如何在字符串中找到字符串，Instr是VB6方式... [英] How Can I find a String within a String, Instr was the VB6 way...

问题描述

大家好，



我想确保我只是读取一个响应的数字部分，它会以字符串1.95e-3mA的形式返回为了得到这个，我一直在使用子串来修剪最后一部分，这个已经回复的板子刚刚回复了1.9e-3mA意味着我的小心

 result = dataBack。子串（ 0 ， 3 ）

有我现在可以看到这一切都崩溃了，因为回复现在是1.9e（打破了计算）而不是1.95。有没有一种简单的方法可以做到这一点，或者我必须重新思考我这样做的方式。 Val（）会解决我的问题吗？...

格伦

Val（）不会因为我需要成为一个字符串我想知道Val它能得到数字和转换它返回一个字符串??? ....

解决方案

如果数字部分后面总是跟着 e ，那么你可以使用 String.Split [ ^ ]将字符串拆分成一个数组，然后取出数组的第一部分：

  Dim 输入作为 字符串 =   1.9e-3mA 
  Dim 结果 As   String  = input.Split（< span class =code-string>  e c）（ 0 ）

但你在评论中说它不会总是 e ，所以你可以使用 Regex.Split [ ^ ]按正则表达式分割：

  Dim 输入 As  字符串 =   1.9e-3mA 
  Dim 结果作为 字符串 = Regex.Split （输入，  [^。\d  - ]）（ 0 ）' 首先添加Imports System.Text.RegularExpressions  

这会将输入字符串拆分为不是数字的字符或一个点，然后它需要数组的第一部分来获取数字部分。

使用正则表达式！ ：笑：

 \d +（\.\ + +）？（[eE]  - ？\d +）？

应该做它 - 它只匹配浮点数部分，并忽略其他任何东西。

 '  Imports System.Text.RegularExpressions  
 
 ' < span class =code-comment>为Visual Basic构建的正则表达式：星期五，2014年4月11日，01：54：44 PM  
 ' 使用Expresso版本：3.0.3634，http：//www.ultrapico.com  
 '   
 ' 对此的描述正则表达式： 
 '   
 ' 任意数字，一次或多次重复 
 '  [1]：编号的捕获组。 [\.\d +]，零次或一次重复 
 '  \。\\ \\ d +  
 '  Literal。 
 ' 任意数字，一次或多次重复 
 '  [2]：编号的捕获组。 [[eE]  - ？\d +]，零或一次重复 
 '  [eE ]  - ？\d +  
 ' 此类中的任何字符：[eE]  
 '   - ，零或一次重复 
 ' 任意数字，一次或多次重复 
 '   
 '   
 
 公开  Dim  regex  As  Regex = 新正则表达式（_ 
   \ + +（\。\ d +）？（[eE]  - ？\d +）？，_ 
 RegexOptions.IgnoreCase _ 
 或 RegexOptions.CultureInvariant _ 
 或 RegexOptions.IgnorePatternWhitespace _ 
 或 RegexOptions.Compiled _ 
）
 
 ' 捕获InputText中的第一个匹配项（如果有） 
  Dim  m  As  Match = regex.Match（InputText）
  Dim  result  As   string  = m.Value 

IndexOf 或 LastIndexOf 在第一个字符串中给出一些其他字符串的位置。

 position = databack.IndexOf（  mA）

显示mA的起始位置。如果它不存在，IndexOf将返回-1。

假设它始终存在，你可以做

 result = databack。子串（ 0 ，databack.IndexOf（  mA< /跨度>））

Hi All,

I want to make sure I am only reading the numeric part of a response that is coming back as a string say "1.95e-3mA " to get this I been using substring to trim off the last part the darned board has just replied with "1.9e-3mA " meaning my careful

result = dataBack.Substring(0,3) 

has broken and now I can see it all falling apart as the reply is now 1.9e (which breaks a calculation) rather than 1.95. Is there a simple way to do this or do I have to rethink the way I am doing this. Would Val() solve my issue?...
Glenn
Val() wont as I need to be a string I wonder Val it to get the number and the convert it back to a string???....

解决方案

If the numeric part would always be followed by an e, then you could use String.Split[^] to split the string into an array and then take the first part of the array:

Dim input As String = "1.9e-3mA"
Dim result As String = input.Split("e"c)(0)

But you said in your comment that it won't always be an e, so you can use Regex.Split[^] to split by a regular expression:

Dim input As String = "1.9e-3mA"
Dim result As String = Regex.Split(input, "[^.\d-]")(0) ' first add Imports System.Text.RegularExpressions

This will split the input string at the chars that are not a digit or a dot, and then it takes the first part of the array to get the numeric part.

Use a Regex! :laugh:

\d+(\.\d+)?([eE]-?\d+)?

Should do it - it matches the floating point number part only, and ignore anything else.

'  Imports System.Text.RegularExpressions

'  Regular expression built for Visual Basic on: Fri, Apr 11, 2014, 01:54:44 PM
'  Using Expresso Version: 3.0.3634, http://www.ultrapico.com
'
'  A description of the regular expression:
'
'  Any digit, one or more repetitions
'  [1]: A numbered capture group. [\.\d+], zero or one repetitions
'      \.\d+
'          Literal .
'          Any digit, one or more repetitions
'  [2]: A numbered capture group. [[eE]-?\d+], zero or one repetitions
'      [eE]-?\d+
'          Any character in this class: [eE]
'          -, zero or one repetitions
'          Any digit, one or more repetitions
'
'

Public Dim regex As Regex = New Regex( _
      "\d+(\.\d+)?([eE]-?\d+)?", _
    RegexOptions.IgnoreCase _
    Or RegexOptions.CultureInvariant _
    Or RegexOptions.IgnorePatternWhitespace _
    Or RegexOptions.Compiled _
    )

' Capture the first Match, if any, in the InputText
Dim m As Match= regex.Match(InputText)
Dim result As string = m.Value

IndexOf or LastIndexOf give the position of some otherstring inside the first string.

position = databack.IndexOf("mA")

shows you where "mA" starts. If it is not present, IndexOf will return -1.
Assuming that it is always present, you could do

result = databack.Substring(0,databack.IndexOf("mA"))

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