无法使用带有codeigniter的ajax post-jquery请求在数据库中插入数据 - Cannot insert data in database using ajax post jquery request with codeigniter

我想使用ajax jquery post方法在数据库中插入行，因为我在Codeigner中使用以下代码，但我的数据没有插入数据库中。请帮我解决一下问题。

视图：

$("#Submit_Course_Goal").on("click", function (e) {
e.preventDefault();
var dataString = $("form#courseGoalForm").serializeArray();
alert("datastring"+dataString);
$.ajax({
    type: "post",
        url: "<?php echo base_url();?>create_course/create_course_goal",
    cache: false,               
    data: dataString,
    success: function(data){
    alert("data"+data);
    },
    error: function(){                      
    alert('Error while request..');
    }
 });
});
<form name="courseGoalForm" id="courseGoalForm" action="" method="post" enctype="multipart/form-data">
<input type="hidden" name="c_id" value="<?=$result;?>" />
<textarea data-focus="false" rows="8" name="description1"> </textarea>
<textarea data-focus="false" rows="8" name="description2">   </textarea>
<textarea data-lang="en" rows="8" name="description3">  </textarea>
<input type="submit" name="submit" value="Save" class="btn btn-primary btn btn-success" id="Submit_Course_Goal" />
</form>

型号：

public function create_course_goal($data,$id) {
   $this->load->database();
   $this->db->where('id', $id);
   $this->db->update('course', $data);
   $course_id=$id;
   if ($this->db->affected_rows() > 0) {
    return $course_id;
   }
   else
   {
   return false;
   }
}

控制器：

public function create_course_goal(){
    $course_goal1=$this->input->post('description1');
    $course_goal2=$this->input->post('description2');
    $course_goal3=$this->input->post('description3');
    $id=$this->input->post('c_id');  
    $data=array('course_goal1'=>$course_goal1,'course_goal2'=>$course_goal2,'course_goal3'=>$course_goal3);
    $result_course = $this->course_model->create_course_goal($data,$id);
    if($result_course!='false')
    {
        return true;
    }
    else
    {
        return false;
    }
}

你试过了吗！

var dataString=$（"#courseGoalForm"）.serialize（）；

而不是

var dataString=$（"form#courseGoalForm"）.serializeArray（）；

试试这些代码。

("#Submit_Course_Goal").on("click", function (e) {
e.preventDefault();
var description1  = $("#description1").val();
var description2  = $("#description2").val();
var description3 = $("#description3").val(); 

$.ajax({
    type: "post",
        url: "<?php echo base_url();?>create_course/create_course_goal",
    cache: false,               
    data: {
        desc1 : description1,
        desc2 : description2,
        desc3 : description3
    },
    success: function(data){
    console.log(data);
    },
    error: function(){                      
    alert('Error while request..');
    }
 });
});
<!-- Form -->
<form name="courseGoalForm" id="courseGoalForm" action="" method="post" enctype="multipart/form-data" onclick="return false">
    <input type="hidden" name="c_id" value="<?=$result;?>" />
    <textarea data-focus="false" rows="8" name="description1" id="description1"> </textarea>
    <textarea data-focus="false" rows="8" name="description2" id="description2">   </textarea>
    <textarea data-lang="en" rows="8" name="description3" id="description3">  </textarea>
    <input type="submit" name="submit" value="Save" class="btn btn-primary btn btn-success" id="Submit_Course_Goal" />
</form>

<!-- Controller -->
<?php 
    public function create_course_goal(){

    $data=array(
        'ID' => $this->input->post('c_id'),
        'course_goal1'=> $this->input->post('desc1'),
        'course_goal2'=> $this->input->post('desc2'),
        'course_goal3'=> $this->input->post('desc3')
        );
    $result = $this->course_model->create_course_goal($data);
    if ($result) {
      echo 'success';
    }else echo 'fail';
}

/*MODEL*/
function create_course_goal($data = array())
  {
    return $this->db->insert('course',$data);
  }
 ?>

试试这个。

控制器。

public function create_course_goal(){

    $data=array(
        'ID' => $this->input->post('c_id'),
        'course_goal1'=> $this->input->post('description1'),
        'course_goal2'=> $this->input->post('description2'),
        'course_goal3'=> $this->input->post('description3')
        );
    $result = $this->course_model->create_course_goal($data);
    if ($result) {
      echo 'success';
    }else echo 'fail';
}

型号

 function create_course_goal($options = array())
  {
    if(isset($options['course_goal1']))
      $this->db->set('course_goal1',$options['course_goal1']);;
    if(isset($options['course_goal2']))
      $this->db->set('course_goal2',$options['course_goal2']);;
    if(isset($options['course_goal3']))
      $this->db->set('course_goal3',$options['course_goal3']);;

    $this->db->where('ID',$options['ID']);
    $this->db->update('course');
    return $this->db->affected_rows();
  }

注：course_goal1、course_gaol2、course_goal3应与数据库中相同。课程应该是数据库表的名称。

这是为了更新数据库，如果你想插入新的数据使用这个模型

function addNewData($data = array())
  {
    return $this->db->insert('course',$data);
  }

注意2：在您的数据库中，"id"应该是主要的并自动递增您的桌子名称应该是course行名称应为'course_goal1'、'course_goal2'、'couse_goal3'

您是否尝试过从表单中删除method='post'并使用ajax 提交数据