因此,我具有此功能来搜索MySQL数据库中的条目：

private function SearchContributors($search)

{

$search_pieces = explode(' ', $search);

if (count($search_pieces) == 1 )

{

$this->db->like('firstname', $search);

$this->db->or_like('lastname', $search);

$result = $this->db->get(); //the line from the error message below

}

//Other stuff for 2 and more pieces

return $result;

}

?>

我有两次使用该功能.

案例A是用户发起的搜索,并从URL(domain.com/contributors/?x=paul)获取搜索查询.这很好.

if (isset($_GET['x']))

{

$x = $_GET['x'];

$result = $this->SearchContributors($x);

}

?>

情况B是用户输入无效的子弹名称(domain.com/contributors/paul代替domain.com/contributors/pauline-surname)并直接获取搜索查询时的备份：

$this->db->where('slug', $slug);

$result = $this->db->get();

if ($result->num_rows() == 0)

{

$x = str_replace('-', ' ', $slug);

$result = $this->SearchContributors($x);

}

?>

这返回了MySQL语法错误：

Error Number: 1064

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘WHERE firstname LIKE ‘%paul%’ OR lastname LIKE ‘%paul%” at line 2

SELECT * WHERE firstname LIKE ‘%paul%’ OR lastname LIKE ‘%paul%’

Filename: /www/htdocs/w00a94ee/c/controllers/contributors.php

Line Number: 23

在这两种情况下,该函数都具有完全相同的字符串保罗,那么为什么它不起作用？

//编辑

function __construct()

{

parent::__construct();

$this->load->database('databasename');

$this->db->from('tablename');

}

解决方法:

您忘记了指定要从中选择哪个表.

$this->db->from('tablename');

编辑：问题是您在构造函数中添加from,然后您调用：

$this->db->where('slug', $slug);

$result = $this->db->get();

在致电SearchContributors之前.这将运行查询并重置变量.

因此,当您调用SearchContributors时,将不再设置FROM.

您需要将$this-> db-> from(‘tablename’);在SearchContributors而不是构造函数中.使模型函数自成一体,而不需要外部函数(例如__construct来调用它们)通常是一个好主意.

标签：codeigniter,mysql,php

来源： https://codeday.me/bug/20191127/2077035.html