查证外键没有建索引的情况代码,PLSQL代码库,德仔网

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【PLSQL】查证外键没有建索引的情况

作者:Dezai.CN / 发布于2013/8/23/ 705


	ps$tkyte@ORA10G> column columns format a30 word_wrappedops$tkyte@ORA10G> column tablename format a15 word_wrappedops$tkyte@ORA10G> column constraint_name format a15 word_wrapped
	select table_name, constraint_name,cname1 || nvl2(cname2,','||cname2,null) ||nvl2(cname3,','||cname3,null) || nvl2(cname4,','||cname4,null) ||nvl2(cname5,','||cname5,null) || nvl2(cname6,','||cname6,null) ||nvl2(cname7,','||cname7,null) || nvl2(cname8,','||cname8,null)columnsfrom ( select b.table_name,b.constraint_name,max(decode( position, 1, column_name, null )) cname1, max(decode( position, 2, column_name, null )) cname2, max(decode( position, 3, column_name, null )) cname3, max(decode( position, 4, column_name, null )) cname4, max(decode( position, 5, column_name, null )) cname5, max(decode( position, 6, column_name, null )) cname6, max(decode( position, 7, column_name, null )) cname7, max(decode( position, 8, column_name, null )) cname8, count(*) col_cnt from (select substr(table_name,1,30) table_name, substr(constraint_name,1,30) constraint_name, substr(column_name,1,30) column_name, position from user_cons_columns ) a, user_constraints b where a.constraint_name = b.constraint_name and b.constraint_type = 'R' group by b.table_name, b.constraint_name ) cons where col_cnt > ALL ( select count(*) from user_ind_columns i where i.table_name = cons.table_name and i.column_name in (cname1, cname2, cname3, cname4, cname5, cname6, cname7, cname8 ) and i.column_position <= cons.col_cnt group by i.index_name ) order by table_name;

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