具有可变大小变量列表的 MySQL 准备语句 [英] MySQL Prepared statements with a variable size variable list

问题描述

你将如何用 PHP 编写一个准备好的 MySQL 语句，每次使用不同数量的参数?此类查询的示例是:

How would you write a prepared MySQL statement in PHP that takes a differing number of arguments each time? An example such query is:

SELECT `age`, `name` FROM `people` WHERE id IN (12, 45, 65, 33)

IN 子句每次运行时都会有不同数量的 id.

The IN clause will have a different number of ids each time it is run.

我有两种可能的解决方案，但想看看是否有更好的方法.

I have two possible solutions in my mind but want to see if there is a better way.

可能的解决方案 1 使语句接受 100 个变量，并用保证不在表中的虚拟值填充其余变量；多次调用超过 100 个值.

Possible Solution 1 Make the statement accept 100 variables and fill the rest with dummy values guaranteed not to be in the table; make multiple calls for more than 100 values.

可能的解决方案 2 不要使用准备好的语句；构建并运行查询检查可能的注入攻击.

Possible Solution 2 Don't use a prepared statement; build and run the query checking stringently for possible injection attacks.

推荐答案

我能想到几个解决方案.

I can think of a couple solutions.

一种解决方案可能是创建一个临时表.将 in 子句中的每个参数插入表中.然后对您的临时表进行简单的连接.

One solution might be to create a temporary table. Do an insert into the table for each parameter that you would have in the in clause. Then do a simple join against your temporary table.

另一种方法可能是做这样的事情.

Another method might be to do something like this.

$dbh=new PDO($dbConnect, $dbUser, $dbPass);
$parms=array(12, 45, 65, 33);
$parmcount=count($parms);   // = 4
$inclause=implode(',',array_fill(0,$parmcount,'?')); // = ?,?,?,?
$sql='SELECT age, name FROM people WHERE id IN (%s)';
$preparesql=sprintf($sql,$inclause);  // = example statement used in the question
$st=$dbh->prepare($preparesql);
$st->execute($parms);

我怀疑但没有证据表明第一个解决方案可能更适用于较大的列表，而后者适用于较小的列表.

I suspect, but have no proof, that the first solution might be better for larger lists, and the later would work for smaller lists.

为了让@orrd 开心，这里有一个简洁的版本.

To make @orrd happy here is a terse version.

$dbh=new PDO($dbConnect, $dbUser, $dbPass);
$parms=array(12, 45, 65, 33);
$st=$dbh->prepare(sprintf('SELECT age, name FROM people WHERE id IN (%s)',
                          implode(',',array_fill(0,count($parms),'?'))));
$st->execute($parms);

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