CODEWITHSUNDEEP
phpjqueryajaxcodeigniter
user3653474
user3653474

Reputation: 3854

Cannot insert data in database using ajax post jquery request with codeigniter

I want to insert row in database using ajax jquery post method for that i am using the below code in Codeigniter, but my data is not inserted in a database. Please help to sort out my issue.

View:

$("#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>

Model:

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;
   }
}

Controller:

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;
    }

}

Upvotes: 0

Views: 1439

Answers (4)

Enes Tufekci
Enes Tufekci

Reputation: 87

Try these codes. 

("#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);

  }

 ?>

Upvotes: 0

Amit Chauhan
Amit Chauhan

Reputation: 682

have you tried this !

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

instead of

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

Upvotes: 1

Ahmed Khan
Ahmed Khan

Reputation: 516

Have you tried removing method='post' from the Form and submit your data with ajax

Upvotes: 0

Enes Tufekci
Enes Tufekci

Reputation: 87

Try this.

Controller.

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';

}

Model 

 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();
  }

Note : course_goal1, course_goal2, course_goal3 should be same as in database. and course should be database table's name.

This was for update database if you want to insert new data use this model

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

  }

Note 2 : in your database 'id' should be primary and auto incrementing your table name should be 'course' and row names should be 'course_goal1', 'course_goal2', 'course_goal3'

Upvotes: 0

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