使用 SQL LIKE 和变量 PHP

Question

我无法将 SQL LIKE 与变量一起使用

$sql = "SELECT * FROM chat WHERE name LIKE 'Motherboard' "; //This works

但是如果使用变量它就不起作用:

$sql = "SELECT * FROM chat WHERE name LIKE '%'+$variable+'%' ";
//or
$sql = "SELECT * FROM chat WHERE name LIKE '$variable' ";

我该如何解决？

Answer 1

在 PHP 中，您使用点连接字符串。

 $sql = "SELECT * FROM chat WHERE name LIKE '" . $variable . "' ";

Answer 2

您也可以在双引号内使用变量：

$sql = "SELECT * FROM chat WHERE name LIKE '$variable' ";

或

$sql = "SELECT * FROM chat WHERE name LIKE '{$variable}' ";

请记住双引号比单引号慢。在你的情况下它将是

$sql = "SELECT * FROM chat WHERE name LIKE '%{$variable}%' ";

Using SQL LIKE with variable PHP