如何在codeigniter中使用ajax在mysql数据库中插入数据？
Posted
技术标签:
【中文标题】如何在codeigniter中使用ajax在mysql数据库中插入数据？【英文标题】：How To insert data in mysql database using ajax in codeigniter? 【发布时间】：2015-09-14 11:53:54 【问题描述】：我正在尝试使用 ajax 验证我的表单并在 mysql 数据库中插入数据。提交验证既未发生，也未插入数据。我在 codeigniter 框架中进行此操作。我是 ajax 的新手。我无法弄清楚哪里出错了。这是我的代码
查看：
<script type="text/javascript">
function validate_name(first_name)
if(first_name.trim() == '' || first_name.length == 0)
$('.first_name').show();
$('.first_name').text('Please enter your name');
return false;
else
$('.first_name').hide();
return true;
function validate_email(email_id)
var pattern = new RegExp(/^[+a-zA-Z0-9._-]+@[a-zA-Z0-9.-]+\.[a-zA-Z]2,4$/i);
if(email_id.trim() == '' || email_id.length == 0)
$('.email-id').show();
$('.email-id').text('Please enter email address');
return false;
else if(!pattern.test(email_id))
$('.email-id').show();
$('.email-id').text('Please enter valid email address');
return false;
else
$('.email-id').hide();
return true;
function validate_inquiry_form(first_name,email_id)
var username_validate = validate_name(first_name);
var email_validate = validate_email(email_id);
if(username_validate == true && email_validate == true)
return true;
else
return false;
$('#submit_enquiry').click(function()
var first_name = $("input[name=first_name]").val();
var last_name = $("input[name=last_name]").val();
var dob = $("input[name=dob]").val();
var gender = $("input[name=gender] :radio:checked").val();
var email_id = $("input[name=email_id]").val();
var password = $("input[name=password]").val();
var address = $("input[name=address]").val();
var phone = $("input[name=phone]").val();
var zipcode = $("input[name=zipcode]").val();
var validate_form = validate_inquiry_form(first_name,email_id);
if(validate_form == true)
$.ajax(
url:'<?php echo base_url(); ?>member/register',
type:'POST',
data:
first_name : first_name ,last_name : last_name ,dob : dob ,male : male ,female : female , email_id : email_id ,password : password ,phone : phone , address : address , zipcode : zipcode
,
success: function(data)
console.log(data);
);
else
return false;
return false;
);
</script>
<form id="registration-form">
<div class="register">
<div class="row">
<div class="col1">
<label for="first_name">First Name<span>*</span></label> <br/>
<input type="text" name="first_name"/>
<li class="first_name error"></li>
</div>
<div class="col2">
Last Name<br/>
<input type="text" name="last_name"/>
</div>
</div>
<div class="row">
<div class="col1">
Date Of Birth <br/>
<input type="text" name="dob"/>
</div>
<div class="col2">
Gender
<br/>
<input type="radio" name="gender" value="Male" /> Male
<input type="radio" name="gender" value="Female" /> Female
</div>
</div>
<div class="row">
<div class="col1">
Email<br/>
<input type="text" name="email_id"/>
<li class="email-id error"></li>
</div>
<div class="col2">
Password<br/>
<input type="password" name="password"/>
</div>
</div>
<div class="row">
<div class="col">
Address<br/>
<textarea name="address" rows="2" ></textarea>
</div>
</div>
<div class="row">
<div class="col1">
Zipcode<br/>
<input type="text" name="zipcode"/>
</div>
<div class="col2">
Phone<br/>
<input type="text" name="phone"/>
</div>
</div>
<div class="row">
<div class="col3">
<input class="" type="button" id="submit_enquiry" name="submit_enquiry" value="Submit Enquiry" />
</div>
</div>
</div>
</form>
控制器：
public function register_user()
$register_user = $this->member_model->add_user();
if($register_user)
return true;
else
return false;
型号：
public function add_user()
$add_user = array(
'mem_name'=> $this->input->post('first_name'),
'mem_lastname'=> $this->input->post('last_name'),
'mem_dob'=> $this->input->post('dob'),
'mem_gender'=> $this->input->post('gender'),
'mem_email'=> $this->input->post('email_id'),
'mem_address'=> $this->input->post('address'),
'mem_zipcode'=> $this->input->post('zipcode'),
'mem_phone'=> $this->input->post('phone'),
'mem_password'=> $this->input->post('password'),
);
$insert = $this->db->insert('membership', $add_user);
$insert_id = $this->db->insert_id();
return $insert_id;
请帮帮我....
【问题讨论】：
提交表单时，我会在控制器端使用 form_validation， 什么确切不起作用？ AJAX 请求本身？在 PHP 中处理它？将数据写入数据库？ 【参考方案1】：在你的ajax中改变这个添加撇号（'）
data:
'first_name' : first_name ,'last_name' : last_name ,'dob' : dob , etc...
,
将此添加到您的控制器成员中
function register()
$add_user = array(
'mem_name'=> $this->input->post('first_name'),
'mem_lastname'=> $this->input->post('last_name'),
'mem_dob'=> $this->input->post('dob'),
'mem_gender'=> $this->input->post('gender'),
'mem_email'=> $this->input->post('email_id'),
'mem_address'=> $this->input->post('address'),
'mem_zipcode'=> $this->input->post('zipcode'),
'mem_phone'=> $this->input->post('phone'),
'mem_password'=> $this->input->post('password'),
);
$this->member_model->add_user($add_user);
在您的 member_model 中
function add_user($add_user)
$insert = $this->db->insert('membership', $add_user);
【讨论】：
我尝试了您的解决方案，但它不起作用.. db @ Arif 中的表单验证和数据都没有输入 什么表单验证？我这里没有看到那个代码，只是js表单验证【参考方案2】：您可以在提交处理程序上使用 jquery validate 和 ajay。请尝试以下代码：
$(function ()
$("#FormId").validate(
errorElement: 'span', errorClass: 'help-block',
rules:
Tag:
required: true
,
Table:
required: true
,
Field:
required: true
,
,
messages:
Tag:
required: "Tag is required."
,
Table:
required: "Table is required."
,
Field:
required: "Field is required."
,
,
highlight: function (element)
$(element).closest('.form-group').addClass('has-error');
,
unhighlight: function (element)
$(element).closest('.form-group').removeClass('has-error');
,
success: function (element)
$(element).closest('.form-group').removeClass('has-error');
$(element).closest('.form-group').children('span.help-block').remove();
,
errorPlacement: function (error, element)
error.appendTo(element.closest('.form-group'));
,
submitHandler: function (form)
$("#saveButton").button('loading');
$.post('', $("#FormId").serialize(), function (data)
//Write your code here
);
);
);```
【讨论】：
以上是关于如何在codeigniter中使用ajax在mysql数据库中插入数据？的主要内容，如果未能解决你的问题，请参考以下文章
Codeigniter 如何在控制器中接收 ajax post 数据
当数据在javascript codeigniter php中使用ajax成功功能时，如何禁用页面重新加载？
如何创建ajax请求以使用DataTable和Codeigniter并在同一文件中返回