这在Python中是有效的:

import base64

def IsBase64(str):
    try:
        base64.b64decode(str)
        return True
    except Exception as e:
        return False

if IsBase64("ABC"):
    print("ABC is Base64-encoded and its result after decoding is: " + str(base64.b64decode("ABC")).replace("b'", "").replace("'", ""))
else:
    print("ABC is NOT Base64-encoded.")

if IsBase64("QUJD"):
    print("QUJD is Base64-encoded and its result after decoding is: " + str(base64.b64decode("QUJD")).replace("b'", "").replace("'", ""))
else:
    print("QUJD is NOT Base64-encoded.")

IsBase64("string here")如果这里的字符串是base64编码的则返回true，如果这里的字符串不是base64编码的则返回false。

对于Java风格，我实际上使用以下正则表达式:

"([A-Za-z0-9+]{4})*([A-Za-z0-9+]{3}=|[A-Za-z0-9+]{2}(==){0,2})?"

在某些情况下，==也是可选的。

最好!

你可以:

检查长度是否为4个字符的倍数 检查每个字符都在A-Z, A-Z, 0-9， +， /集合中，除了末尾的填充为0,1或2 '='字符

如果你期望它是base64，那么你可以使用平台上可用的任何库来尝试将它解码为字节数组，如果它不是有效的base64则抛出异常。当然，这取决于你的平台。

试试这个:

public void checkForEncode(String string) {
    String pattern = "^([A-Za-z0-9+/]{4})*([A-Za-z0-9+/]{4}|[A-Za-z0-9+/]{3}=|[A-Za-z0-9+/]{2}==)$";
    Pattern r = Pattern.compile(pattern);
    Matcher m = r.matcher(string);
    if (m.find()) {
        System.out.println("true");
    } else {
        System.out.println("false");
    }
}

没有办法区分字符串和base64编码，除非字符串在您的系统中有一些特定的限制或标识。

Function Check_If_Base64(ByVal msgFile As String) As Boolean
Dim I As Long
Dim Buffer As String
Dim Car As String

Check_If_Base64 = True

Buffer = Leggi_File(msgFile)
Buffer = Replace(Buffer, vbCrLf, "")
For I = 1 To Len(Buffer)
    Car = Mid(Buffer, I, 1)
    If (Car < "A" Or Car > "Z") _
    And (Car < "a" Or Car > "z") _
    And (Car < "0" Or Car > "9") _
    And (Car <> "+" And Car <> "/" And Car <> "=") Then
        Check_If_Base64 = False
        Exit For
    End If
Next I
End Function
Function Leggi_File(PathAndFileName As String) As String
Dim FF As Integer
FF = FreeFile()
Open PathAndFileName For Binary As #FF
Leggi_File = Input(LOF(FF), #FF)
Close #FF
End Function