c# - 尝试构建时出现CS1003和CS1026语法错误

Question

我不确定为什么会收到此错误。我正在尝试在Visual Studio上构建它。我是新手，但我需要构建此代码。
有我的错误和代码:
(30,19):错误CS1003:语法错误，预期为'('
(30,88):错误CS1026:)
(31,19):错误CS1003:语法错误，'('预期
(31,51):错误CS1026:)
(36,23):错误CS1003:语法错误，预期为'('
(36,63):错误CS1026:)
(37,23):错误CS1003:语法错误，预期为'('
(37,156):错误CS1026:)

namespace MelonLoader.AssemblyGenerator
{
    public static class DownloaderAndUnpacker
    {
        public static void Run(string url, string targetVersion, string currentVersion, string destinationFolder, string tempFile)
        {
            if (targetVersion == currentVersion)
            {
                Logger.Log($"{destinationFolder} already contains required version, skipping download");
                return;
            }
            
            Logger.Log($"Cleaning {destinationFolder}");
            foreach (var entry in Directory.EnumerateFileSystemEntries(destinationFolder))
            {
                if (Directory.Exists(entry))
                    Directory.Delete(entry, true);
                else
                    File.Delete(entry);
            }

            Logger.Log($"Downloading {url} to {tempFile}");
            Program.webClient.DownloadFile(url, tempFile);
            Logger.Log($"Extracting {tempFile} to {destinationFolder}");
            
/*line 30*/ using var stream = new FileStream(tempFile, FileMode.Open, FileAccess.Read);
            using var zip = new ZipArchive(stream);
            
            foreach (var zipArchiveEntry in zip.Entries)
            {
                Logger.Log($"Extracting {zipArchiveEntry.FullName}");
                using var entryStream = zipArchiveEntry.Open();
                using var targetStream = new FileStream(Path.Combine(destinationFolder, zipArchiveEntry.FullName), FileMode.OpenOrCreate, FileAccess.Write);
                entryStream.CopyTo(targetStream);
            }
        }
    }
}

Answer 1

这似乎是C#8“使用声明”；可能有可用的最新编译器，但您的csproj配置为使用下层C#。有可能编辑cspoj以更新或添加:

<LangVersion>8</LangVersion>

(或更高版本，在 <PropertyGroup>元素内)
会解决它。
如果您不能使用C#8，则:

using (var stream = new FileStream(tempFile, FileMode.Open, FileAccess.Read))
using (var zip = new ZipArchive(stream))
{

    foreach (var zipArchiveEntry in zip.Entries)
    {
        Logger.Log($"Extracting {zipArchiveEntry.FullName}");
        using (var entryStream = zipArchiveEntry.Open())
        using (var targetStream = new FileStream(Path.Combine(destinationFolder, zipArchiveEntry.FullName), FileMode.OpenOrCreate, FileAccess.Write))
        {
            entryStream.CopyTo(targetStream);
        }
    }
}