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mysqli_fetch_assoc()期望参数/调用成员函数bind_param()错误。 如何获取并修复实际的mysql错误？ 1个答案

mysqli还是死了，它必须死吗？ 4个答案

我正在使用以下脚本来处理将信息添加到我的网站的表单。 我遇到的问题是，当我提交表单时，什么也没有提交到数据库，并且没有错误。 如何将错误报告添加到查询中？

if (isset($_POST['itemdescription'])) {$itemdescription = $_POST['itemdescription'];}else {$itemdescription = '';}

if (isset($_POST['itemnumber'])) {$itemnumber = $_POST['itemnumber'];}else {$itemnumber = '';}

if (isset($_POST['sellerid'])) {$sellerid = $_POST['sellerid'];}else {$sellerid = '';}

if (isset($_POST['purchasedate'])) {$purchasedatepre = $_POST['purchasedate'];$date = DateTime::createFromFormat("D F d, Y", $purchasedatepre);$purchasedate = date('Y-m-d',strtotime($purchasedatepre));}else {$purchasedatepre = ''; $purchasedate = '';}

if (isset($_POST['otherinfo'])) {$otherinfo = $_POST['otherinfo'];}else {$otherinfo = '';}

if (isset($_POST['numberofitems'])) {$numberofitems = $_POST['numberofitems'];}else {$numberofitems = '';}

if (isset($_POST['numberofitemsused'])) {$numberofitemsused = $_POST['numberofitemsused'];}else {$numberofitemsused = '';}

if (isset($_POST['isitdelivered'])) {$isitdelivered = $_POST['isitdelivered'];}else {$isitdelivered = '';}

if (isset($_POST['price'])) {$price = $_POST['price'];}else {$price = '';}

$itemdescription = str_replace("'", "", "$itemdescription");

$itemnumber = str_replace("'", "", "$itemnumber");

$sellerid = str_replace("'", "", "$sellerid");

$otherinfo = str_replace("'", "", "$otherinfo");

include("connectmysqli.php");

mysqli_query($db,"INSERT INTO stockdetails (`itemdescription`,`itemnumber`,`sellerid`,`purchasedate`,`otherinfo`,`numberofitems`,`isitdelivered`,`price`) VALUES ('$itemdescription','$itemnumber','$sellerid','$purchasedate','$otherinfo','$numberofitems','$numberofitemsused','$isitdelivered','$price')");

// header('Location: stockmanager.php?&key='.$key);

?>