请问如何精确替换指定的字符

[lcfno1]

我正在开发的一个会员系统，希望看到再线人员列表，把在线人员姓名都放在了application("onlineuser") 中,以逗号分开.
现在如果某一人退出需要用空格替换掉application("onlineuser")中的该人姓名的字符.
比如现在application("onlineuser")的值为 "张三,李四,王五,张三丰,张无忌",可是如果"张三"退出, 执行语句:
application("onlineuser")=replace(application("onlineuser"),"张三","")
就会造成同时替换掉"张三"和"张三丰"中的"张三"从而application("onlineuser")的值为 ",李四,王五,丰,张无忌", 把不该替换的张三丰替换了一部分.
现在请问各位高人有没有什么办法解决这个问题

[Primer2002cn]

试试看：
dim strUser,strLeaver,strTmp,i
strLeaver="张三"
strUser="张三,李四,王五,张三丰,张无忌"
strTmp=split(strUser,",")
for i=lbound(strTmp) to ubound(strTmp)
if trim(strTmp(i))=trim(strLeaver) then
strTmp(i)=""
end if
strUser=trim(strTmp(i)) & ","
next
strUser=left(strUser,len(strUser)-1)

[Primer2002cn]

错了,应该这样：
dim strUser,strLeaver,strTmp,i
strLeaver="张三"
strUser="张三,李四,王五,张三丰,张无忌"
strTmp=split(strUser,",")
strUser=""
for i=lbound(strTmp) to ubound(strTmp)
if trim(strTmp(i))=trim(strLeaver) then
strTmp(i)=""
else
strUser=strUser & trim(strTmp(i)) & ","
end if
next
strUser=left(strUser,len(strUser)-1)

[Primer2002cn]

dim strUser,strLeaver,strTmp,i
strLeaver="张三"
strUser="张三,李四,王五,张三丰,张无忌"
strTmp=split(strUser,",")
strUser=""
for i=lbound(strTmp) to ubound(strTmp)
if trim(strTmp(i))<>trim(strLeaver) then
strUser=strUser & trim(strTmp(i)) & ","
end if
next
strUser=left(strUser,len(strUser)-1)

[wsqsoft]

application("onlineuser")=application("onlineuser")&","
application("onlineuser")=replace(application("onlineuser"),"张三,","")
application("onlineuser")=left(application("onlineuser"),len(application("onlineuser"))-1)

[kirc]

补充一下楼上的，最后一句：
If application("onlineuser")<>"" Then
application("onlineuser")=left(application("onlineuser"),len(application("onlineuser"))-1)
End If

[Haoye]

楼上两位的都可以实现，后者的更为简便。