Return from a jQuery $.each loop

Question

If a given array doesn't contain a given value, I wish to open a confirm dialog. The following works, however, my use of intermediate variable t seems a little excessive and I expect there is a more elegant way to do so. Could I return from the $.each loop and cause the upstream anonymous function to return false?

$(function(){
    myArr=[['de'],['df','de'],['df','dz'],['de']];
    if((function(){
        var t=true;
        $.each(myArr, function() {
            console.log($.inArray('de', this)=='-1');
            if($.inArray('de', this)=='-1') {t=false;return false;};    //Doesn't return true to parent
        })
        return t;
        })() || confirm("Continue even though one of the choices doesn't contain 'de'?") ){
        console.log('proceed');
    }
});

have a look at underscore. theres a lot of nice functions in ther! — Daniel A. White
– Daniel A. White, Commented Mar 14, 2015 at 15:36
myArr seems to looks just a normal javascript. Why don't you use javascript function ? myArray.forEach(function(element){ if(element.indexOf('de')) != -1 { confirm("....") } }); — Arnaud Bertrand
– Arnaud Bertrand, Commented Mar 14, 2015 at 15:39

dfsq · Accepted Answer · 2015-03-14 15:40:14Z

3

You can use Array.prototype.some method, it will make code more comprehensive and simpler:

var myArr=[['de'],['df','de'],['df','dz'],['de']];

if (myArr.some(function(el) {
    return el.indexOf('de') === -1;
}) && confirm("Continue even though one of the choices doesn't contain 'de'?")) {
    document.write('proceed');
}

answered Mar 14, 2015 at 15:40

dfsq

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2 Comments

user1032531 Over a year ago

Thanks dfsq. Never knew about some before.

dfsq Over a year ago

@DavidThomas I think that some is better, because it stops iterating once the first return el.indexOf('de') === -1 is true. At this point it doesn't matter anyway if other elements has de or not.

user488187 · Accepted Answer · 2015-03-14 15:44:54Z

0

You could use grep instead, filtering out values that include 'de' and then counting the remaining:

$(function(){
    var myArr=[['de'],['df','de'],['df','dz'],['de']];
    var notDe = $.grep(myArr, function(item, index) {
            return ($.inArray('de', this)=='-1');
        });
    if(notDe.length == 0 || confirm("Continue even though one of the choices doesn't contain 'de'?") ){
        console.log('proceed');
    }
});

answered Mar 14, 2015 at 15:44

user488187

Comments

Tomanow · Accepted Answer · 2015-03-14 15:45:00Z

0

Another more readable solution:

$(function () {
    myArr = [
        ['de'],
        ['df', 'de'],
        ['df', 'dz'],
        ['de']
    ];

    var t = 0;
    $.each(myArr, function (k, v) {
        if ($.inArray('de', v) === -1) {
            t++;
        }
    });

    if (t > 0) {
        if (confirm("Continue even though " + t + " of the choices do not contain 'de'?")) {
            console.log('proceed');
        }
    }
});

answered Mar 14, 2015 at 15:45

Tomanow

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