dongxu2398 2016-03-04 17:34
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使用Codeigniter更新MySql中的数据不起作用

MySql database is not updating with the new data entered in view. With following the MVC architecture.

My Controller :

public function saveEdit() {
            $this->load->helper('form');

            $id = $this->uri->segment(3); 

            $save=array(
            'userstory' => $this->input->post('userstoryarea'),
            'datetocomplete' => $this-> input->post('datecomplete'),
            'name'=> $this->input->post('developer')
            );

            $this->load_userStory->saveEdited($id,$save);
            $this->viewUStory();
        }

My Model

public function saveEdited($id,$save){

    $this->db->where('devid', $id);
    $this->db->update('developer', $save);


}

In some of the places i have called some javascript methods which will return some values to display.

My View

<?php foreach ($story as $UserStory): ?>                
        <form method="post" action="<?php echo base_url() ."addstories/saveEdit"?>" >
        <p> Select Devoloper to add : 
        <select class="form-control" id="developer" name ="developer" style="width:200px;"> </p>

         <?php 
            $val2 = $row['name'];
            foreach($developers as $row)
            { 
             echo '<option value="'.$row->full_name.'" >'.$row->full_name.'</option>';
            }
            ?>
            </select>
            <p>

            <p>
            Select a date to complete the project :   <i>

             <input type="date" id="datecomplete"  name ="datecomplete" value="<?php echo $UserStory->datetocomplete;?>" onchange="calculate()">
            </i>    </p>


            <p>             
            <input type="textarea" name="userStory" id="userStory" value=" <?php echo $UserStory->userstory; ?> ">
            </p>

            <input type="submit" name ="dsubmit" value="Save" id="submit">
            </form>
            <?php endforeach; ?>
  • 写回答

1条回答 默认 最新

  • drrqwokuz71031449 2016-03-04 17:40
    关注

    pass the third value id in action like i pass $id

    <form method="post" action="<?php echo base_url() ."addstories/saveEdit/".$id?>" >
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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