将 IN 运算符与 pg_prepare 一起使用


Using IN operator with pg_prepare

我想将pg_prepare与 IN 运算符一起使用

SELECT id FROM people WHERE name IN $1;

我还想将其与 postgres 数组一起使用,我似乎也无法正常工作,也无法在文档中找到它。

感谢您的帮助,

马克

在这里他们谈论您正在寻找的内容: http://www.php.net/manual/en/function.pg-prepare.php#62675

$result = pg_prepare($dbconn, "my_query", 'SELECT id FROM people WHERE name IN($1,$2,$3)'); 
$result = pg_execute($dbconn, "my_query", array("foo", "bar", "test"));