以下是引用ycvf在2021-4-1 17:44:23的发言：

这个函数GETWORDNUM(jmss1, lnj, ",")没见过，有知道的吗、

以下是引用厨师王德榜在2021-4-1 17:15:35的发言：

我的代码,按照要求不用任何Select ... 语句 ,另辟蹊径用了字典对象,供大家参考:
IF NOT USED('sp') THEN
    USE "C:\sp.dbf" IN 0 EXCLUSIVE  &&  改成你的路径
ENDIF
LOCAL icnt as Integer ,ii as Integer
LOCAL ckey as String ,str1 as String
LOCAL time11 as Datetime ,time22 as Datetime

IF  USED('result1') THEN
    USE IN result1
ENDIF

LOCAL  oDic As Object
oDic = CreateObject("Scripting.Dictionary")
time11 = SECONDS()
SELECT sp
GO TOP
SCAN
    str1 = ALLTRIM(sp.jmss1)
    icnt = ALINES(arr2,str1,1,",")
    FOR ii =1 TO icnt
        ckey = LTRIM(STR(ii)) - "_" - arr2(ii)
        IF odic.Exists(ckey) THEN
            odic.Item(ckey) = odic.Item(ckey) + 1
        ELSE
            odic.Add( ckey,1) &&
        ENDIF     
    ENDFOR

ENDSCAN

DIMENSION arrKey(1) as String
arrKey =odic.Keys

CREATE CURSOR result1 (第几题 c(4) , 分值 c(4),次数 i )
FOR ii = 1 TO odic.Count  &&
*!*        tmp11 = LEFT(arrKey(ii), AT('_',arrKey(ii)) -1)
*!*        tmp22 = SUBSTR(arrKey(ii), AT('_',arrKey(ii)) +1)
*!*        tmp33 = odic.Item(arrKey(ii))
   
    INSERT INTO result1 (第几题  , 分值 ,次数  ) ;
        VALUES ( LEFT(arrKey(ii), AT('_',arrKey(ii)) -1),  ;
                SUBSTR(arrKey(ii), AT('_',arrKey(ii)) +1), ;
                odic.Item(arrKey(ii)) )
ENDFOR
time22 = SECONDS()
odic.RemoveAll
RELEASE oDic
SELECT result1
INDEX on 第几题 + 分值 TO 'C:\ind12'   &&  改成你的路径

BROWSE TITLE "总耗时" +  LTRIM(STR(time22 - time11 ,14,4 ))

在我的电脑上运行结果:

以下是引用吹水佬在2021-4-1 21:07:26的发言：

SUBSTR() + AT() 就可以，但字符串不长，与GETWORDNUM()差别不大。ALINES()因要分配内存和读写操作会耗费一些。

DIMENSION az[100]
USE sp
t1 = SECONDS()
n = GETWORDCOUNT(jmss1, ",")
FOR i=1 TO n
    STORE 0 TO az
    SCAN
        j = (VAL(GETWORDNUM(jmss1,i,","))+1) * 10
        az[j] = az[j]+1
    ENDSCAN
ENDFOR
? SECONDS() - t1

FOR i=1 TO 100
    IF az[i] > 0
        ? i/10-1, az[i]
    ENDIF
ENDFOR

T1 = SECONDS()
CLOSE DATABASES 
USE 选择\sp ALIAS aa
ln = OCCURS(",", jmss1) + 1 && 获取题目的数量
FOR lnj = 1 TO ln
    lnCnt = 0
    RELEASE la
    LOCAL la[10, 2]
    SCAN
        lnfz = VAL(GETWORDNUM(jmss1, lnj, ",")) &&VAL(SUBSTR(jmss1, 2 *lnj - 1, 1))
        IF ASCAN(la, lnfz, 1, ALEN(la, 1), 1) = 0
            lnCnt = lnCnt + 1
            la[lnCnt, 1] = lnfz
            la[lnCnt, 2] = 1
        ELSE 
            lnRow = (ASCAN(la, lnfz, 1, ALEN(la, 1), 1) + 1) / 2
            la[lnRow, 2] = la[lnRow, 2] + 1
        ENDIF 
    ENDSCAN 
ENDFOR 
MESSAGEBOX(SECONDS() - T1)
FOR lnj = 1 TO ALEN(la, 1)
    IF ASCAN(la, .F., 1, ALEN(la, 1), 1) > 0
        lnCnt = (ASCAN(la, .F., 1, ALEN(la, 1), 1) + 1) / 2
        EXIT 
    ENDIF 
ENDFOR
DIMENSION la(lnCnt - 1, 2)
ASORT(la)
LIST MEMORY LIKE la && 最后一题的数组结果

以下是引用ycvf在2021-4-1 21:57:55的发言：

能举例说明吗？