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  • 处理回车事件是否有既兼容ie又兼容火狐的方法

                    document.onkeypress = function (e) {
                            var code;
                            if (!e) {
                                var e = window.event;
                            }
                            if (e.keyCode) {
                                code = e.keyCode;
                            }
                            else if (e.which) {
                                code = e.which;
                            }
                            if (code == 13) {
                                document.getElementById("searchButton").click();
                                return false;
                            }
                        }

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  • 原文地址:https://www.cnblogs.com/zgaspnet/p/2134622.html
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