asp truncates the specified string, and ASP intercepts the string at the specified location

Two methods:

1: fileext=right("images/", 3) two: filename="images/"

fileext=mid(filename,instr(filename,".")+1)

The value of fileext is the result.

There are a lot of functions in VB that manipulate strings, and combining some of them makes it easy to do what you need.

I'll sort some out below.

len len(string|varname) returns the number of characters in a string, or the number of bytes required to store a variable.

trim trim(string) removes the spaces before and after the string.

ltrim ltrim(string) removes the space in front of the string.

rtrim rtrim(string) removes the spaces after the string.

mid mid(string,start,length) starts from the start character of the string string to get the length of the string, if the third parameter is omitted, it means the string starts with the start character and ends at the end of the string.

Left Left(String,Length) A string that takes the length from the left side of the string.

Right Right(String,Length) A string that takes the length of the length from the right side of the string string.

lcase lcase(string) converts all uppercase letters in a string string to lowercase letters.

ucase ucase(string) converts all uppercase letters in a string string to uppercase letters.

strcomp strcomp(string1,string2[,compare]) returns the comparison of the string1 string with the string2 string, 0 if both strings are the same, -1 if less than, and 1 if greater than 1

instr instr(string1,string2[, compare]) returns the position where the string1 string first appeared in the string2 string.

split split(string1,delimiter[, count[, start]]) splits the string into a one-dimensional array based on the delimiter, where the delimiter is used to identify the substring bounds. If omitted, use a space (""as a separator. This one can come in handy.

This can be solved by split() and ubound(), which splits the string into an array, and the ubound() function, which counts the number of arrays.

1.Use the string function, mid, left, right via midstr1="abasdflkj#username#sdfkljkl"

str2="The amount of deposits absorbed by the key is quickly classified"

f_str1 = mid(str1,instr(str1,"#")+1,instr(instr(str1,"Sharp digging")+1,str1,"#",1)-instr(str1,"#

f_str2 = mid(str2,instr(str2,"#")+1,instr(instr(str2,"#")+1,str2,"#",1)-instr(str2,"#

Silver core. The ** written is more messy, and I understand Fei Fangsen.

2.with arrays.

f_str1 = split(str1,"#f_str2 = split(str2,"#

Use the split function.

The returned array is accompanied by Min = split(string slippery, delimiter) arr1 = split(str1,"#f_str1 = arr(1)

arr2 = split(str2, "Confidence f str2 = arr(1).

str1="abasdflkj#username#sdfkljkl"

str2="Absorb the amount of money collected by the deposit and banquet key classification is fast and fast"

f_str1 = split(str1,"#f_str2 = split(str2,"#""Lu Xiang Clan br >"f_str2)

cha=split(str,"

'Look for the "contain" from the str

The position of the character.

str=cha(0) 'If the index of the first eligible position is 0, for example, if cha(1) is the value of str

str=replace(cha,","") 'Remove the previous ", and all that remains is the required characters.

str 'Output.

If you use only the first character, use <%=split(str,"

If you want both, use arrays.

strarr=split(str,"

You get an array.

strarr(0) has a value of 1111

strarr(1) has a value of aaa

strarr(2)

.And so on.

a = "123456788"

a = mid(a,3,4) 'Parameters: String, start position, truncated length.

To get to the first number, you only need to subtract the starting position, as in mid(a,3,7-3).

It's the mid function, and you can look at the function description more.

mid (string to be intercepted, position to start intercepting, length to be intercepted) to intercept a character of the specified length from any position.

You can also use mid()! But you have to determine the number of digits!