0

我的输出应该是这样的:(例如数据)

myArray = [ [otherArray0], [otherArray1], [otherArrayN] ]

我的尝试:

var myArray = [];
var num = 50;
for (i=0; i<num;i++)
{

    var a = 0;
    $('#divID .class select[value!=""]').each(function() //get values from a select
    {
        var otherArray+i = []; //each for cycle is a new "sub array"
        otherArray+i[a] = $(this).val(); //store in array the values from select
        a++
     })
}

我想好了吗?

4

4 回答 4

7

首先

otherArray[i] = []; // create a new array

然后

otherArray[i][a] = $(this).val();

但是您的代码可以变得更简单

var myArray = [];
var num = 50;
for (i=0; i<num;i++)
{
    myArray.push($.map($('#divID .class select[value!=""]'),function(e){
        return e.val();
    }));
}
于 2013-01-31T17:08:02.893 回答
2

设置一个空数组[]pushpush

var myArray = [];
for (int i=0; i<50; i++)
{
    var otherArray = [];
    $(/*...*/).each(function()
    {
       var val = /*...*/;
       otherArray.push(val);
    });
    myArray.push(otherArray);
}
于 2013-01-31T17:12:14.957 回答
1

在很多行之间阅读,我认为这就是你想要的:

var myArray = [];
$("#divID .class select").each(function() {
    var subarray = [];
    $("option[value!='']", $(this)).each(function() {
        subarray.push($(this).val());
    });
    myArray.push(subarray);
}

myArray然后将包含:

[[sel1Opt1, sel1Opt2, sel1Opt3, ...], [sel2Opt1, sel2Opt2, sel3Opt3, ...], ...]
于 2013-01-31T17:43:49.730 回答
0
var num = 50;
for (i=0; i<num;i++)
{
var myArray[i] = array ();
var a = 0;
$('#divID .class select[value!=""]').each(function() //get values from a select
{
    myArray[i].push( $(this).val()); //store in array the values from select

})
}
于 2013-01-31T17:12:59.180 回答