可以用函数 Instr(,,"?"),查找出“?”的位置N,然后把N走边的数给一个数组,并判断是否为空,为空就去掉。 dim i as integer,j as integer dim strn as string,a() as string,B() strn==""?bueskhy.kdk??di.fsdkfjsdl??kdfjlskdfj.lsdk??klsdjf.iohje?" j=instr(strn,"?") A(I)=LEFT(STRN,J-1) STRN=RIGHT(STRN,J+1) DO WHILE J>0 J=INSTR(STRN,"?") I=I+1 REDIM PRESERVE A(I) A(I)=LEFT(STRN,J-1) STRN=RIGHT(STRN,J+1) LOOP REDIM PRESERVE A(I) A(I)=STRN J=0 FOR I=LBOUND(A) TO UBOUND(A) IF LEN(A(i)) > O THEN J=J+1 REDIM PRESEVE B(J) B(J)=A(I) END IF NEXT I 这样就可分离 你要的类容了
ok!
dim i as integer,j as integer
dim strn as string,a() as string,B()
strn==""?bueskhy.kdk??di.fsdkfjsdl??kdfjlskdfj.lsdk??klsdjf.iohje?"
j=instr(strn,"?")
A(I)=LEFT(STRN,J-1)
STRN=RIGHT(STRN,J+1)
DO WHILE J>0
J=INSTR(STRN,"?")
I=I+1
REDIM PRESERVE A(I)
A(I)=LEFT(STRN,J-1)
STRN=RIGHT(STRN,J+1)
LOOP
REDIM PRESERVE A(I)
A(I)=STRN
J=0
FOR I=LBOUND(A) TO UBOUND(A)
IF LEN(A(i)) > O THEN
J=J+1
REDIM PRESEVE B(J)
B(J)=A(I)
END IF
NEXT I
这样就可分离 你要的类容了