Question

ajax form表单提交，父页面提交子页面form出错，求大神指导

var fm_account = doc.getElementById('fm_account');获取子页面formid
$.ajax({
cache: true,
type: "POST",
url:'../text.php',
data: fm_account.serialize(),form值序列出错，Uncaught TypeError: undefined is not a function
async: false,
error: function(request) {
alert("Connection error");
},
success: function(data) {

if (data==1){
//刷新列表
alert('保存成功');
}else{
//保存失败
alert('保存失败');
}
//关闭窗口
}
});

Answer 1

fm_account.ajaxSubmit({

})

建议你改成这种形式提交表单