doutan5337 2019-07-04 10:15
浏览 67
已采纳

php codeigniter不要将变量更新为mysql

I have a problem with my PHP project with CodeIgniter.

I have a form in which I pass variables to the Database and execute an UPDATE Query by modifying the data on the Database.

The problem is that when I insert the new data it does not give me error, it shows me the view in which it shows the data of the table but the data are not modified.

Can anyone help me understand please?

MODELS Hello_Model.php

function displayrecordsById($id)
 {
 $query=$this->db->query("SELECT * FROM Todolist WHERE id='".$id."'");
 return $query->result();
 }

 function updaterecords($testo,$stato,$id)
 {
     $this->db->query("update Todolist SET testo='$testo',stato='$stato' WHERE id='".$id."'");

 } 

Views update_records.php

<html>
<head>
<title>Registration form</title>
</head>

<body>
 <?php
  //$i=1;
  foreach($data as $row)
  {
  ?>
    <form method="post">
        <table width="600" border="1" cellspacing="5" cellpadding="5">
  <tr>
    <td width="230">Id </td>
    <td width="329"><input type="text" name="id" value="<?php echo $row->id; ?>"/></td>
  </tr>
  <tr>
    <td>Testo </td>
    <td><input type="text" name="testo" value="<?php echo $row->testo; ?>"/></td>
  </tr>
  <tr>
    <td>Stato</td>
    <td><input type='text' name='stato' value= "<?php echo $row->stato; ?> "/></td>
  </tr>
  <tr>
    <td colspan="2" align="center">
    <input type="submit" name="update" value="Update Records"/></td>
  </tr>
</table>
    </form>
    <?php } ?>
</body>
</html>
</html>

Controllers Hello.php

public function updatedata()
 {
 $id=$this->input->get('id');
 $result['data']=$this->Hello_Model->displayrecordsById($id);
 $this->load->view('update_records',$result); 

 if($this->input->post('update'))
 {
 $n=$this->input->post('testo');
 $e=$this->input->post('stato');
 $this->Hello_Model->updaterecords($id,$n,$e);
 redirect("http://simone.fabriziolerose.it/index.php/Hello/dispdata");
        echo "Success!";

 }
 }

I took the code from this tutorial

  • 写回答

1条回答 默认 最新

  • douzhan5058 2019-07-04 10:24
    关注

    i think problem in calling model function check:

    change this:

    $this->Hello_Model->updaterecords($id,$n,$e);
    

    To

     $this->Hello_Model->updaterecords($n,$e,$id);
    

    and you change your custom query to CI query like:

    For Update

    $this->db->where("id",$id);
    $this->db->update("Todolist",array("testo"=>$testo,"stato"=>$stato));
    

    For Select:

    $this->db->where("id",$id);
    $query=$this->db->get("Todolist");
    $row=$query->row();
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 乌班图ip地址配置及远程SSH
  • ¥15 怎么让点阵屏显示静态爱心,用keiluVision5写出让点阵屏显示静态爱心的代码,越快越好
  • ¥15 PSPICE制作一个加法器
  • ¥15 javaweb项目无法正常跳转
  • ¥15 VMBox虚拟机无法访问
  • ¥15 skd显示找不到头文件
  • ¥15 机器视觉中图片中长度与真实长度的关系
  • ¥15 fastreport table 怎么只让每页的最下面和最顶部有横线
  • ¥15 R语言卸载之后无法重装,显示电脑存在下载某些较大二进制文件行为,怎么办
  • ¥15 java 的protected权限 ,问题在注释里