Question

请查看下面的代码..

                <div id="click_me">Save</div> 


<div id="blocks_sortable">
    <div id="block_1">
        <h2>Block 1</h2>

        <div class="items_sortable connectedSortable">
            <div id="item_1"> 
                <span>Item 1</span></div>   
            <div id="item_2"> 
                <span>Item 2</span></div>
            <div id="item_3"> 
                <span>Item 3</span></div>
        </div>
    </div>  
    <div id="block_2">
        <h2>Block 2</h2>

        <div class="items_sortable connectedSortable">
            <div id="item_4"> 
                <span>Item 4</span></div>   
            <div id="item_5"> 
                <span>Item 5</span></div>
            <div id="item_6"> 
                <span>Item 6</span></div>
        </div>
    </div>
</div>

<script>
    $("#click_me").click(function() {
            var result = $("#blocks_sortable > div");

            for(var i=0; i<result.length; i++){
                var str = ""; 
                var str2 = "";
                var block = result[i]; //div object

                //retrieve block id and create the string with id
                str += "block="+$(block).attr("id")+"&items=";

                //trying to select all the items of the current Block
                var result2 = $(block+" > div");                        
                    for(var j=0; j<result2.length; j++){
                        var item = result2[j];

                        str2 += $(item).attr("id")+",";
                    } //end for items

                str = str+str2;
                // looking for a final string in the format of .. block=block_1&items=item_1,item_2,item_3 for loop 1
                alert(str);
            }
    });
</script>

它不起作用。有没有更容易和有效的解决方案来获得这些ID。请帮忙。

谢谢。

Answer 1

jQuery map函数是你的朋友

$.map( 
    $("#blocks_sortable > div"),
    function(elementOfArray){
        var block = $(elementOfArray);
        var children = $.map(
            $("div[id]", block),
            function( eoa ){
                return eoa.id;
            }).join(",");

        return elementOfArray.id + "=" + children;
    }
).join(",");

将返回"block_1=item_1,item_2,item_3,block_2=item_4,item_5,item_6"，这会让你更接近你想去的地方（我想）。

编辑： Solution和source此处。

Answer 2

我认为您在致电str += "block="+$(block).attr("id")+"&items=";

时收到错误消息

.attr无效，因为block是dom元素而不是jquery对象。你也有与var项目相同的问题。

工作解决方案是here。演示是here。

这个javascript修复了它：

   $("#click_me").click(function() {
            var result = $("#blocks_sortable > div");

            for(var i=0; i<result.length; i++){
                var str = "";
                var str2 = "";
                var block = result[i]; //div object

                //retrieve block id and create the string with id
                str += "block="+ block.id +"&items=";

                //trying to select all the items of the current Block
                var result2 = $("#" + block.id + "> div > div");
                    for(var j=0; j<result2.length; j++){
                        var item = result2[j];

                        str2 += item.id +",";
                    } //end for items

                str = str+str2;
                // looking for a final string in the format of .. block=block_1&items=item_1,item_2,item_3 for loop 1
                alert(str);
            }
    });

Answer 3

    $("#click_me").click(function() {
        var result = $("#blocks_sortable > div"); 

        for(var i=0; i<result.length; i++){
            var str = ""; 
            var str2 = "";
            var block = result[i].id; //div object

            //retrieve block id and create the string with id
            //str += "block="+$(block).attr("id")+"&items=";
            str += "block="+ block + "&";
            //trying to select all the items of the current Block
            var inner_span = "#" + block + " > div";
            var result2 = $(inner_span).children();

                for(var j=0; j<result2.length; j++){
                    var item = result2[j];

                    str2 += $(item).attr("id")+",";
                } //end for items

            str = str+str2;
            // looking for a final string in the format of .. block=block_1&items=item_1,item_2,item_3 for loop 1
            alert(str); // output block=block_1&item_1,item_2,item_3 
                        //        block=block_2&item_4,item_5,item_6

        }

Answer 4

尝试没有“＆gt;”

var result = $("#blocks_sortable div");

使用jquery检索子div的id？

4 个答案: