at 19:24

score 0 · Answer 1 · answered Aug 22 '13 at 19:24

0

You can keep the selected languages in an array. Then before you append to the html, check whether the value already exists:

var selected_lang = new Array();
function addLanguage() {
//your code here..
    if (!$.inArray(language_display, selected_lang)) {
        selected_lang.push(language_display)
        // rest of your code to append html
    }
}

answered Aug 22 '13 at 19:24

PiSquared

173
6

thanks for the response. I don't think this is a bad idea, but a friend was telling me since these languages are attached to airport routes (ie you could have different/same languages per different routes), an empty array would not be best in this situation. – Nubtacular Aug 23 '13 at 15:33

score 0 · Answer 2 · answered Aug 26 '13 at 15:42

This is the solution I came up with (feel free to offer constructive criticism):

function addLanguage()
{
        var languages = $("#languages_dd").val();
        language_display = languages.split("-");
        var units = $("#units_dd").val();
        var unit_display = $("#units_dd :selected").text();
        $(".none_class").hide();
        var shouldAdd = "YES";
        $("#summary li").each(function(){
            var matches = 0;
            $(this).find('input:hidden').each(function(){
                var stringVal= $(this).val();
                console.log(stringVal);
                if(stringVal.indexOf(languages) != -1){
                    matches++;
                }
                if(stringVal.indexOf(units) != -1){
                    matches++;
                }
            });
            if(matches == 2){
                shouldAdd = "NO";
            }
        });
        if(shouldAdd == "YES") {
            $("#summary").append("<li><input type='radio' name='language_item'> <span class='route_summary_field_big'>"+language_display[0]+"</span>"+unit_display+"<input type='hidden' name='languages[]' value='"+languages+"'><input type='hidden' name='units[]' value='"+units+"'></li>");
        }
        $("#error_msg").html("");
}

Javascript: If you add a selection, don't allow that selection to be added again (no duplicates)

2 Answers2