1樓：匿名使用者
private sub command1_click()
dim stra() as string, strb() as string, i, j, n as integer
stra = split("123,211,234324,43,4,123,43,211", ",") 'stra賦值bai
duredim strb(0)
strb(0) = stra(0) 'strb的第乙個元素賦值
for i = 1 to ubound(stra) '遍歷stra
for j = 0 to ubound(strb) '查詢zhi
是否已經包含在strb中
if stra(i) = strb(j) then exit for
next
if j > ubound(strb) then '如果沒dao包含
n = n + 1
redim preserve strb(n) '擴大數回組
strb(n) = stra(i)
end if
next
for i = 0 to ubound(strb) '顯示strb的所有答元素
print strb(i)
next
end sub
2樓：匿名使用者
function removeduplication(byref sourcearray()) as variant()
..dim returns()
..dim uboundreturns as integer:uboundreturns = -1
..dim i as integer, j as integer
..for i=0 to ubound(sourcearray)
....dim found as boolean:found=false
....for j=0 to uboundreturns - 1
......if (sourcearray(i) = returns(j)) then found = true
....next
....if (not found) then
......uboundreturns = uboundreturns + 1
......redim preserve returns(uboundreturns)
......returns(uboundreturns) = sourcearray(i)
....endif
..next
....removeduplication = returns
end function
3樓：匿名使用者
1、字典可以很好的解決問題。內
2、**如容
下：private sub command1_click()
on error resume next
dim stra(), strb(), dic as object, i as integer
stra = array("123", "123", "211", "234324", "43", "4", "43", "211")
set dic = createobject("scripting.dictionary")
for i = 0 to ubound(stra)
dic.add stra(i), i
next
redim strb(dic.count - 1)
i = 0
for each x in dic.keys
strb(i) = x
i = i + 1
next
msgbox "stra:" & join(stra, ",") & chr(10) & "strb:" & join(strb, ",")
end sub
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