如何检查多对多集合的唯一性？ [英] How to check uniqueness of many-to-many collection?

问题描述

1. 用户提交了一个集合，这是我的PHP应用程序所需的流程（是的，它含糊不清，但它更容易）比方说，通过整数ID大约5个对象。 （它实际上更像15，但为了方便起见，我们假设为5）。
   
2. 应用程序检查此集合是否已经提交过，并将它保存在MySQL数据库中（如果不是
li>

3. 应用程序将这些对象保存到数据库中（如果它们之前未保存）并且集合是多对多的，所以有一个对象表，一个集合表和一个关联这两者的表。）
   
   
   

   一对样本流： p>
   
   

   ---

   
   
   
   
   
   1. 用户提交111,112,113,114
      
   2. 这套是新的！
      
   3. 我们已经看到了对象111和112，但是为113和114提取并保存数据，因为我们还没有。
   
   

   
   
   

   ---

   
   
   
   
   
   1. 另一个用户提交111,112,113,114
   
   
   2. 我们以前看过这个集合。不要打扰储蓄。
   
   
   3. 由于我们已经看到了这个集合，我们很明显看到了这些对象。不要打扰储蓄。
   
   

   
   
   

   ---

   
   
   

   步骤1和3很简单。第2步是我不知道如何继续。看起来不必要的数据库繁重的是要查询包含这些确切ID的集合的关系，所以我即将发布一些明显的解决方案，例如简单的ID列表和散列，但我也想知道是否还有更多理想的解决方案。

   
   
   

   谢谢！

   解决方案

   ，我会这样做：
   
   

   
   
   1. 在用户提交后，在进入数据库之前，我会将数据库中的数据提取到数组中。在你的例子中，我将有 $ collection = array（'111'，'112'，'113'，'114'）;
   
   

   2. 我将分两步检查新的用户输入。第一步是查看它是否已经存在于数据库中。如果不是，则插入。否则忽略：
      
      $ fore $（$输入为$输入）
      {
      if（！in_array（$ input，$ collection））
      {
      //在此处插入
      }
      }

   
   

   然后在第二个循环中，我反过来删除未被用户选中的数据。

    foreach（$ collection as $ data ）
    {
    if（！in_array（$ data，$ inputs））
    {
    // do delete here 
   } 
   } 

   就你而言，你可能需要也可能不需要第二个循环。我需要这个，因为我把输入作为复选框，用户可以选择激活/停用，因此我把它翻译为插入和删除。

   
   

   Here's the desired flow of my PHP app (yes, it's vague, but it's easier that way):

   1. User submits a set of, let's say, about 5 objects by integer IDs. (It'll really be more like 15, but let's say 5 for ease.)
   2. App checks if this collection has been submitted before, and saves it in a MySQL database if not
   3. App saves these objects in the database, if they haven't been saved before

   (Objects and collections are many-to-many, so there is an objects table, a collections table, and a table relating the two.)

   A couple sample flows:

   ---

   1. User submits 111, 112, 113, 114
   2. This set is new! The collection is saved.
   3. We've seen objects 111 and 112, but fetch and save the data for 113 and 114, since we haven't.

   ---

   1. Another user submits 111, 112, 113, 114
   2. We've seen this collection before. Don't bother saving.
   3. Since we've seen the collection, we've obviously seen the objects. Don't bother saving.

   ---

   Steps 1 and 3 are simple. Step 2 is where I'm not sure how to proceed. It seems unnecessarily database-heavy to be querying the relationship for sets containing those exact IDs, so I'm about to post a few obvious solutions such as a simple ID list and hashing, but I'd also like to know if there are more ideal solutions out there.

   Thanks!

   解决方案

   In my application, I will do this step:

   1. After user submit, and before entering to database, I will fetch the data in the database into an array. In your example above, I will have $collection = array('111', '112', '113', '114');

   2. I will check the new user input in two step. First step is to see if it already in the database or not. If it not, then insert. Otherwise ignore:

      foreach ( $inputs as $input )
      {
        if ( ! in_array($input, $collection) )
        {
          //do insert here
        }
      }

   Then in second loop, I do it in reverse, to delete the data that not selected by user.

   foreach ( $collection as $data )
   {
     if ( ! in_array($data, $inputs) )
     {
       //do delete here
     }
   }

   In your case, you might or might not need the second loop. I needed this since I make the input as checkboxes, that the user can choose to activate / deactivate, thus I translate it as insert and delete.

   这篇关于如何检查多对多集合的唯一性？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！