Reputation: 599
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 = $statement1->fetchAll();
So the execute line gives me the string conversion error. I have printed my array and the question marks and they output fine. The sql statement is also fine, I have tried it on phpMyAdmin and it works just fine so clearly I am not using the execute properly but I am not sure how to change the way I have executed it.
Can someone explain it to me?
Upvotes: 4
Views: 18793
Reputation: 76656
execute()
method expects a single array. From the documentation:
Execute the prepared statement. If the prepared statement included parameter markers, you must either:
- (snip)
- pass an array of input-only parameter values
With array($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
)
)
This is certainly not what it expects. It needs a single-dimensional array containing the values to be inserted.
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");
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);
... and then modify the execute()
call like so:
$statement1->execute($array);
Upvotes: 6
Reputation: 227260
->execute()
expects you to send it an array with each element representing each ?
in the query.
You are only sending it 2 elements, so it's gonna try to use the entire array as the 2nd ?
, which is why it's trying to convert it to a string. And then it will probably complain about not having enough parameters.
Try it like this:
$dataArray = $array;
array_unshift($dataArray, $id);
$statement1->execute($dataArray);
Upvotes: 1