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Javascript 为什么插件jqueryajax选择器不工作?_Javascript_Jquery_Jquery Plugins - Fatal编程技术网

Javascript 为什么插件jqueryajax选择器不工作?

Javascript 为什么插件jqueryajax选择器不工作?,javascript,jquery,jquery-plugins,Javascript,Jquery,Jquery Plugins,我的插件: (function ($) { $.fn.Conniction_With_DB = function () { this.focusout(function () { var Val = this.val(); alert(Val); }; })(jQuery); 调用我的插件: <script src="../js/jquery.js" type="text/javascript"></script>

我的插件:

 (function ($) {
 $.fn.Conniction_With_DB = function () {
 this.focusout(function () {
        var Val = this.val();
         alert(Val);
 };
 })(jQuery);
调用我的插件:

 <script src="../js/jquery.js" type="text/javascript"></script>        // jquery API
 <script src="../js/JQ_Plug_Ins.js" type="text/javascript"></script>   // jquery Plugin

 <script type="text/javascript">
    $(function () {
         $('input:text').Conniction_With_DB();
    });
 </script>
//jquery API
//jquery插件
$(函数(){
$('input:text')。与_DB()的连接;
});
问题是:为什么我的插件在我调用它时没有响应

PS:我认为插件中的错误在这一部分(
this.focusout


或者在这部分的调用插件中(
$('input:text')
)..

出现语法错误,您忘记了一些大括号(第6行),您必须使用
$(this.val()
,而不是
this.val()
将是
focusout
回调中的DOM对象,而不是jQuery对象

(function ($) {
    $.fn.Conniction_With_DB = function () {
        this.focusout(function () {
            var Val = $(this).val();
            alert(Val);
        });
    };
})(jQuery);

jquery的哪个版本。1.4中增加了focusout