Question

我正在构建移动验证功能。具有相同类别的四个输入值。

    function isNumberKeyNext(evt){
    
        var charCode = (evt.which) ? evt.which : event.keyCode;
        if (charCode != 46 && charCode > 31 && (charCode < 48 || charCode > 57))
        {
            return false;
        }
        else
        {
                if ((charCode == 8 || charCode == 46)) {
                    $(this).prev('input').focus();
                } else {
                    $(this).next().focus();
                }
        }
    }

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<input type="number" class="mobile-verify pass" maxlength="1" name="code[]"  onkeypress="return isNumberKeyNext(event)" />
    <input type="number" class="mobile-verify pass" maxlength="1" name="code[]"  onkeypress="return isNumberKeyNext(event)" />
    <input type="number" class="mobile-verify pass" maxlength="1" name="code[]"  onkeypress="return isNumberKeyNext(event)" />
    <input type="number" class="mobile-verify pass" maxlength="1" name="code[]"  onkeypress="return isNumberKeyNext(event)" />

HTML

我只想输入NUMBERS，如果是正确的数字，请关注下一个字段，如果按下退格键，则应该关注前一个字段。我怎样才能做到这一点？我在哪里犯了错误？提前谢谢。

Answer 1

使用此fiddle

HTML：

<div class="container">
<input type="number" id="1" class="mobile-verify pass" maxlength="1" name="code[]"  onkeyup="return isNumberKeyNext(event,this)" maxlength="1"/>
    <input type="number" id="2" class="mobile-verify pass" maxlength="1" name="code[]"  onkeyup="return isNumberKeyNext(event,this)" maxlength="1" />
    <input type="number" id="3" class="mobile-verify pass" maxlength="1" name="code[]"  onkeyup="return isNumberKeyNext(event,this)" maxlength="1"/>
    <input type="number" id="4" class="mobile-verify pass" maxlength="1" name="code[]"  onkeyup="return isNumberKeyNext(event,this)" maxlength="1"/>
    </div>

JS：

function isNumberKeyNext(evt,$this){

        var charCode = (evt.which) ? evt.which : event.keyCode;
        if (charCode != 46 && charCode > 31 && (charCode < 48 || charCode > 57))
        {
            $($this).val('');
            return false;
        }
        else
        {
                if ((charCode == 8 || charCode == 46)) {
                    $($this).prev('input').focus();
                } else {
                    $($this).next('input').focus();
                }
        }
    }

JQuery函数onKeyUp焦点next / previous

1 个答案: