ajax有什么问题提交电话?

时间:2015-02-13 05:31:25

标签: jquery ajaxsubmit

$scope.submitTheForm = function(htmlcode){  
        var idOfForm = "formOfCalCPDF";
        var dataPassed = $.param({'htmlcode':htmlcode,'mode':'getpdf'});
        alert("coming here");
        $(idOfForm).ajaxSubmit({
            url: 'ggs.erm.payrollJava.Taxsummary',
            type: "POST",
             data: dataPassed,
            error:function (){},
            success: function (data){}
        });
    };

当我调用此函数时,虽然正在执行警报,但没有POST请求,您是否看到任何类型的问题?

1 个答案:

答案 0 :(得分:2)

您忘了id选择器试试这个: -

$scope.submitTheForm = function(htmlcode){  
    var idOfForm = "formOfCalCPDF";
    var dataPassed = $.param({'htmlcode':htmlcode,'mode':'getpdf'});
    alert("coming here");
    $('#'+idOfForm).ajaxSubmit({
        url: 'ggs.erm.payrollJava.Taxsummary',
        type: "POST",
         data: dataPassed,
        error:function (){},
        success: function (data){}
    });
};