Question

我有这个sql代码，我知道我只有一个结果，但查询结果是一个数组。如何获取字符串或int值？

$MEMBER_ID = $this->db
     ->select ('MEMBER_ID', false)
     ->from($this->table_member)   
     ->where('MEMBER_EMAIL, $this->input->post('InputEmail'))
     ->get()
     ->result();

使用CodeIgniter。

谢谢，

Answer 1

如果您确切知道该查询只获得一个结果，那么从第一个数组元素中取字段MEMBER_ID：

$MEMBER_ID = $this->db
             ->select ('MEMBER_ID', false)
             ->from($this->table_member)   
             ->where('MEMBER_EMAIL, $this->input->post('InputEmail'))
             ->get()
             ->result()[0]['MEMBER_ID'];

或者像这样使用更简单的查询：

$MEMBER_ID = $this->db->query("Select MEMBER_ID from table_member where id = $id")->row()->MEMBER_ID;

Answer 2

对于这样一个简单的查询，您不需要所有这些Query Builder调用。这样可以正常工作。

$id = $this->input->post('InputEmail');

$MEMBER_ID = $this->db
     ->query("Select MEMBER_ID from {$this->table_member} where id = $id")
     ->row()
     ->MEMBER_ID;

但如果你坚持不懈的话

$MEMBER_ID = $this->db
     ->select ('MEMBER_ID', false)
     ->from($this->table_member)   
     ->where('MEMBER_EMAIL, $this->input->post('InputEmail'))
     ->get()
     ->row()
     ->MEMBER_ID;

Answer 3

如果您知道查询应该只返回1个结果，那么请不要使用ressult()或result_array()

检查一下： Generating Query Results

所以在你的代码上你可以试试这个

{p}在row()例外：

$MEMBER_ID = $this->db
     ->select ('MEMBER_ID', false)
     ->from($this->table_member)   
     ->where('MEMBER_EMAIL, $this->input->post('InputEmail'))
     ->get()
     ->row()->MEMBER_ID;
return $MEMBER_ID;

row_array()

中的

$MEMBER_ID = $this->db
         ->select ('MEMBER_ID', false)
         ->from($this->table_member)   
         ->where('MEMBER_EMAIL, $this->input->post('InputEmail'))
         ->get()
         ->row_array()['MEMBER_ID'];
return $MEMBER_ID;

如何将箭头转换为Int / String值PHP / CodeIgniter

3 个答案: