开发者

Why this jquery selector gives me error? uncaught exception: Syntax error, unrecognized expression: ''

开发者 https://www.devze.com 2023-03-26 00:18 出处:网络
I am using this code ti dynamically select attributes of a element but it gives me an error in firebug

I am using this code ti dynamically select attributes of a element but it gives me an error in firebug

Error:

uncaught exception: Syntax error, unrecognized expression: ''

Here is my code:

jQuery('.mydata').click(function(){

    var current_id=jQuery(this).attr('id');
    var current_datatype=jQuery(this).attr('datatype');

    var next_id=parseInt(current_id);

    next_id=next_id+1;

    next_id="'#"+next_id+"'";

    var next_datatype=jQuery(next_id).attr('datat开发者_高级运维ype'); //this line gives error

});


What if you do

next_id = "#" + next_id; instead of next_id="'#"+next_id+"'";


when selecting an id you don't need the quotes if you assign it to a variable

change:

next_id="'#"+next_id+"'";
var next_datatype=jQuery(next_id).attr('datatype'); //this line gives error

into:

next_id="#"+next_id;
var next_datatype=jQuery(next_id).attr('datatype'); //this line gives error


I think you don't need the extra quotes here

next_id="'#"+next_id+"'";

should be read

next_id="#"+next_id;


change this ...

next_id="'#"+next_id+"'";

to this ...

next_id="#"+next_id;


Your selector becomes something like '#2' instead of #2. You need to remove the extra '. Also you should never use parseInt without passing in a radix, like parseInt(currentId, 10).

0

精彩评论

暂无评论...
验证码 换一张
取 消

关注公众号