How to replace substring in mongodb document

问题

I have a lot of mongodb documents in a collection of the form:

{
....
\"URL\":\"www.abc.com/helloWorldt/...\"
.....
}

I want to replace helloWorldt with helloWorld to get:

{
....
\"URL\":\"www.abc.com/helloWorld/...\"
.....
}

How can I achieve this for all documents in my collection?

回答1:

db.media.find({mediaContainer:"ContainerS3"}).forEach(function(e,i) {
    e.url=e.url.replace("//a.n.com","//b.n.com");
    db.media.save(e);
});

回答2:

Starting Mongo 4.2, db.collection.update() can accept an aggregation pipeline, finally allowing the update of a field based on another field:

// { URL: "www.abc.com/helloWorldt/..." }
// { URL: "www.abc.com/HelloWo/..." }
db.collection.update(
  { URL: { $regex: "/helloWorldt/" } },
  [{
    $set: { URL: {
      $concat: [
        { $arrayElemAt: [ { $split: [ "$URL", "/helloWorldt/" ] }, 0 ] },
        "/helloWorld/",
        { $arrayElemAt: [ { $split: [ "$URL", "/helloWorldt/" ] }, 1 ] }
      ]
    }}
  }],
  { multi: true }
)
// { URL: "www.abc.com/helloWorld/..." }
// { URL: "www.abc.com/HelloWo/..." }

This is a bit verbose since there isn't a proper string $replace operator yet.

• The first part { URL: { $regex: "/helloWorldt/" } } is the match query, filtering which documents to update (the ones with hello world wrongly spelled).

• The second part [{ $set: { URL: ... } }] is the update aggregation pipeline (note the squared brackets signifying the use of an aggregation pipeline). $set is a new aggregation operator which in this case creates/replaces a field. The new value is weirdly computed using a mix of $concat and $split by lack of a proper string $replace operator. Note how URL is modified directly based on the its own value ($URL).

• Don't forget { multi: true }, otherwise only the first matching document will be updated.

回答3:

Currently, you can't use the value of a field to update it. So you'll have to iterate through the documents and update each document using a function. There's an example of how you might do that here: MongoDB: Updating documents using data from the same document

回答4:

nodejs. Using mongodb package from npm

db.collection('ABC').find({url: /helloWorldt/}).toArray((err, docs) => {
  docs.forEach(doc => {
    let URL = doc.URL.replace('helloWorldt', 'helloWorld');
    db.collection('ABC').updateOne({_id: doc._id}, {URL});
  });
});

回答5:

To replace ALL occurrences of the substring in your document use:

db.media.find({mediaContainer:"ContainerS3"}).forEach(function(e,i) {
var find = "//a.n.com";
var re = new RegExp(find, 'g');
e.url=e.url.replace(re,"//b.n.com");
db.media.save(e);
});

回答6:

The formatting of my comment to the selected answer (@Naveed's answer) has got scrambled - so adding this as an answer. All credit goes to Naveed.

----------------------------------------------------------------------

Just awesome. My case was - I have a field which is an array - so I had to add an extra loop.

My query is:

db.getCollection("profile").find({"photos": {$ne: "" }}).forEach(function(e,i) {
    e.photos.forEach(function(url, j) {
        url = url.replace("http://a.com", "https://dev.a.com");
        e.photos[j] = url;
    });
    db.getCollection("profile").save(e);
    eval(printjson(e));
})

回答7:

Now you can do it!

We can use Mongo script to manipulate data on the fly. It works for me!

I use this script to correct my address data.

Example of current address: "No.12, FIFTH AVENUE,".

I want to remove the last redundant comma, the expected new address ""No.12, FIFTH AVENUE".

var cursor = db.myCollection.find().limit(100);

while (cursor.hasNext()) {
  var currentDocument = cursor.next();

  var address = currentDocument['address'];
  var lastPosition = address.length - 1;

  var lastChar = address.charAt(lastPosition);

  if (lastChar == ",") {

    var newAddress = address.slice(0, lastPosition);


    currentDocument['address'] = newAddress;

    db.localbizs.update({_id: currentDocument._id}, currentDocument);

  }
}

Hope this helps!

回答8:

Just in case if you are using examples from the answers here and get "Updated 0 existing records" when running your replace script, check whether your client is connected to the primary MongoDB node that allows you to store/write changes.

回答9:

Using mongodump,bsondump and mongoimport.

Sometimes the mongodb collections can get little complex with nested arrays/objects etc where it would be relatively difficult to build loops around them. My work around is kinda raw but works in most scenarios regardless of complexity of the collection.

1. Export The collection using mongodump into .bson

mongodump --db=<db_name> --collection=<products> --out=data/

2. Convert .bson into .json format using bsondump

bsondump --outFile products.json data/<db_name>/products.bson

3. Replace the strings in the .json file with sed(for linux terminal) or with any other tools

sed -i 's/oldstring/newstring/g' products.json

4. Import back the .json collection with mongoimport with --drop tag where it would remove the collection before importing

mongoimport --db=<db_name>  --drop --collection products <products.json

Alternatively you can use --uri for connections in both mongoimport and mongodump

example

mongodump --uri "mongodb://mongoadmin:mystrongpassword@10.148.0.7:27017,10.148.0.8:27017,10.148.0.9:27017/my-dbs?replicaSet=rs0&authSource=admin" --collection=products --out=data/

来源：https://stackoverflow.com/questions/12589792/how-to-replace-substring-in-mongodb-document