Codeigniter $this->db->get()，如何返回特定行的值? [英] Codeigniter $this->db->get(), how do I return values for a specific row?

问题描述

假设我有一个包含三列的数据库表:ID、Name 和 Age.我需要找到具有特定(唯一)ID 的用户，然后返回年龄.目前，我正在使用以下代码

Say I have a database table with three columns: ID, Name, and Age. I need to find the user with a specific (unique) ID, and then return the age. Currently, I am using the following code

$this->db->where('id', '3');
$q = $this->db->get('my_users_table');

如何获取该用户的年龄?我想我必须使用

How do I go about getting the age for this user? I think I have to use

$q->result()

但不确定如何在一行中使用它.

But not sure how to use it with one row.

推荐答案

解决方案一

$this->db->where('id', '3');
// here we select every column of the table
$q = $this->db->get('my_users_table');
$data = $q->result_array();

echo($data[0]['age']);

解决方案二

// here we select just the age column
$this->db->select('age');
$this->db->where('id', '3');
$q = $this->db->get('my_users_table');
$data = $q->result_array();

echo($data[0]['age']);

解决方案三

$this->db->select('age');
$this->db->where('id', '3');
$q = $this->db->get('my_users_table');
// if id is unique, we want to return just one row
$data = array_shift($q->result_array());

echo($data['age']);

解决方案四(无活动记录)

$q = $this->db->query('SELECT age FROM my_users_table WHERE id = ?',array(3));
$data = array_shift($q->result_array());
echo($data['age']);

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