Question

我有一张桌子：

Building                                        Building
1__bEast                                        1 East
1__bSouth                                       1 South
500__bBldg__d                                   500 Bldg.
501__bBldg__d                                   501 Bldg.
B__u1                                           B-1
B__u2                                           B-2
B__u2__bWest                                    B-2 West
Building__b10__b__PBldg__bTen__p                Building 10 (Bldg Ten)
D__7T__b__PDiagnostic__b__7__bTreatment__p      D & T (Diagnostic & Treatment)
n__fa                                           n/a

左侧的列表显示实际的表格。右边的列表是我希望它们出现在结果中的方式。为了让我这样做，我使用以下嵌套的REPLACE命令：

REPLACE(
    REPLACE(
        REPLACE(
            REPLACE(
                REPLACE(
                    REPLACE(Building,'__P','(')
                ,'__p',')')
            ,'__b',' ')
        ,'__u','-')
    ,'__7','&')
,'__d','.') as [Building]

问题在于“__P”和“__p”被视为相同，结果最终如下：

Building 10 (Bldg Ten(
D & T (Diagnostic & Treatment(

右括号“）”不显示。

我尝试使用：

...REPLACE(REPLACE(Building COLLATE SQL_Latin1_General_CP1_CI_AS,'__P','('),'__p',')')...

但它似乎不适用于嵌套的REPLACE查询。我得到的结果基本相同。

有办法做到这一点吗？

Answer 1

使用COLLATE SQL_Latin1_General_CP1_CS_AS（区分大小写）：

CREATE TABLE #tab(Building VARCHAR(100));

INSERT INTO #tab
SELECT '1__bEast'                                     
UNION ALL SELECT '1__bSouth'                                       
UNION ALL SELECT '500__bBldg__d'                                  
UNION ALL SELECT '501__bBldg__d'                                   
UNION ALL SELECT 'B__u1'                                           
UNION ALL SELECT 'B__u2'                                           
UNION ALL SELECT 'B__u2__bWest'                                    
UNION ALL SELECT 'Building__b10__b__PBldg__bTen__p'              
UNION ALL SELECT 'D__7T__b__PDiagnostic__b__7__bTreatment__p'      
UNION ALL SELECT 'n__fa';

SELECT Building,
REPLACE(
    REPLACE(
        REPLACE(
            REPLACE(
                REPLACE(
                    REPLACE(Building COLLATE SQL_Latin1_General_CP1_CS_AS ,'__P','(')
                ,'__p',')')
            ,'__b',' ')
        ,'__u','-')
    ,'__7','&')
,'__d','.') as [Building] 
FROM #tab

的 LiveDemo

输出：

╔════════════════════════════════════════════╦══════════════════════════════╗ ║ Buidling ║ Replaced ║ ╠════════════════════════════════════════════╬══════════════════════════════╣ ║ 1__bEast ║ 1 East ║ ║ 1__bSouth ║ 1 South ║ ║ 500__bBldg__d ║ 500 Bldg. ║ ║ 501__bBldg__d ║ 501 Bldg. ║ ║ B__u1 ║ B-1 ║ ║ B__u2 ║ B-2 ║ ║ B__u2__bWest ║ B-2 West ║ ║ Building__b10__b__PBldg__bTen__p ║ Building 10 (Bldg Ten) ║ ║ D__7T__b__PDiagnostic__b__7__bTreatment__p ║ D&T (Diagnostic & Treatment) ║ ║ n__fa ║ n__fa ║ ╚════════════════════════════════════════════╩══════════════════════════════╝

根据您的需要，再添加一个REPLACE(..., 'n__fa', 'n/a')或REPLACE(..., '__f', '/')。

如何执行CASE SENSITIVE嵌套REPLACE命令？

1 个答案: