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fixed cyclic reference bug
mysql-test/r/subselect.result: test of cyclic reference mysql-test/t/subselect.test: test of cyclic reference sql/share/czech/errmsg.txt: new error message sql/share/danish/errmsg.txt: new error message sql/share/dutch/errmsg.txt: new error message sql/share/english/errmsg.txt: new error message sql/share/estonian/errmsg.txt: new error message sql/share/french/errmsg.txt: new error message sql/share/german/errmsg.txt: new error message sql/share/greek/errmsg.txt: new error message sql/share/hungarian/errmsg.txt: new error message sql/share/italian/errmsg.txt: new error message sql/share/japanese/errmsg.txt: new error message sql/share/korean/errmsg.txt: new error message sql/share/norwegian-ny/errmsg.txt: new error message sql/share/norwegian/errmsg.txt: new error message sql/share/polish/errmsg.txt: new error message sql/share/portuguese/errmsg.txt: new error message sql/share/romanian/errmsg.txt: new error message sql/share/russian/errmsg.txt: new error message sql/share/serbian/errmsg.txt: new error message sql/share/slovak/errmsg.txt: new error message sql/share/spanish/errmsg.txt: new error message sql/share/swedish/errmsg.txt: new error message sql/share/ukrainian/errmsg.txt: new error message
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@ -8,6 +8,8 @@ SELECT (SELECT 1) UNION SELECT (SELECT 2);
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SELECT (SELECT (SELECT 0 UNION SELECT 0));
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(SELECT (SELECT 0 UNION SELECT 0))
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0
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SELECT (SELECT 1 FROM (SELECT 1) HAVING a=1) as a;
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Cyclic reference on subqueries
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drop table if exists t1,t2,t3,t4,t5,attend,clinic,inscrit;
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create table t1 (a int);
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create table t2 (a int, b int);
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@ -46,7 +48,7 @@ a b
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1 7
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2 7
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3 8
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select * from t2 where t2.b=(select a from t3 order by 1 desc limit 1)
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select * from t2 where t2.b=(select a from t3 order by 1 desc limit 1)
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union (select * from t4 where t4.b=(select max(t2.a)*4 from t2) order by a);
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a b
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1 7
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@ -1,6 +1,8 @@
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select (select 2);
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SELECT (SELECT 1) UNION SELECT (SELECT 2);
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SELECT (SELECT (SELECT 0 UNION SELECT 0));
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-- error 1243
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SELECT (SELECT 1 FROM (SELECT 1) HAVING a=1) as a;
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drop table if exists t1,t2,t3,t4,t5,attend,clinic,inscrit;
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create table t1 (a int);
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create table t2 (a int, b int);
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@ -18,7 +20,7 @@ insert into t3 values (6),(7),(3);
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select * from t2 where t2.b=(select a from t3 order by 1 desc limit 1);
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select * from t2 where t2.b=(select a from t3 order by 1 desc limit 1)
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union (select * from t4 order by a limit 2) limit 3;
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select * from t2 where t2.b=(select a from t3 order by 1 desc limit 1)
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select * from t2 where t2.b=(select a from t3 order by 1 desc limit 1)
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union (select * from t4 where t4.b=(select max(t2.a)*4 from t2) order by a);
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explain select * from t2 where t2.b=(select a from t3 order by 1 desc limit 1)
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union (select * from t4 where t4.b=(select max(t2.a)*4 from t2) order by a);
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