Question

我有问题当我从数据库中获取记录时

mysql_select_db($database_Photo_con, $Photo_con);
$query_rsPhoto = "SELECT * FROM photographs WHERE visible = 1 ORDER BY id ASC";

这显示了所有可见= 1

的记录

如果搜索

mysql_select_db($database_Photo_con, $Photo_con);
$query_rsPhoto = sprintf("SELECT * FROM photographs WHERE caption LIKE %s OR caption_2 LIKE %s  AND visible = 1 ", GetSQLValueString("%" . $colname_rsPhoto . "%", "text"),GetSQLValueString("%" . $colname_rsPhoto . "%", "text"));

它告诉我所有人有谁知道为什么会这样？

Answer 1

这是因为运算符优先级（AND的优先级高于OR），所以你应该使用括号。

$query_rsPhoto = sprintf("SELECT * FROM photographs WHERE (caption LIKE %s OR caption_2 LIKE %s)  AND visible = 1 ", GetSQLValueString("%" . $colname_rsPhoto . "%", "text"),GetSQLValueString("%" . $colname_rsPhoto . "%", "text"));

如果GetSQLValueString没有返回带引号的字符串，则必须在查询中添加它们

$query_rsPhoto = sprintf("SELECT * FROM photographs WHERE (caption LIKE '%s' OR caption_2 LIKE '%s')  AND visible = 1 ", GetSQLValueString("%" . $colname_rsPhoto . "%", "text"),GetSQLValueString("%" . $colname_rsPhoto . "%", "text"));

搜索visible = 1的记录

1 个答案: