UNION ALL的数组绑定问题

如何解决UNION ALL的数组绑定问题

我正在尝试使用UNION ALL联接多个查询。

我尝试了以下代码。出现错误Invalid parameter number: number of bound variables does not match number of tokens。

$code = '1,2,3,4';
$codeArray = explode(',',$code);
$inQuery = implode(',array_fill(0,count($codeArray),'?'));    
$full_dt = date('Y-m-d H:i:s');
$start_date = "2020-08-28 14-44-23";
$medication = "OD";
    $query = "SELECT SUM(counts) AS allcounts FROM
  (SELECT COUNT(b.id) AS counts
   FROM pat_info a
   INNER JOIN pat_medication b ON a.id = b.pat_id
   WHERE a.status != 2
     AND b.status != 2
     AND b.directions = '$medication'
     AND b.last_med_time < '$start_date'
     AND '$full_dt' BETWEEN b.start_date AND b.end_date
     AND a.location_code IN($inQuery)
   UNION ALL SELECT COUNT(d.id) AS counts
   FROM pat_info c
   INNER JOIN prn_medication d ON c.id = d.pat_id
   WHERE c.status != 2
     AND d.status != 2
     AND d.dose_frequency = '$medication'
     AND d.last_med_time < '$start_date'
     AND '$full_dt' BETWEEN d.start_date AND d.end_date
     AND c.location_code IN($inQuery) ) x ";
    $statement = $conn->prepare($query);
    $codeArray = array_merge($codeArray,$codeArray);
    $statement->execute($codeArray);

有print_r($inQuery);个结果?,??,?

使用print_r($codeArray);

在array_merge($codeArray,$codeArray);之前

结果Array ( [0] => 1 [1] => 2 ) Array ( [0] => 1 [1] => 2 [2] => 1 [3] => 2 )

使用print_r($codeArray);

在array_merge($codeArray,$codeArray);

之后

结果Array ( [0] => 1 [1] => 2 [2] => 1 [3] => 2 ) Array ( [0] => 1 [1] => 2 [2] => 1 [3] => 2 [4] => 1 [5] => 2 [6] => 1 [7] => 2 )

解决方法

我认为在这种情况下，最好将bindParam与命名参数一起使用，例如：

$code = '1,2,3,4';
$full_dt = date('Y-m-d H:i:s');
$start_date = "2020-08-28 14-44-23";
$medication = "OD";
    $query = "SELECT SUM(counts) AS allcounts FROM
  (SELECT COUNT(b.id) AS counts
   FROM pat_info a
   INNER JOIN pat_medication b ON a.id = b.pat_id
   WHERE a.status != 2
     AND b.status != 2
     AND b.directions = :medication
     AND b.last_med_time < :start_date
     AND :full_dt BETWEEN b.start_date AND b.end_date
     AND FIND_IN_SET(a.location_code,:code)>0
   UNION ALL SELECT COUNT(d.id) AS counts
   FROM pat_info c
   INNER JOIN prn_medication d ON c.id = d.pat_id
   WHERE c.status != 2
     AND d.status != 2
     AND d.dose_frequency = :medication
     AND d.last_med_time < :start_date
     AND :full_dt BETWEEN d.start_date AND d.end_date
     AND FIND_IN_SET(c.location_code,:code)>0 ) x ;";
    $statement = $conn->prepare($query);
    $statement->bindParam(':code',$code,PDO::PARAM_STR);
    $statement->bindParam(':full_dt',$full_dt,PDO::PARAM_STR);
    $statement->bindParam(':start_date',$start_date,PDO::PARAM_STR);
    $statement->bindParam(':medication',$medication,PDO::PARAM_STR);
    $statement->execute();

问题应该是数字“？”与您传递的参数数量不匹配。

此处在“ $ statement-> execute（$ codeArray）”中，$ codearray的元素数应与逗号分隔的“？”数相同。在$ inQuery中。与“ With print_r（$ inQuery）; result？，??，？”共享的内容看起来很有趣，不是逗号分隔的问号序列吗？

绑定所有参数

看看$full_dt = date('Y-m-d H:i:s');

让我们回显查询，并仔细看一下这一部分：

...
...
AND '2020-09-16 18:57:31' BETWEEN d.start_date AND d.end_date
...
...

然后我们转到文档：https://www.php.net/manual/en/pdostatement.bindparam.php#refsect1-pdostatement.bindparam-parameters

参数标识符。对于使用命名占位符的准备好的语句，这将是格式为:name的参数名。对于使用问号占位符的准备好的语句，这将是参数的1索引位置。

您能否看到来自:57的{{1}}和:31的两个不请自来的来宾:i和:s？

预计它们也将绑定到某些数据。

INNER JOIN在每个表中都有数据时返回结果。这意味着返回行将是在每个表上至少共享一个a.id，b.pat_id和c.pat_id的条目。

也许您对LEFT JOIN更加感兴趣，这将返回所有扩展了pat_info和pat_medication的{{1}}条目，无论您在其中有数据还是没有数据表。

https://dev.mysql.com/doc/refman/5.7/en/join.html

选择* 从表A 内部联接表B b 在a.common = b.common 内部联接表C c 在b.common = c.common

UNION ALL的数组绑定问题

如何解决UNION ALL的数组绑定问题

解决方法

绑定所有参数

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