我需要 lastInsertID 的值來運行帶有最新插入的新查詢，但是挑戰(zhàn)在于我的插入是 php 函數(shù)的一部分。我不知道返回 lastInsertID 的正確方法。輸入表單重定向到 index.php，其中我有一個查找“addClient”案例的開關(guān)，獲取數(shù)據(jù)，然后調(diào)用該函數(shù)。最后一行被插入兩次--update-- 我確定在將函數(shù)分配給 var 時，會創(chuàng)建第二個條目。有什么建議？$displayLastID = addClient($clientOrgName, $clientContactFName, ...etc...) 創(chuàng)建第二個條目。//here's my functionfunction addClient($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);}