PDO将数组绑定到Where IN [英] PDO bind array to Where IN

问题描述

我想将字符串数组绑定到SQL命令的WHERE IN部分，然后在SQL Server上运行该命令.问题可能是我尝试绑定字符串数组而不是整数数组.

I want to bind a array of Strings to the WHERE IN part of a SQL command, which I want to run afterwards on a SQL Server. The problem is probably that I try to bind an array of Strings and not an array of integers.

$totalCount = 
"SELECT referral, COUNT(username) AS cnt FROM accounts
WHERE referral IN ($refIdsPartial) GROUP BY referral";

$ps_totalCounts = $dbh->prepare($totalCount);
$ps_totalCounts->execute();

//loop over total counts
foreach($ps_totalCounts as $row){
    echo "Test<br>";
}

我为您回应了$ refIdsPartial，所以您知道这是什么:

54469c27c687b332339627,54469ba0dec3e703865612,54469c77945c7091266617

它只是字符串/varchars的内爆数组. 我已经使用Managementstudio测试了SQL命令，并且可以确保此SQL命令有效，只要我对每个String/Varchar使用引号即可.示例:

Its just an imploded array of strings/varchars. I tested the SQL command with my Managementstudio and I can ensure that this SQL command works, as long das I use the quote signs for each String/Varchar. Example:

SELECT referral, COUNT(username) AS cnt FROM accounts
WHERE referral IN ('54469c27c687b332339627','54469ba0dec3e703865612') GROUP BY referral

我的问题:

在上面的代码中，它永远不会进入foreach，因此查询的结果似乎为空.那里出了什么问题(当然，我只测试了应该有结果的查询)?

In the code above it never goes into the foreach, so the result of the Query seems to be empty. What is wrong there (Ofcourse I tested only queries which should have results)?

推荐答案

您可以使用一些字符串操作.

You could use some string manipulation.

您可以使用str_repeat("?", count(explode(",", $refIdsPartial)))来count您需要的?数.这将创建您的占位符.

You can count the number of ? you'd need by using str_repeat("?", count(explode(",", $refIdsPartial))). This will create your placeholders.

$totalCount = 
"SELECT referral, COUNT(username) AS cnt FROM accounts
WHERE referral IN (". str_repeat("?,", count(explode(",", $refIdsPartial))-1) . "?) GROUP BY referral";

现在占位符就位了，您可以从字符串中展开,并执行

Now that the placeholders are in place, you can explode the , from the string and execute

$ps_totalCounts->execute( explode(",", $refIdsPartial) );

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