Question

const string Pattern = @"(?si)<([^\s<]*totalWork[^\s<]*)>.*?</\1>";
var filter = Builders<JobInfoRecord>.Filter.Regex(x => x.SerializedBackgroundJobInfo,
                                            new BsonRegularExpression(Pattern, "i"));

var documents = await records.Find(filter).ToListAsync();

====

在我得到documents后，我处理了我身边的每个文件。

const string EmptyTag = "<$1></$1>";
var updatedJobInfo = Regex.Replace(document.SerializedBackgroundJobInfo, Pattern, EmptyTag);

我怎样才能在mongo方面做Regex.Replace？或者那只能发生在客户端？

以下Replace是否适用于Mongo方面？

using (var cursor = await jobInfoDocuments.FindAsync<JobInfoRecord>(filter))
{
      while (await cursor.MoveNextAsync())
      {
             var batch = cursor.Current;
             foreach (var document in batch)
             {
                 var newInfo = Regex.Replace(document.SerializedBackgroundJobInfo, regex, EmptyTag);

                  // Applying several operations within the one request.
                  operationList.Add(new UpdateOneModel<JobInfoRecord>(Builders<JobInfoRecord>.Filter.Eq("_id", document.JobId),
                                                                       Builders<JobInfoRecord>.Update.Set("SerializedBackgroundJobInfo", newInfo)));
              }

Answer 1

您可以使用javascript执行此操作，但请确保修复filter以使用mongo shell

db.records.find(filter).forEach(function (doc) {
  var pattern = /<([^\s<]*totalWork[^\s<]*)>[\s\S]*?</\1>/i;
  var EmptyTag = "<$1></$1>";
  
  doc.SerializedBackgroundJobInfo = doc.SerializedBackgroundJobInfo.replace(pattern, EmptyTag);
  
  db.records.save(doc);
})

如何在Mongo端更新文档？

1 个答案: