Question

我正在尝试使用mysql LIKE从数据库中获取结果但是在wordpress中它不起作用的是我正在尝试的代码

//this is what i am putting in where clause.
$state = $_POST['state'];
//table name.
$table_name = $wpdb->prefix . 'userprofile';
//trying but this is returning empty
$q = 'SELECT * FROM ' . $table_name . 'WHERE state LIKE \'%' . esc_sql( like_escape( $state ) ) . '%\'';
echo $q;
$result = $wpdb->get_results($q);
if (empty($result)) {
    echo "the result is empty";
}
//returns empty array.
print_r($result);

Answer 1

就像Marc B.说的那样，有一些缺失的引号和不必要的引用..改变你的查询行，对此：

<?php
trait AbstractTrait
{
    public function concreteMethod()
    {
        return $this->abstractMethod();
    }

    public abstract function abstractMethod();
}

class TraitClassTest extends PHPUnit_Framework_TestCase
{
    public function testConcreteMethod()
    {
        $mock = $this->getMockForTrait('AbstractTrait');

        $mock->expects($this->any())
             ->method('abstractMethod')
             ->will($this->returnValue(TRUE));

        $this->assertTrue($mock->concreteMethod());
    }
}
?>

和你的POST行：

$q = "SELECT * FROM $table_name
      WHERE state LIKE '%". esc_sql( like_escape( $state ) ) . "%'
      AND WHERE city LIKE '%". esc_sql( like_escape( $city ) ) . "%'
      AND WHERE session LIKE '%". esc_sql( like_escape( $session ) ) . "%'
      OR WHERE another LIKE '%". esc_sql( like_escape( $another ) ) . "%' ";

Answer 2

你错过了一个空格：

$q = 'SELECT * FROM ' . $table_name . 'WHERE[..snip..]
                                       ^---here

表示你正在制作

SELECT * FROM whateveruserprofileWHERE

这是无效的SQL。

Answer 3

这样做

$ state = esc_sql（$ state）;

$ state = like_escape（$ state）;

$ state =＆＃39;％＆＃39; 。 $ state。＆＃39;％＆＃39 ;;

$ q =＆＃39; SELECT * FROM＆＃39; 。 $ table_name。＆＃39; WHERE状态LIKE＆＃39; $ state＆＃39;;

Wordpress自定义数据库查询

3 个答案: