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Store the each column from a multi column result set into a separate variable.

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msvensson@neptunus.(none)
2005-12-20 15:34:58 +01:00
parent b38a183d62
commit 90bbcc9b4d
3 changed files with 109 additions and 14 deletions
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19
mysql-test/r/mysqltest.result
View File

@ -222,6 +222,25 @@ mysqltest: At line 1: Missing arguments to let
mysqltest: At line 1: Missing variable name in let
mysqltest: At line 1: Variable name in =hi does not start with '$'
mysqltest: At line 1: Missing assignment operator in let
var1
hi 1 hi there
hi
1
hi there
var2
2
var2 again
2
2
var3 two columns with same name
1 2 3
2
2
3
mysqltest: At line 1: Missing file name in source
mysqltest: At line 1: Could not open file ./non_existingFile
mysqltest: In included file "./var/tmp/recursive.sql": At line 1: Source directives are nesting too deep

36
mysql-test/t/mysqltest.test
View File

@ -539,6 +539,42 @@ echo $novar1;
--error 1
--exec echo "let hi;" | $MYSQL_TEST 2>&1
# ----------------------------------------------------------------------------
# Test to assign let from query
# let $<var_name>=`<query>`;
# ----------------------------------------------------------------------------
echo var1;
let $var1= `select "hi" as "Col", 1 as "Column1", "hi there" as Col3`;
echo $var1;
echo $var1_Col;
echo $var1_Column1;
echo $var1_Col3;
echo var2;
let $var2= `select 2 as "Column num 2"`;
echo $var2;
echo $var2_Column num 2;
echo $var2_Column;
echo var2 again;
let $var2= `select 2 as "Column num 2"`;
echo $var2;
echo $var2_Column num 2;
echo $var2_Column_num_2;
echo $var2_Column;
echo var3 two columns with same name;
let $var3= `select 1 as "Col", 2 as "Col", 3 as "var3"`;
echo $var3;
echo $var3_Col;
echo $var3_Col;
echo $var3_var3;
#echo failing query in let;
#--error 1
#--exec echo "let $var2= `failing query;`" | $MYSQL_TEST 2>&1
# ----------------------------------------------------------------------------
# Test source command
# ----------------------------------------------------------------------------