collections

问题 What is the best way to drop a collection in MongoDB? I am using the following: db.collection.drop() As described in the manual: db.collection.drop() Removes a collection from the database. The method also removes any indexes associated with the dropped collection. The method provides a wrapper around the drop command. But how can I drop it from the command line? 回答1: So either of these are valid ways to do it: mongo <dbname> --eval 'db.<collection>.drop()' # ^^^^^^^^ ^^^^^^^^^^^^ db.

问题 What is the best way to drop a collection in MongoDB? I am using the following: db.collection.drop() As described in the manual: db.collection.drop() Removes a collection from the database. The method also removes any indexes associated with the dropped collection. The method provides a wrapper around the drop command. But how can I drop it from the command line? 回答1: So either of these are valid ways to do it: mongo <dbname> --eval 'db.<collection>.drop()' # ^^^^^^^^ ^^^^^^^^^^^^ db.

问题 What is the best way to drop a collection in MongoDB? I am using the following: db.collection.drop() As described in the manual: db.collection.drop() Removes a collection from the database. The method also removes any indexes associated with the dropped collection. The method provides a wrapper around the drop command. But how can I drop it from the command line? 回答1: So either of these are valid ways to do it: mongo <dbname> --eval 'db.<collection>.drop()' # ^^^^^^^^ ^^^^^^^^^^^^ db.

问题 I was reading the comments on this answer and I saw this quote. Object instantiation and object-oriented features are blazing fast to use (faster than C++ in many cases) because they're designed in from the beginning. and Collections are fast. Standard Java beats standard C/C++ in this area, even for most optimized C code. One user (with really high rep I might add) boldly defended this claim, stating that heap allocation in java is better than C++'s and added this statement defending the

问题 This question already has answers here : Java - Iterating over every two elements in a list (11 answers) Closed 4 years ago . I have an ArrayList and I want to iterate over it each 2 elements, like this: ([1,2], [3, 4], [5, 6]) Currently I'm able to iterate the list like this: ([1,2], [2, 3], [3, 4]) but this is not the result I want. Here is my code: Iterator<MyObject> iterator = myList.iterator(); if (iterator.hasNext()) { MyObject o1 = iterator.next(); while (iterator.hasNext()) { final

问题 I'm trying to public function show(Product $product) { return ProductDetailResource::collection($product); } Call to undefined method App\ Model ::mapInto(), exception: BadMethodCallException file vendor\laravel\framework\src\Illuminate\Support\Traits\ForwardsCalls.php then google and cant find any results 回答1: just need to replace public function show(Product $product) { return ProductDetailResource::make($product); } instead of: public function show(Product $product) { return

问题 I want to use Laravel 5.1 Collection's Unique method to filter unique IDs from nested objects. Given the data structure { "key1": [ {"id": 1}, {"id": 1} ], "key2": [ {"id": 1}, {"id": 2} ] } I want to return the same datastructure with duplicate id 1 removed from "key 1". I wanted to use $unique = $collection->unique('id'); , but this doesn't seem to apply to a nested datastructure as I have. So I thought to use $collection $input = $request->all(); $collection = collect($input); $collection-

问题 I want to use Laravel 5.1 Collection's Unique method to filter unique IDs from nested objects. Given the data structure { "key1": [ {"id": 1}, {"id": 1} ], "key2": [ {"id": 1}, {"id": 2} ] } I want to return the same datastructure with duplicate id 1 removed from "key 1". I wanted to use $unique = $collection->unique('id'); , but this doesn't seem to apply to a nested datastructure as I have. So I thought to use $collection $input = $request->all(); $collection = collect($input); $collection-

问题 Suppose there is a list say List<String> myList = new ArrayList<String>(); myList.add("okay"); myList.add("omg"); myList.add("kk"); I am doing this: List<String> fianllist = myStream.map(item -> item.toUpperCase()).filter(item ->item.startsWith("O")).collect(Collectors.toList()); My question is what the difference between map and filter as both can take a lambda expression as a parameter. Can some one please explain? 回答1: By using map , you transform the object values. The map operation

问题 Suppose there is a list say List<String> myList = new ArrayList<String>(); myList.add("okay"); myList.add("omg"); myList.add("kk"); I am doing this: List<String> fianllist = myStream.map(item -> item.toUpperCase()).filter(item ->item.startsWith("O")).collect(Collectors.toList()); My question is what the difference between map and filter as both can take a lambda expression as a parameter. Can some one please explain? 回答1: By using map , you transform the object values. The map operation