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

Question

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;
    }

}

Value for datastring is [object Object],[object Object],[object Object],[object Object] — user3653474
– user3653474, Commented Jan 1, 2016 at 11:43

Amit Chauhan · Accepted Answer · 2016-01-01 12:46:26Z

1

have you tried this !

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

instead of

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

answered Jan 1, 2016 at 12:46

Amit Chauhan

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1 Comment

user3653474 Over a year ago

Yes used but not found the solution

Enes Tufekci · Accepted Answer · 2016-01-01 12:48:10Z

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);

  }

 ?>

Enes Tufekci · Accepted Answer · 2016-01-01 12:20:23Z

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'

public function testCourse(){ $data=array( 'ID' => '1', 'course_goal1'=> 'test1', 'course_goal2'=> 'test2', 'course_goal3'=> 'test3' ); $result = $this->course_model->create_course_goal($data); if ($result) { echo 'success'; }else echo 'fail'; } what is the result
btw are you triying to set a new record to database or update.
If i fill the field value manually i get the record inserted in the database.
like $data=array('course_goal1'=>'record1','course_goal2'=>'record2','course_goal3'=>'record3');
Enes- I found the problem actually its textarea i am getting value from textbox but cannot get value from textarea how to solve this

Ahmed Khan · Accepted Answer · 2016-01-01 12:33:05Z

0

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

answered Jan 1, 2016 at 12:33

Ahmed Khan

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5 Comments

user3653474 Over a year ago

I have to submit it using post method

Ahmed Khan Over a year ago

YOu are already doing it by ajax why are you adding it two times?

user3653474 Over a year ago

I am getting value from textbox but cannot get value using textarea

Ahmed Khan Over a year ago

have you used network tab in web console to see whether the value is submitting via post or not?

Ahmed Khan Over a year ago

This <input type="hidden" name="c_id" value="<?=$result;?>" /> Should be like this <input type="hidden" name="c_id" value="<?php echo $result;?>" />

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