如何在Mongo端更新文件？

Anatoly 发表于 Dev

安那托利

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();

====

得到后，我将documents处理我身边的每个文档。

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

我该怎么办Regex.Replace？还是只能在客户中发生？

Replace在Mongo方面有以下作品吗？

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)));
              }

汤姆

您可以使用它，javascript但请确保该修复程序filter可与mongo shell一起使用

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);
})

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