MongoDB drop every database

I would like to know if there're a command to drop every databases from my MongoDB?

I know if I want to drop only one datatable, I just need to type the name of the database like th开发者_开发知识库e code below but I dont want to have to specify it.

mongo DB_NAME --eval 'db.dropDatabase();'

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you can create a javascript loop that do the job and then execute it in the mongoconsole.

var dbs = db.getMongo().getDBNames()
for(var i in dbs){
    db = db.getMongo().getDB( dbs[i] );
    print( "dropping db " + db.getName() );
    db.dropDatabase();
}

save it to dropall.js and then execute:

mongo dropall.js

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Try this command:

mongo --quiet --eval 'db.getMongo().getDBNames().forEach(function(i){db.getSiblingDB(i).dropDatabase()})'

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You can also do this with a simple mongo command:

db.adminCommand("listDatabases").databases.forEach( function (d) {
    if (d.name != "local" && d.name != "admin"  && d.name != "apiomat"  && d.name != "config")
        db.getSiblingDB(d.name).dropDatabase();
 })

---

db
  .getMongo()
  .getDBNames()
  .filter(n => !['admin','local','config'].includes(n))
  .forEach(dname =>
    db
      .getMongo()
      .getDB(dname)
      .dropDatabase()
  )
;

One-liner

db.getMongo().getDBNames().filter(n => !['admin','local','config'].includes(n)).forEach(dname => db.getMongo().getDB(dname).dropDatabase());

This drops all DBs but preserves MongoDB's internal collections.

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Save this to drop_all_dbs.js:

var databases = db.getMongo().getDBNames()
for(var i in databases){
    db = db.getMongo().getDB( databases[i] );
    if(db.getName() == "admin" || db.getName() == "local"){
        print("skipping db " + db.getName())
        continue
    }
    print( "dropping db " + db.getName() );
    db.dropDatabase();
}

Now you can execute:

mongo drop_all_dbs.js

and all databases (except for admin and local) will be dropped.

This answer is a copy of ALoR's one, just fix drop of system dbs

---

Adding to @ALoR's answer, for convenience you can put the following in ~/.mongorc.js

function dropDatabases(){
    var mongo = db.getMongo();

    var dbNames = mongo.getDBNames();
    for (var i = 0; i < dbNames.length; i++) {
        var db = mongo.getDB( dbNames[i] );

        print( "Dropping database " + db.getName() + "..." );
        db.dropDatabase();
    }
}

Then at the mongo shell you can simply do

dropDatabases()

From the docs:

Mongo will read the .mongorc.js file from the home directory of the user invoking mongo. In the file, users can define variables, customize the mongo shell prompt, or update information that they would like updated every time they launch a shell.

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You can do it easy through c# official driver:

var _mongoServer = MongoServer.Create("mongodb://localhost:27020");

var names = _mongoServer.GetDatabaseNames();
foreach (var name in names)
{
   _mongoServer.DropDatabase(name);
}

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Here is a oneliner that allows you to filter for specific names:

for (const dbName of db.getMongo().getDBNames().filter(dbName => dbName.startsWith('test-'))) {console.log(dbName); const dbToDelete = db.getMongo().getDB( dbName ); dbToDelete.dropDatabase()}

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This is what worked best for me.

Save this to a file called drop_most_databases.js:

const databasesToSkip = ['admin', 'local', 'config'];
db.getMongo()
  .getDBNames()
  .forEach((dbname) => {
    if (databasesToSkip.includes(dbname)) {
      print(`Skipping ${dbname}`);
    } else {
      print(`Dropping ${dbname}...`);
      db.getMongo().getDB(dbname).dropDatabase();
    }
  });
print(`Now, the only databases that remain are:`);
db.getMongo()
  .getDBNames()
  .forEach((dbname) => {
    print(dbname);
  });

From Terminal, run mongo drop_most_databases.js.

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It is as easy as

mongo --eval 'db.dropDatabase()'

Or, you can start a mongo session on your terminal and write

db.dropDatabase()

Which is exactly the same.