Question

假设我有两张桌子;表A和表格如下所示：

IL_0000: ldarg.0
IL_0001: ldfld class [System.Windows.Forms]System.Windows.Forms.Button ClassLibrary1.Test::_btn
IL_0006: ldc.i4.1
IL_0007: callvirt instance void [System.Windows.Forms]System.Windows.Forms.Control::set_Enabled(bool)
IL_000c: ldarg.0
IL_000d: ldfld class [System.Windows.Forms]System.Windows.Forms.Button ClassLibrary1.Test::_btn
IL_0012: call valuetype [System.Drawing]System.Drawing.Color [System.Drawing]System.Drawing.Color::get_Red()
IL_0017: callvirt instance void [System.Windows.Forms]System.Windows.Forms.Control::set_ForeColor(valuetype [System.Drawing]System.Drawing.Color)
IL_001c: ldarg.0
IL_001d: ldfld class [System.Windows.Forms]System.Windows.Forms.Button ClassLibrary1.Test::_btn
IL_0022: ldarg.0
IL_0023: ldfld class [System.Windows.Forms]System.Windows.Forms.Button ClassLibrary1.Test::_btn
IL_0028: callvirt instance class [System.Drawing]System.Drawing.Font [System.Windows.Forms]System.Windows.Forms.Control::get_Font()
IL_002d: callvirt instance class [System.Drawing]System.Drawing.FontFamily [System.Drawing]System.Drawing.Font::get_FontFamily()
IL_0032: ldarg.0
IL_0033: ldfld class [System.Windows.Forms]System.Windows.Forms.Button ClassLibrary1.Test::_btn
IL_0038: callvirt instance class [System.Drawing]System.Drawing.Font [System.Windows.Forms]System.Windows.Forms.Control::get_Font()
IL_003d: callvirt instance float32 [System.Drawing]System.Drawing.Font::get_Size()
IL_0042: ldc.i4.1
IL_0043: ldc.i4.2
IL_0044: newobj instance void [System.Drawing]System.Drawing.Font::.ctor(class [System.Drawing]System.Drawing.FontFamily, float32, valuetype [System.Drawing]System.Drawing.FontStyle, valuetype [System.Drawing]System.Drawing.GraphicsUnit)
IL_0049: callvirt instance void [System.Windows.Forms]System.Windows.Forms.Control::set_Font(class [System.Drawing]System.Drawing.Font)
IL_004e: ret

如果我尝试使用完全连接加入它们，结果将取决于我在select语句中使用的表。例如，以下查询将生成以下结果

A
Color   ID
Blue    1
Green   2
Red     3

B
Color   ID
Blue    1
Brown   2
Red     3

如果我决定在select语句中使用B.color而不是A.color，我会得到以下结果：

select A.color, count(*)
from A 
full join B on a.color = B.color
group by 1
order by 1

color   count
Blue    1
Green   1
Red     1
        1

如何让结果集包含所有颜色值。我知道我可以使用unionall完成，并且我可以在select语句中使用case语句，当另一个为null时使用一个case语句，但是还有另一种更简洁的方法来实现它吗？

谢谢，本

Answer 1

如果值存在于一个表而不是另一个表中，请使用coalesce从另一个表中获取值。

select coalesce(A.color,B.color) as color, count(*)
from A 
full join B on a.color = B.color
group by 1
order by 1

如何在完全加入时不丢失记录

1 个答案: