Question

我想要做的是当我从键盘返回100按下右键时再按一次返回200然后再返回300等等...

function looptest() {
    for (i = 100; i < 1000; i+=100) {
       result+=i;
    }
}

$(document).keydown(function (e) {

    switch (e.which) {
        case 37: // left
            // do something
            break;
        case 38: // up
            //do something
            break;
        case 39: // right
           console.log(looptest());
            break;
        case 40: // down
            //do something
            break;
        default:
            return; // exit this handler for other keys
    }
    e.preventDefault(); // prevent the default action (scroll / move caret)
});

这是我的代码，它每次返回100，我的代码中出了什么问题？

Answer 1

在全局空间中指定K = 0

 $(document).keydown(function (e) {

    switch (e.which) {
        case 37: // left
            // do something
            break;
        case 38: // up
            //do something
            break;
        case 39: // right
           K+=100
            break;
        case 40: // down
            //do something
            break;
        default:
            return; // exit this handler for other keys
    }
    e.preventDefault(); // prevent the default action (scroll / move caret)
});

Answer 2

每次按键盘右键，此代码都会向i添加100。

var i = 0;

$(document).keydown(function (e) {

    switch (e.which) {
        case 37: // left
            // do something
            break;
        case 38: // up
            //do something
            break;
        case 39: // right
            i = i+100;
            console.log(i);
            break;
        case 40: // down
            //do something
            break;
        default:
            return; // exit this handler for other keys
    }
    e.preventDefault(); // prevent the default action (scroll / move caret)
});

Answer 3

var result = 0; 
   function looptest() {
        for (i = 100; i < 1000; i+=100) {
           result =(result + i);
        }
    }

    $(document).keydown(function (e) {

        switch (e.which) {
            case 37: // left
                // do something
                break;
            case 38: // up
                //do something
                break;
            case 39: // right
               console.log(looptest());
                break;
            case 40: // down
                //do something
                break;
            default:
                return; // exit this handler for other keys
        }
        e.preventDefault(); // prevent the default action (scroll / move caret)
    });

全局声明结果虽然添加它是concating所以请在里面添加（）

Answer 4

只需做一些像

这样的事情

var count= 100;

$(document).keydown(function (e) {

    switch (e.which) {
        case 37: // left
            // do something
            break;
        case 38: // up
            //do something
            break;
        case 39: // right

            console.log(count)
          count=count+100;

            break;
        case 40: // down
            //do something
            break;
        default:
            return; // exit this handler for other keys
    }
    e.preventDefault(); // prevent the default action (scroll / move caret)
});

或者如果您想使用looptest()，请更改您的功能，如下所示

count=0;
function looptest() {
   count=count+100;
}

如何每次从for循环返回值

4 个答案: