我正在使用Lumenworks csv阅读器读取csv文件。下面是一个示例记录
"001-0000265-003"|"Some detail"|"detal1"|"detail2"|"detal3"|"detail4"|"detail5"|"detail6"我创建了一个具有下面构造函数的类来读取这个文件
using (var input = new CsvReader(stream, true, '|'))
{
//logic to create an xml here
}在没有双引号的情况下,这很好。但是当那些骗子像这样
"001-0000265-003"|"Some " detail"|"detal1"|"detail2"|"detal3"|"detail4"|"detail5"|"detail6"读者抛出异常
An unhandled exception of type 'LumenWorks.Framework.IO.Csv.MalformedCsvException' occurred in LumenWorks.Framework.IO.dll然后我使用了CsvReader构造函数,它包含7个参数,
CsvReader(stream, true, '|', '"', '"', '#', LumenWorks.Framework.IO.Csv.ValueTrimmingOptions.All))但我还是会犯同样的错误。请提供任何建议。
我读了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"发布于 2014-10-15 12:22:18
我已经用你的样本数据对它进行了测试,很难修复这个格式错误的行(F.E。来自Catch-block)。因此,我不会使用引号字符,而是使用管道分隔符,然后通过csv[i].Trim('"')删除csv[i].Trim('"')。
下面是一个解析文件并返回所有行字段的方法:
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;
}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]https://stackoverflow.com/questions/26381067
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