VB character truncation character problem, VB truncation string

s = "I want it.

The same goes for solar days.

dim re

set re = createobject("")= true true

for each i in

msgbox

next

Hehe, according to the supplement, it turned out to be to extract these.

Try my multi-purpose module.

Let's say your source data is placed in text1 and output to text1.

private sub command1_click()

text1 = join(findstrmulti(text1, "aspx"">", ", 0), vbcrlf)

end sub

In this module, an array is returned, the first parameter is the original data, the second is the prefix, the third is the suffix, and the fourth parameter is whether to include the prefix and the prefix, (0 means not included, 1 means included).

public function findstrmulti(strall$, firststr$, endstr$, findmod&) as string()

dim i&, j&, n&, tmp$()

redim tmp$(10000)

doi = instr(i + 1, strall, firststr)

if i = 0 then exit do

if findmod = 0 then i = i + len(firststr)

j = instr(i, strall, endstr)

if j = 0 then exit do

if findmod = 1 then j = j + len(endstr)

tmp(n) = mid(strall, i, j - i)

n = n + 1

loopif n > 0 then

redim preserve tmp$(n - 1)

findstrmulti = tmp

end if

end function

flag=0 'Records whether the decimal point has been read.

for i=1 to len(text(0))

if mid(text(0),i,1) <"." then

text(i-flag)=mid(text(0),i,1)

elseflag=1

end if

The next program is displayed in different text controls, which is an array of text controls.

text(0) is used to input data, and text(1) to text(n) each display a single digit.

mid function.

Returns a variant (string) that contains the specified number of characters in the string.

Syntax mid(string, start[, length]).

start parameter is required. is the position of the character in the part that is taken out. If start exceeds the number of characters in the string, mid returns a zero-length string ("")。

length optional parameter; is variant (long). The number of characters to be returned. If you omit or length exceeds the number of characters in the text, including the characters at start, all characters in the string from start to end are returned.

It is possible that the value of your start variable is greater than the length of the string, resulting in an empty string, or the value of the variable is 0, resulting in an error.

Mine is super simple.

private sub command1_click()dim a as string

a = 'Must contain 2 decimal places for the amount input.

if a <>"" then

right(a, 1) 'points = left(right$(a, 2), 1).'Angle = left(right$(a, 5), 2).'Meta = left$(a, len(a) -5).'$100 end if

end sub

The x-string variable can be implemented with the mid function.

a=mid（x，1，3）

b=mid（x，5，3）

s = "13011301776 denim trousers 776 40, 38, 36, 35, 34, 33, 32, 31, fall 2011 The first wave"

dim aa() as string

aa = split(s, vbtab)

a=aa(0)

b=aa(1)

c=aa(2)

d=aa(3)

e=aa(4)

f=aa(5)

private sub command1_click() 'Truncate strings.

dim tmpstr as string

for i = 0 to split(text1, vblf)tmpstr = split(split(i, ",")(0), ":")(1)

mid(tmpstr, 1, len(tmpstr) -2)tmpstr = split(split(i, ",")(1), ":")(1)

mid(tmpstr, 1, len(tmpstr) -3)end sub 'Not Tested Please notify if there is an error.