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VBScript InStr Function

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vbscript – Classic ASP InStr() Evaluates True on Empty Comparison String – Stack Overflow

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vbscript - Classic ASP InStr() Evaluates True on Empty Comparison String - Stack Overflow
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ASP Function InStr

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ASP Function InStr
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ASP Func Instr

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VBScript InStr Function

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Top 20 asp classic instr mới nhất 2022

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1 VBScript InStr Function – W3Schools

2 VBScript InStrRev Function – W3Schools

3 Classic ASP InStr() Evaluates True on Empty Comparison String

4 ASP Function InStr – Virtual Splat Software Pvt Ltd

5 ASP Tutorials – Using the InStr function to find a substring

6 VBScript Functions InStr DevGuru

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Top 20 asp classic instr mới nhất 2022
Top 20 asp classic instr mới nhất 2022

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ASP Tutorials, Script and Code Examples – Using InStr to extract sub strings

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ASP Tutorials, Script and Code Examples - Using InStr to extract sub strings
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Sams Teach Yourself Active Server Pages 3.0 in 21 Days – Scott Mitchell, James Atkinson – Google Sách

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Sams Teach Yourself Active Server Pages 3.0 in 21 Days - Scott Mitchell, James Atkinson - Google Sách
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VBScript InStr Function

VBScript InStr Function

The InStr function returns the position of the first occurrence of one string within another.

The InStr function can return the following values:

If string1 is “” – InStr returns 0

If string1 is Null – InStr returns Null

If string2 is “” – InStr returns start

If string2 is Null – InStr returns Null

If string2 is not found – InStr returns 0

If string2 is found within string1 – InStr returns the position at which match is found

If start > Len(string1) – InStr returns 0

Tip: Also look at the InStrRev function

Syntax

InStr([start,]string1,string2[,compare])

Parameter Description start Optional. Specifies the starting position for each search. The search begins at the first character position (1) by default. This parameter is required if compare is specified string1 Required. The string to be searched string2 Required. The string expression to search for compare Optional. Specifies the string comparison to use. Default is 0 Can have one of the following values: 0 = vbBinaryCompare – Perform a binary comparison

1 = vbTextCompare – Perform a textual comparison

Examples

Example 1 <% txt="This is a beautiful day!" response.write(InStr(txt,"beautiful")) %> The output of the code above will be: 11 Show Example »

Example 2 Finding the letter “i”, using different starting positions: <% txt="This is a beautiful day!" response.write(InStr(1,txt,"i") & "
“)

response.write(InStr(7,txt,”i”) & “
“)

%> The output of the code above will be: 3

16 Show Example »

Classic ASP InStr() Evaluates True on Empty Comparison String

I ran into an issue with the Classic ASP VbScript InStr() function. As shown below, the second call to InStr() returns 1 when searching for an empty string in a non empty string. I’m curious why this is happening.

‘ InStr Test Dim someText : someText = “So say we all” Dim emptyString : emptyString = “” ” I expect this to be true If inStr(1,someText,”so”,1) > 0 Then Response.write ( “I found “”so””
” ) End If ” I expect this to be false If inStr(1, someText, emptyString, 1) > 0 Then Response.Write( “I found an empty string
” ) End If

EDIT: Some additional clarification: The reason for the question came up when debugging legacy code and running into a situation like this:

Function Go(value) If InStr(1, “Option1|Option2|Option3”, value, 1) > 0 Then ‘ Do some stuff End If End Function

In some cases function Go() can get called with an empty string. The original developer’s intent was not to check whether value was empty, but rather, whether or not value was equal to one of the piped delimited values (Option1,Option2, etc.).

Thinking about this further it makes sense that every string is created from an empty string, and I can understand why a programming language would assume a string with all characters removed still contains the empty string.

What doesn’t make sense to me is why programming languages are implementing this. Consider these 2 statements:

InStr(“so say we all”, “s”) ” evaluates to 1 InStr(“so say we all”, “”) ” evaluates to 1

The InStr() function will return the position of the first occurrence of one string within another. In both of the above cases, the result is 1. However, position 1 always contains the character “s”, not an empty string. Furthermore, using another string function like Len() or LenB() on an empty string alone will result in 0, indicating a character length of 0.

It seems that there is some inconsistency here. The empty string contained in all strings is not actually a character, but the InStr() function is treating it as one when other string functions are not. I find this to be un-intuitive and un-necessary.

ASP Function InStr

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The InStr function returns the position of the first occurrence of one string within another. The Function will return the position of the first occurence of String To Find within String To Search. The Function will return 0 if String To Find is not found. Gives the position of the first occurrence of one string within another. The InStr function returns the numeric position of the first occurrence of a specified substring within a specified string when starting from the beginning (left end) of the string. You can have the search for the substring be sensitive to the case (upper versus lower), or not. The default is to be case sensitive (binary comparison). An output of zero indicates no match. The Syntax of the InStr function is :- InStr([start,]string1,string2[,compare]) Its Description is :- Parameter Description start Optional. Specifies the starting position for each search. The search begins at the first character position by default. This parameter is required if compare is specified string1 Required. The string to be searched string2 Required. The string expression to search for compare Optional. Specifies the string comparison to use. Default is 0

Can have one of the following values: 0 = vbBinaryCompare – Perform a binary comparison

1 = vbTextCompare – Perform a textual comparison The Compare Argument can have the following values :- Constant Value Description vbBinaryCompare 0 Perform a binary comparison. vbTextCompare 1 Perform a textual comparison. vbDatabaseCompare 2 Perform a comparison based upon information contained in the database where the comparison is to be performed.

The Instr function returns the following values :- If Instr returns string1 is zero-length 0 string1 is Null Null string2 is zero-length start string2 is Null Null string2 is not found 0 string2 is found within string1 Position at which match is found start > Len(string2) 0 Example#1 :- Code :-

dim txt,pos

txt=”This is a beautiful day!”

pos=InStr(txt,”his”)

document.write(pos)

Output :-

2 Example#2 :- Code :-

dim txt,pos

txt=”This is a beautiful day!”

‘A textual comparison starting at position 4

pos=InStr(4,txt,”is”,1)

document.write(pos)

Output :-

6 Example#3 :- Code :-

dim txt,pos

txt=”This is a beautiful day!”

‘A binary comparison starting at position 1

pos=InStr(1,txt,”B”,0)

document.write(pos)

Output :-

0 Example#4 :- Code :-

<% String1="It's taken me 4 years of work" String2="t" Location= Instr(String1,String2) response.write Location %>

Output :-

2 Note :- you can set the starting position for each search. If omitted, search begins at the first character position.

Example#5 :- Code :-

<% String1="It's taken me 4 years of work" String2="t" Location= Instr(3,String1,String2) response.write Location %>

Output :-

6

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