为什么我的 jquery 代码不起作用？

Question

<script>
        var i=0;
        $(document).ready(function(){
           $("#tolevel_" + (i+1) ).click(function(){
              $("#level_" + i).hide(500,'swing', function(){
                  $("#level_" +(i+1)).show(500, 'swing', function(){
                      i=i+1;
                  });

              });
           });
            $("#backtolevel_" + (i-1) ).click(function(){
                $("#level_" + i).hide(500,'swing', function(){
                    $("#level_" +(i-1)).show(500, 'swing', function(){
                        i=i-1;
                    });
                });
            });
        });
    </script>

Ok here is my jquery. The level ids are divs and tolevel and backtolevel are two buttons in each. for example if the div id is "level_1" the ids of the buttons in that div are "backtolevel_0" and "tolevel_2". Ok it only works the first time. it doesn't matter on what level I am. I have tested it on different levels but it only works once. don't know why! can you guys help me? I think somewhat the incrementing or decrementing are not working.

Answer 1

Try with following code:

<script>

        $(document).ready(function(){
           $("[id^=tolevel_]").click(function(){
                var currentID = $(this).attr("id");
                var number = currentID.replace("tolevel_","");//get the number in string format
                number = parseInt(number);
                var i = number -1;
                $("#level_" + i).hide(500,'swing', function(){
                  $("#level_" +(i+1)).show(500, 'swing', function(){

                  });
                });
           });

            $("[id^=backtolevel_]").click(function(){
                var currentID = $(this).attr("id");
                var number = currentID.replace("backtolevel_","");//get the number in string format
                number = parseInt(number);
                var i = number + 1;
                $("#level_" + i).hide(500,'swing', function(){
                    $("#level_" +(i-1)).show(500, 'swing', function(){

                    });
                });
            });
        });
    </script>

It user `jquery's Attribute Starts With Selector. See more about it at https://api.jquery.com/attribute-starts-with-selector/ . Really no need to write code for each and every element. This should work.