MySQL：加入（2个表）与单个查询（1个表）

In PHP, I have an array of 11 persons where just the ID of each person is given:

在PHP中，我有一个11人的数组，其中只给出了每个人的ID：

$persons = array(1, 3, 39, 72, 17, 20, 102, 99, 77, 2, 982);

In my MySQL database, there's a table containing detailed information about each person:

在我的MySQL数据库中，有一个表格，其中包含有关每个人的详细信息：

TABLE personInfo
 - ID (integer)
 - name (string)
 - birth date (timestamp)
 - weight (decimal)
 - ...

PROBLEM:

问题：

So now, I want to select the matching name for each ID in the PHP array. I can only imagine two solutions to do this:

所以现在，我想为PHP数组中的每个ID选择匹配的名称。我只能想象两种解决方案：

1. for-loop:

1. for-loop：

foreach ($persons as $person) {
    $result = mysql_query("SELECT name FROM personInfo WHERE id = ".$person);
}

2. logical operator OR

2.逻辑运算符OR

$result = mysql_query("SELECT name FROM personInfo WHERE id = 1 OR id = 3 OR id = 39 OR ...");

Both solutions are slow, aren't they?

两种解决方案都很慢，不是吗？

But if I had another MySQL table containing the IDs of the PHP array ...

但是，如果我有另一个MySQL表包含PHP数组的ID ...

TABLE ids
 - ID (integer)

... I could use a join to make a really fast MySQL query, right?

...我可以使用连接来进行非常快速的MySQL查询，对吧？

$result = mysql_query("SELECT a.ID, b.name FROM ids AS a JOIN personInfo AS b ON a.ID = b.ID");

QUESTION:

题：

Is all this correct so far? If yes: Why is this so? The MySQL query is faster if I have a second table? With only one table it is incredibly slow? What is the fastest way to solve my problem (selecting the names matching the IDs of the PHP array)?

到目前为止这一切都正确吗？如果是：为什么会这样？如果我有第二个表，MySQL查询会更快吗？只有一张桌子，它非常慢？解决我的问题的最快方法是什么（选择与PHP数组的ID匹配的名称）？

2 个解决方案

#1

If you have lots of ids (several hundreds or more), feeding the values into a temporary table and joining it is actually faster.

如果你有很多id（几百或更多），将值输入临时表并加入它实际上更快。

You may want to read this article:

您可能想阅读这篇文章：

Passing parameters in MySQL: IN list vs. temporary table
在MySQL中传递参数：IN列表与临时表

#2

IF you have ID's you should just query the Id's you have. So

如果你有ID，你应该只查询你的ID。所以

 $result = mysql_query("SELECT name FROM personInfo WHERE id = 1 OR id = 3 OR id = 39 OR ...");

would be right, although I would use

是的，虽然我会用

$result = mysql_query("SELECT name FROM personInfo WHERE id IN (1,2,3,4....) )

#1