GREPARRAY(TABLES("EAPDB"),ITEM!=SQL("EAPDB","SELECT TABLE_NAME FROM all_tables WHERE OWNER <>'EAP' ORDER BY TABLE_NAME",1))
我想把SQL函数取出来的数据去掉，出来的结果还是返回原数组，而且item的使用方法不知道对不对
GREPARRAY(TABLES("EAPDB"),INARRAY(ITEM,SQL("EAPDB","SELECT TABLE_NAME FROM all_tables WHERE OWNER <>'EAP' ORDER BY TABLE_NAME",1))=0)
这个写法有问题
SQL("EAPDB","SELECT TABLE_NAME FROM all_tables WHERE OWNER <>'EAP' ORDER BY TABLE_NAME",1)
出来是一串数组
ITEM是单个值，肯定是不会匹配的
GREPARRAY(TABLES("EAPDB"),inarray(ITEM,SQL("EAPDB","SELECT TABLE_NAME FROM all_tables WHERE OWNER <>'EAP' ORDER BY TABLE_NAME",1) )<=0 )
GREPARRAY(array，fn):函数(返回true或者false)是条件，过滤此数组，最后形成一个新数组。
示例：
GREPARRAY([3,4,2,3,6,8,7], item != 3)等于[4,2,6,8,7].
公式搞复杂了
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