CODEWITHSUNDEEP
phpjqueryhtmlformspost
jjclarkson
jjclarkson

Reputation: 5954

Post array of multiple checkbox values

Why is only one value of the "db" checkbox values array being sent to the server side script?

JQUERY:

$(".db").live("change", function() {
    $(this).add($(this).next("label")).add($(this).next().next("br")).remove().insertAfter(".db:last + label + br"); 
    var url = "myurl.php";
    var db = [];
    $.each($('.db:checked'), function() {
        db.push($(this).val()); 
    });
    if(db.length == 0) { 
        db = "none"; 
    }       
    $.post(url, {db: db}, function(response) {
        $("#dbdisplay").html(response); 
    });
    return true;
});

HTML:

<input type="checkbox" name="db[]" class="db" value="track"/><label for="track">track</label></br>
<input type="checkbox" name="db[]" class="db" value="gps"/><label for="gps">gps</label></br>
<input type="checkbox" name="db[]" class="db" value="accounting"/><label for="accounting">accounting</label></br>

Edit: I ended up answering my own question, but does anyone have documentation (or an explanation) of why this is necessary? It was difficult for me to find the exact answer (thus the posthumous post).

Upvotes: 9

Views: 49580

Answers (4)

Guest
Guest

Reputation: 51

var checkeditems = $('input:checkbox[name="review[]"]:checked')
                       .map(function() { return $(this).val() })
                       .get()
                       .join(",");
$.ajax({
    type: "POST",
    url: "/index.php/openItems/",   
    data: "ids=" + checkeditems,                                        
    success: function(msg) { $(".mainContainer").html(msg); }
});

Upvotes: 5

user223748
user223748

Reputation:

$('input[name="mycheckboxes"]:checked').map(function(){ return $(this).val(); }).get().join(",");

then explode in PHP $mycheckboxes = explode(',',$_GET['mycheckboxes']);

Upvotes: 1

jjclarkson
jjclarkson

Reputation: 5954

You need to have the square brackets to specify an array [] on the submitted variable name.

{'db[]': db}

$(".db").live("change", function() {
    $(this).add($(this).next("label")).add($(this).next().next("br")).remove().insertAfter(".db:last + label + br"); 
    var url = "myurl.php";
    var db = [];
    $.each($('.db:checked'), function() {
        db.push($(this).val()); 
    });
    if(db.length == 0) { 
        db = "none"; 
    }       
    $.post(url, {'db[]': db}, function(response) {
        $("#dbdisplay").html(response); 
    });
    return true;
});

Upvotes: 5

karim79
karim79

Reputation: 342795

I agree with @jjclarkson. Just to add, instead of pushing your ids to an array, you can use $.map:

$(".db").live("change", function() {
    $(this).add($(this).next("label")).add($(this).next().next("br")).remove().insertAfter(".db:last + label + br"); 
    var url = "myurl.php";

    var db = $('.db:checked').map(function(i,n) {
        return $(n).val();
    }).get(); //get converts it to an array

    if(db.length == 0) { 
        db = "none"; 
    }       
    $.post(url, {'db[]': db}, function(response) {
        $("#dbdisplay").html(response); 
    });
    return true;
});

Upvotes: 15

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