在pg_prepare中使用LIKE通配符

Question

我一直在尝试在预准备语句中使用LIKE,但由于使用通配符%导致语法错误,php不会执行该语句.

这是代码

$query = pg_prepare($conn, "MyStatement", 
    'SELECT "Query" from "MyTable" 
     WHERE "Query" LIKE $1% 
     ORDER BY "MyColumn" DESC;');

$result = pg_execute($conn, "MyStatement", array($my_param));

问题是,php在第二行中向我显示了一个声明语法错误的警告.

非常感谢你提前!!!

Answer 1

我使用PDO适配器遇到了相同的问题绑定参数.解决方案是使用变量传递"%":

$query = pg_prepare($conn, "MyStatement", 
'SELECT "Query" from "MyTable" 
 WHERE "Query" LIKE $1 
 ORDER BY "MyColumn" DESC;');

$result = pg_execute($conn, "MyStatement", array($my_param."%"));

如果你需要

...LIKE '%param%' ...

然后你的查询将是:

$result = pg_execute($conn, "MyStatement", array("%".$my_param."%"));