Question

如果在数据库中存在codeigniter，我想插入替换这是我的模特

public function upload($id_akun,$data){
               $query = $this->db->query("SELECT * FROM akun WHERE
id_akun = '{$data['id_akun']}' ");
$result = $query->result_array();
$count = count($result);

if (empty($count)) {

     $this->db->insert('foto', $data);  
}
elseif ($count == 1) {
   $this->db->where('id_akun', $data['id_akun']);
                    $this->db->update('foto', $data);  
}

我成功替换（更新）数据（如果存在）但不插入数据if（empty（$ count））

有条件的（$ count == 1）---＆gt;没关系 condtional（空（$ count））---＆gt;问题？

Answer 1

试试这个

在模型中

public function upload($id_akun,$table){
    $query = $this->db->query("SELECT * FROM akun WHERE id_akun = '$id_akun' ");
    $result = $query->result_array();
    $count = count($result);

    if (empty($count)){
        $this->db->insert('foto', $table);
        return 0; 
    }
    elseif ($count == 1) {
        $this->db->where('id_akun', $id_akun);
        $this->db->update('foto', $table);  
        return 1;
    }
    elseif (($count >= 1) {
        return 2;
    }
}

在控制器中

$id_akun = 25;
$table = 'table_name';

$result = $this->model_name->upload($id_akun,$table);

switch ($result) {
    case 0:
        $output = 'Data Inserted'
        break;
    case 1:
        $output = 'Record Inserted'
        break;
    case 2:
        $output = 'Multiple Record Found'
        break;
}

Answer 2

你可以尝试这样..

public function upload($id_akun,$data){       
   $this->db->where('id_akun',$data['id_akun']);
   $this->db->from('akun');
   $query = $this->db->get();
   $rowcount = $query->num_rows();

  if ($rowcount==0) {

    $this->db->insert('foto', $data);  
  }
 else {
   $this->db->where('id_akun', $data['id_akun']);
                $this->db->update('foto', $data);  
  }

Answer 3

您的问题是，您正在检查count()的结果是否为空。这将总是评估为假。

public function upload($id_akun, $data)
{
    //use active record methods, otherwise, you need to sanitize input
    $this->db->select('*');
    $this->db->where('id_akun', $data['id_akun']);
    $query = $this->db->get('akun'); 

    if($query->num_rows() == 0){
        $this->db->update...
    }else{
         ....
    }
}

Answer 4

解释

如果在抓取记录中找不到任何内容，则 result_array（）函数会导致空数组。因此，当您获取记录时，它会返回空数组，并且您正在计算这个空数组，它返回零长度，该长度不等于空，因此您的条件始终保持为假。

您可以使用以下代码。

<强>代码

     function upload($id_akun, $data) {
       //use active record methods, otherwise, you need to sanitize input
       $this->db->select('*');
       $this->db->where('id_akun', $data['id_akun']);
       $query = $this->db->get('akun');
       if ($query->num_rows() == 0) {

           //write your Insert query here

       } elseif ($query->num_rows() > 0) {

          //write your Update query here 

       }
     }

Answer 5

使用Active Record（或查询生成器很快就会调用它）有几个好处：

安全
代码清洁度
数据库无关的

所以，所有时间都尝试使用活动记录。

<?php
public function upload($id_akun,$data){ 
    $query = $this->db->select('*')->where('id_akun', $data['id_akun'])->from('akun')->get();
    $count = $query->num_rows();
    $result = $query->result_array();

    if ($count == 0)
        return $this->db->insert('foto', $data);
    }
    elseif ($count > 1) {
        $this->db->where('id_akun', $data['id_akun']);
        return $this->db->update('foto', $data);  
    }
}

在codeigniter中插入或替换

5 个答案: