How to replace substring in mongodb document

Question

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?

Answer 1

db.filetranscoding.updateMany({ profiles: { $regex: /N_/ } },[{$set: { profiles: {$$replaceAll: { input: "$profiles", find:"N_",replacement: "" }},"status":"100"}}])

1. filetranscoding -- Collection Name

2. profiles -- ColumnName in which you want to update

3. /N_/ -- String which you are searching (where Condition )

4. find:"N_",replacement: "" -- N_ which u want to remove "" from which you want to remove here we are taking blank String

Answer 2

If you want to search for a sub string, and replace it with another, you can try like below,

    db.collection.find({ "fieldName": /.*stringToBeReplaced.*/ }).forEach(function(e, i){
        if (e.fieldName.indexOf('stringToBeReplaced') > -1) {
          e.content = e.content.replace('stringToBeReplaced', 'newString');
          db.collection.update({ "_id": e._id }, { '$set': { 'fieldName': e.fieldName} }, false, true);
        }
    }) 

Answer 3

Nowadays,

• starting Mongo 4.2, db.collection.updateMany (alias of db.collection.update) can accept an aggregation pipeline, finally allowing the update of a field based on its own value.
• starting Mongo 4.4, the new aggregation operator $replaceOne makes it very easy to replace part of a string.

// { URL: "www.abc.com/helloWorldt/..." }
// { URL: "www.abc.com/HelloWo/..." }
db.collection.updateMany(
  { URL: { $regex: /helloWorldt/ } },
  [{
    $set: { URL: {
      $replaceOne: { input: "$URL", find: "helloWorldt", replacement: "helloWorld" }
    }}
  }]
)
// { URL: "www.abc.com/helloWorld/..." }
// { URL: "www.abc.com/HelloWo/..." }

• The first part ({ URL: { $regex: /helloWorldt/ } }) is the match query, filtering which documents to update (the ones containing "helloWorldt") and is just there to make the query faster.
• The second part ($set: { URL: {...) is the update aggregation pipeline (note the squared brackets signifying the use of an aggregation pipeline):
  • $set is a new aggregation operator (Mongo 4.2) which in this case replaces the value of a field.
  • The new value is computed with the new $replaceOne operator. Note how URL is modified directly based on the its own value ($URL).

---

Before Mongo 4.4 and starting Mongo 4.2, due to the lack of a proper string $replace operator, we have to use a bancal mix of $concat and $split:

db.collection.updateMany(
  { URL: { $regex: "/helloWorldt/" } },
  [{
    $set: { URL: {
      $concat: [
        { $arrayElemAt: [ { $split: [ "$URL", "/helloWorldt/" ] }, 0 ] },
        "/helloWorld/",
        { $arrayElemAt: [ { $split: [ "$URL", "/helloWorldt/" ] }, 1 ] }
      ]
    }}
  }]
)

Answer 4

This can be done by using the Regex in the first part of the method replace and it will replace the [all if g in regex pattern] occurrence(s) of that string with the second string, this is the same regex as in Javascript e.g:

const string = "www.abc.com/helloWorldt/...";
console.log(string);
var pattern = new RegExp(/helloWorldt/)
replacedString = string.replace(pattern, "helloWorld");
console.log(replacedString);

Since the regex is replacing the string, now we can do this is MongoDB shell easily by finding and iterating with each element by the method forEach and saving one by one inside the forEach loop as below:

> db.media.find()
{ "_id" : ObjectId("5e016628a16075c5bd26fbe3"), "URL" : "www.abc.com/helloWorld/" }
{ "_id" : ObjectId("5e016701a16075c5bd26fbe4"), "URL" : "www.abc.com/helloWorldt/" }
> 
> db.media.find().forEach(function(o) {o.URL = o.URL.replace(/helloWorldt/, "helloWorld"); printjson(o);db.media.save(o)})
{
    "_id" : ObjectId("5e016628a16075c5bd26fbe3"),
    "URL" : "www.abc.com/helloWorld/"
}
{
    "_id" : ObjectId("5e016701a16075c5bd26fbe4"),
    "URL" : "www.abc.com/helloWorld/"
}
> db.media.find()
{ "_id" : ObjectId("5e016628a16075c5bd26fbe3"), "URL" : "www.abc.com/helloWorld/" }
{ "_id" : ObjectId("5e016701a16075c5bd26fbe4"), "URL" : "www.abc.com/helloWorld/" }
>

Answer 5

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/

Answer 6

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);
});

Answer 7

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));
})

Answer 8

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});
  });
});

Answer 9

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

Answer 10

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!

Answer 11

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