目标返回 mysql lastInsertId 但是插入是在创建挑战的 php 函数中

Question

我需要 lastInsertID 的值来运行带有最新插入的新查询，但是挑战在于我的插入是 php 函数的一部分。我不知道返回 lastInsertID 的正确方法。

输入表单重定向到 index.php，其中我有一个查找“addClient”案例的开关，获取数据，然后调用该函数。最后一行被插入两次

--update-- 我确定在将函数分配给 var 时，会创建第二个条目。有什么建议？

$displayLastID = addClient($clientOrgName, $clientContactFName, ...etc...) 创建第二个条目。

//here's my function

function addClient($clientOrgName, $clientContactFName, $clientContactLName, $clientContactEmail, $clientContactPhone, 

                    $clientOnsiteFName, $clientOnsiteLName, $clientOnsiteEmail, $clientOnsitePhone, $clientAddress1,                    

                    $clientAddress2, $clientCity, $clientState, $clientZipCode, $clientTaxExempt, $clientAccountNumber) {

    global $db; 

    $query = 'INSERT INTO client

                (client_org_name, client_contact_first_name, client_contact_last_name, client_contact_email,                

                client_contact_phone, client_onsite_first_name, client_onsite_last_name, client_onsite_email,               

                client_onsite_phone, client_address_1, client_address_2, client_city, client_state, client_zipcode,                 

                client_tax_exempt, client_account_number)

            VALUES

                (:clientOrgName,:clientContactFName,:clientContactLName,:clientContactEmail,

                :clientContactPhone,:clientOnsiteFName,:clientOnsiteLName,:clientOnsiteEmail,           

                :clientOnsitePhone,:clientAddress1,:clientAddress2,:clientCity,:clientState,:clientZipCode,

                :clientTaxExempt,:clientAccountNumber )';

    $statement = $db->prepare($query);

    $statement->bindValue(':clientOrgName',$clientOrgName );

    $statement->bindValue(':clientContactFName',$clientContactFName );

    $statement->bindValue(':clientContactLName', $clientContactLName);

    $statement->bindValue(':clientContactEmail', $clientContactEmail);

    $statement->bindValue(':clientContactPhone', $clientContactPhone);

    $statement->bindValue(':clientOnsiteFName', $clientOnsiteFName);

}