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handle ExpressionAnalyzer exceptions, while trying to optimize order by, in case of complicated queries

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This commit is contained in:
CurtizJ 2019-07-27 03:02:04 +03:00
parent b1d981ec3a
commit cb85721cba
1 changed files with 13 additions and 12 deletions
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25
dbms/src/Interpreters/InterpreterSelectQuery.cpp
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@ -660,19 +660,10 @@ static SortingInfoPtr optimizeSortingWithPK(const MergeTreeData & merge_tree, co
const auto & sorting_key_columns = merge_tree.getSortingKeyColumns();
size_t prefix_size = std::min(order_descr.size(), sorting_key_columns.size());
auto is_virtual_column = [](const String & column_name)
{
return column_name == "_part" || column_name == "_part_index"
|| column_name == "_partition_id" || column_name == "_sample_factor";
};
auto order_by_expr = query.orderBy();
auto syntax_result = SyntaxAnalyzer(context).analyze(order_by_expr, merge_tree.getColumns().getAll());
auto syntax_result = SyntaxAnalyzer(context).analyze(order_by_expr, merge_tree.getColumns().getAllPhysical());
for (size_t i = 0; i < prefix_size; ++i)
{
if (is_virtual_column(order_descr[i].column_name))
break;
/// Read in pk order in case of exact match with order key element
/// or in some simple cases when order key element is wrapped into monotonic function.
int current_direction = order_descr[i].direction;
@ -680,8 +671,18 @@ static SortingInfoPtr optimizeSortingWithPK(const MergeTreeData & merge_tree, co
prefix_order_descr.push_back(order_descr[i]);
else
{
const auto & ast = query.orderBy()->children[0];
auto actions = ExpressionAnalyzer(ast->children.at(0), syntax_result, context).getActions(false);
const auto & ast = query.orderBy()->children[i];
ExpressionActionsPtr actions;
try
{
ExpressionAnalyzer(ast->children.at(0), syntax_result, context).getActions(false);
}
catch (const Exception &)
{
/// Can't analyze order expression at this stage.
/// May be some actions required for order will be executed later.
break;
}
const auto & input_columns = actions->getRequiredColumnsWithTypes();
if (input_columns.size() != 1 || input_columns.front().name != sorting_key_columns[i])