Isu*_*ana 4 .net c# csv lumenworks
我正在使用Lumenworks csv阅读器阅读csv文件.以下是一个示例记录
"001-0000265-003"|"Some detail"|"detal1"|"detail2"|"detal3"|"detail4"|"detail5"|"detail6"
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我用下面的构造函数创建了一个类来读取这个文件
using (var input = new CsvReader(stream, true, '|'))
{
//logic to create an xml here
}
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当细节内没有双引号时,这很好用.但是当scinarios像这样
"001-0000265-003"|"Some " detail"|"detal1"|"detail2"|"detal3"|"detail4"|"detail5"|"detail6"
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读者抛出异常
An unhandled exception of type 'LumenWorks.Framework.IO.Csv.MalformedCsvException' occurred in LumenWorks.Framework.IO.dll
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那么我使用了带有7个参数的CsvReader构造函数,
CsvReader(stream, true, '|', '"', '"', '#', LumenWorks.Framework.IO.Csv.ValueTrimmingOptions.All))
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但我仍然得到同样的错误.请提供任何建议.
我正在阅读一些复杂的文件如下,
"001-0000265-003"|"ABC 33"X23" CDE 32'X33" AAA, BB'C"|"detal1"|"detail2"|"detal3"|"detail4"|"detail5"|"detail6"
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我用你的样本数据对它进行了测试,很难修复这个格式不正确的行(来自Catch-block).所以我不会使用引号字符,而只是使用管道分隔符并删除"后来的通道csv[i].Trim('"').
这是一个解析文件并返回所有行的字段的方法:
private static List<List<string>> GetAllLineFields(string fullPath)
{
List<List<string>> allLineFields = new List<List<string>>();
var fileInfo = new System.IO.FileInfo(fullPath);
using (var reader = new System.IO.StreamReader(fileInfo.FullName, Encoding.Default))
{
Char quotingCharacter = '\0'; // no quoting-character;
Char escapeCharacter = quotingCharacter;
Char delimiter = '|';
using (var csv = new CsvReader(reader, true, delimiter, quotingCharacter, escapeCharacter, '\0', ValueTrimmingOptions.All))
{
csv.DefaultParseErrorAction = ParseErrorAction.ThrowException;
//csv.ParseError += csv_ParseError; // if you want to handle it somewhere else
csv.SkipEmptyLines = true;
while (csv.ReadNextRecord())
{
List<string> fields = new List<string>(csv.FieldCount);
for (int i = 0; i < csv.FieldCount; i++)
{
try
{
string field = csv[i];
fields.Add(field.Trim('"'));
} catch (MalformedCsvException ex)
{
// log, should not be possible anymore
throw;
}
}
allLineFields.Add(fields);
}
}
}
return allLineFields;
}
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使用包含示例数据的文件进行测试和输出:
List<List<string>> allLineFields = GetAllLineFields(@"C:\Temp\Test\CsvFile.csv");
foreach (List<string> lineFields in allLineFields)
Console.WriteLine(string.Join(",", lineFields.Select(s => string.Format("[{0}]", s))));
[001-0000265-003],[Some detail],[detal1],[detail2],[detal3],[detail4],[detail5],[detail6]
[001-0000265-003],[Some " detail],[detal1],[detail2],[detal3],[detail4],[detail5],[detail6]
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