5.0.0 - 普通 - 已关闭
例如我写了如下语句 $subQuery1=$databa
$result['rows'] = $databa
return json ($result);
最终返回的结果类似
"SELECT T1.* FROM (SELECT thinkphp.*, ROW_NUMBER() OVER ( ORDER BY rand()) AS ROW_NUMBER FROM (SELECT [fnumber] FROM ( SELECT T1.* FROM (SELECT thinkphp.*, ROW_NUMBER() OVER ( ORDER BY rand()) AS ROW_NUMBER FROM (SELECT [fnumber] FROM [t_ICItem]) AS thinkphp) AS T1 ) a) AS thinkphp) AS T1"
难道不应该返回最终查询数据库结果集，怎么返回构建的查询数据库语句？
就是只要我代码任何地方有写->buildSql()；无论其他地方是否有利用子查询，最终返回的结果都是完整的SQL语句文本，而不是理论上查询数据库内容的数据。
把$subQuery1=$databa
我的环境是微软SQL SERVER 链接。PHP 7.1 。另外相同的代码在THINKPHP5.0版本可以正常读取数据，thinkphp5.1最新版本却不能。
