Php PDO绑定错误

我尝试从db中获取一些值

$client_ids = array('client_id' => $this->arParams['client_id']);
 print_r($client_ids);
 $client_ids_in = implode(',', array_fill(0, count($client_ids), '?'));
 $query = "SELECT odc.curr_id FROM office.dictionary_currency AS odc LEFT JOIN office.adwords_clients_google AS oacg ON odc.curr_code = oacg.client_currency WHERE oacg.client_id IN ($client_ids_in)";
        $google_currency = $this->DB->prepare($query);
        $google_currency->execute($client_ids);
        $google_currency->setFetchMode(PDO::FETCH_ASSOC);
        $google_currency = $google_currency->fetch();
        $google_currency = $google_currency['curr_id'];

$client\u id

看起来像

Array
(
    [client_id] => 15087
)

$query

SELECT odc.curr_id FROM office.dictionary_currency AS odc LEFT JOIN office.adwords_clients_google AS oacg ON odc.curr_code = oacg.client_currency WHERE oacg.client_id IN (?)

我得到一个错误

PHP Warning:  PDOStatement::execute(): SQLSTATE[HY093]: Invalid parameter number: parameter was not defined in /var/www/instruments/reports/report.php on line 531

这意味着，行中有错误

$google_currency = $this->DB->prepare($query);

出了什么问题？如何修复？

更改代码

 $client_ids = array('client_id' => $this->arParams['client_id']);

变成

 $client_ids = array($this->arParams['client_id']);

因此，

$client\u id

就像

Array
(
   0 => 15087
)

---

如果使用通用占位符

？

，则需要传递以执行非关联数组。 因此，您可以从数组中删除

'client\u id'

键，或者使用命名占位符

可能的解决方案

$client_ids = array($this->arParams['client_id']);

---

添加了$query，非常感谢您的帮助！非常感谢你的帮助！