jQuery的形式没有提交
问题描述:
这是我写过的提交表单后验证并接收PHP文件中的值使用$ _POST的脚本,但我没有获得值在PHP文件。当我尝试回声一个尊重其显示空白php.Please指引,我是新来的jQueryjQuery的形式没有提交
<script>
$("#changepassform").validate({
rules: {
old_password: "required",
password: "required",
password2: {
equalTo: "#password"
},
},
messages: {
old_password: "Please enter old password",
password: "Please enter new password",
password2: " Enter Confirm Password Same as Password"
},
submitHandler: function(form) {
var current_password = $("#current_password").val();
var new_password = $("#password").val();
var comfirm_password = $("#password2").val();
var id = $("#id").val();
var dataString = 'newpassword1=' + new_password + '&id1=' + id;
$.ajax({
type: "POST",
url: "changepassword.php",
data: "dataString",
success: function(response) {
$("#status").html(response);
}
});
}
//form.submit();
//return false;
});
// required to block normal submit since you used ajax
//form.submit();
</script>
答
错误在发送数据..
<script>
$("#changepassform").validate({
rules: {
old_password: "required",
password: "required",
password2: {
equalTo: "#password"
},
},
messages: {
old_password: "Please enter old password",
password: "Please enter new password",
password2: " Enter Confirm Password Same as Password"
},
submitHandler: function(form) {
var current_password = $("#current_password").val();
var new_password = $("#password").val();
var comfirm_password = $("#password2").val();
var id = $("#id").val();
var dataString = 'newpassword1=' + new_password + '&id1=' + id;
$.ajax({
type: "POST",
url: "changepassword.php",
data: dataString,
success: function(response) {
$("#status").html(response);
}
});
}
//form.submit();
//return false;
});
// required to block normal submit since you used ajax
//form.submit();
</script>
使用serialize()
method.The序列化值可以是在进行AJAX请求时在URL查询字符串中使用。
更改数据:“dataString”为data:dataString, –
而不是data:“dataString”,update为dataString,并尝试解析表单。 –
谢谢我的朋友prakash,但什么是不引用dataString后面的原因...将不胜感激 – Hemant