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例如，这是字符串：
“你好，这很有挑战性\n”+“你认为这很容易吗？\n”+ variableName +“3 + 4 = 7\n”

Dim example = """Hello, this is challenging\n"" + ""you think it is easy?\n"" + variableName + "" 3 + 4 = 7\n"""

我想用编程的方式来排列字符串变成：
“你好，这很有挑战性”+换行符+“你觉得这很容易吗？” + 换行符 + 变量名 + " 3 + 4 = 7" + 换行符

Dim output = """Hello, this is challenging"" + newline + ""you think it is easy?"" + newline + variableName + "" 3 + 4 = 7"" + newline"

如您所见，它涉及获取引号内的文本
所以我在想：
1. 使用正则表达式来获取引号，但是正如您所看到的，我们将省略变量名
2. 我正在考虑使用 + 号拆分，但如您所见，“3 + 4 = 7”中会有误报

告诉我你怎么看，容易吗？还有其他步骤吗？

更新的示例和输出：

Dim example2 = """Hello, this \nis challenging\n"" + ""you think it is easy?\n"" + variableName + "" 3 + 4 = 7\n"""

Dim output2 = """Hello, this "" + newline + ""is challenging"" + newline + ""you think it is easy?"" + newline + variableName + "" 3 + 4 = 7"" + newline"
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2 回答 2

1

这个单线对我有用：

Dim example = """Hello, this is challenging\n"" + ""you think it is easy?\n"" + variableName + "" 3 + 4 = 7\n"""
Dim output = """Hello, this is challenging"" + newline + ""you think it is easy?"" + newline + variableName + "" 3 + 4 = 7"" + newline"

Dim result = String.Join("""", example.Split(""""c).Select(Function(x, n) If(n Mod 2 = 1, x.Replace("\n", """ + newline"), x))).Replace("newline""", "newline")

我和你的输出一样。

这是更新的示例工作正常：

Dim example2 = """Hello, this \nis challenging\n"" + ""you think it is easy?\n"" + variableName + "" 3 + 4 = 7\n"""
Dim output2 = """Hello, this "" + newline + ""is challenging"" + newline + ""you think it is easy?"" + newline + variableName + "" 3 + 4 = 7"" + newline"
Dim result2 = String.Join("""", example2.Split(""""c).Select(Function(x, n) If(n Mod 2 = 1, x.Replace("\n", """ + newline + """), x))).Replace("newline + """"", "newline")

我"Hello, this " + newline + "is challenging" + newline + "you think it is easy?" + newline + variableName + " 3 + 4 = 7" + newline按照你的得到output2。

这是发生了什么result2：

Dim splitOnQuotes = example2.Split(""""c)
'splitOnQuotes = { "", "Hello, this \nis challenging\n", " + ", "you think it is easy?\n", " + variableName + ", " 3 + 4 = 7\n", "" }

所有的双引号都被拆分了。

Dim replaceSlashNOnOddLines = splitOnQuotes.Select(Function(x, n) If(n Mod 2 = 1, x.Replace("\n", """ + newline + """), x))
'replaceSlashNOnOddLines = { "", "Hello, this " + newline + "is challenging" + newline + "", " + ", "you think it is easy?" + newline + "", " + variableName + ", " 3 + 4 = 7" + newline + "", "" }

在我们用 替换的每个奇数元素\n上" + newline + "。

Dim joinOnQuotes = String.Join("""", replaceSlashNOnOddLines)
'joinOnQuotes = "Hello, this "" + newline + ""is challenging"" + newline + """" + ""you think it is easy?"" + newline + """" + variableName + "" 3 + 4 = 7"" + newline + """""

然后加入备份的部分 "。

Dim result2 = joinOnQuotes.Replace("newline + """"", "newline")

但是我们有额外的双引号，形式为newline + ""，所以我们只需将它们替换为newline。

于 2020-05-12T00:02:22.363 回答
0

标准方法是遍历字符串计数字符/翻转布尔值，指示您是否在字符串内

Dim inSideAString = false
Dim quoteChar = "'"c
Dim escapeChar = "\"c
Dim lookForChar  "+"

Dim lastChar = " "c

For i = 0 to theString.Length - 1

  Dim c = theString(i)

  Dim prevChar = If(i > 0, theString(i-1), " "c) 'make it not the escape char

  If c = quoteChar AndAlso prevChar <> escapeChar Then 
    insideAString = (Not insideAString)
    Continue For
  End If

  If c = lookForChar Then
    If insideAString Then 
      Console.Write($"Found a {lookForChar} inside a string at position {i}!")
    Else
      Console.Write($"Found a {lookForChar} outside a string at position {i}!")
    End If
  End If

Next i
于 2020-05-11T23:42:11.623 回答