PDO执行数组字符串转换错误 [英] PDO execute array to string conversion error

问题描述

我得到一个数组字符串转换错误，当我尝试运行PDO执行。的执行必须输入一个是正常串与其他的阵列。我的继承人code：

I am getting an array to string conversion error when I am trying to run a PDO execute. The execute has to inputs one being a normal string and the other an array. Heres my code:

$id = "1";    
$array = array("a",  "b",  "c");
$in  = str_repeat('?,', count($array) - 1) . '?';

$sql1 = "SELECT * FROM tbl1 AS tb1 RIGHT JOIN tbl2 AS tb2 ON (tb1.e_id = tb2.e_id)LEFT JOIN tbl3 AS tb3 ON (tb2.m_id = tb3.m_id) WHERE tb1.u_id = ? AND tb1.letters IN ($in)";


$statement1 = $db_handle->prepare($sql1);
$statement1->setFetchMode(PDO::FETCH_ASSOC);
$statement1->execute(array($id,$array));

$res = $statement1->fetchAll();

因此​​，执行行给我的字符串转换错误。我有我的打印阵列和问号，他们输出的罚款。 SQL语句还不错，我已经试过它的phpMyAdmin，这我没有使用正确的执行，但我不知道如何改变我已执行的方式工作得很好，那么清楚。

So the execute line gives me the string conversion error. I have printed my array and the question marks and they output fine. The sql statement is also fine, I have tried it on phpMyAdmin and it works just fine so clearly I am not using the execute properly but I am not sure how to change the way I have executed it.

有人能解释一下？

推荐答案

的execute（）方法需要一个单一的阵列。从文档：

execute() method expects a single array. From the documentation:

执行prepared声明。如果prepared声明包含参数标记，则必须：

Execute the prepared statement. If the prepared statement included parameter markers, you must either:

  
• （SNIP）

  
• 通过输入唯一的参数值的数组

  

使用阵列（$ ID，$阵列）你会传递一个多维数组，看起来像这样：

With array($id,$array) you'd be passing a multi-dimensional array that looks like this:

Array
(
    [0] => 1
    [1] => Array
        (
            [0] => a
            [1] => b
            [2] => c
        )

)

这当然不是什么期望。它需要被插入含有的值的一维数组。

This is certainly not what it expects. It needs a single-dimensional array containing the values to be inserted.

要解决这个问题，你必须修改阵列格式。该ID添加到阵列的开始，像这样：

To fix this issue, you have to modify the array format. Add the ID to the start of the array, like so:

$array = array(1, "a",  "b",  "c");

如果是动态生成的ID变量，那么你可以手动将其添加到阵列中开始使用 array_unshift（）功能：

If the ID variable is generated dynamically, then you can manually add it to the array start using array_unshift() function:

$id = "1";    
$array = array("a",  "b",  "c");
$array = array_unshift($array, $id);

...然后修改的execute（）调用像这样：

$statement1->execute($array);

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