如何顯示MySQLi查詢的錯誤？我使用以下腳本來處理表單以向我的網站添加信息。我遇到的問題是，當我提交表單時，沒有任何內容被提交到數據庫，并且沒有錯誤。如何向查詢添加錯誤報告？<?phpif (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='.$key);?>
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TA貢獻1821條經驗 獲得超5個贊
只需or die(mysqli_error($db));在查詢結束時添加，就會打印出mysqli錯誤。
mysqli_query($db,"INSERT INTO stockdetails (`itemdescription`,`itemnumber`,`sellerid`,`purchasedate`,`otherinfo`,`numberofitems`,
`isitdelivered`,`price`) VALUES ('$itemdescription','$itemnumber','$sellerid','$purchasedate','$otherinfo','$numberofitems','
$numberofitemsused','$isitdelivered','$price')") or die(mysqli_error($db));作為旁注，我會說你有風險mysql injection，請點擊這里如何在PHP中阻止SQL注入？。您應該使用準備好的語句來避免任何風險。
狐的傳說
TA貢獻1804條經驗 獲得超3個贊
mysqli_error()
如：
$sql = "Your SQL statement here";$result = mysqli_query($sql) or trigger_error("Query Failed! SQL: $sql - Error: ".mysqli_error(),
E_USER_ERROR);觸發錯誤比die更好，因為你可以將它用于開發和生產，它是永久的解決方案。
largeQ
TA貢獻2039條經驗 獲得超8個贊
出于開發目的，您可以在行中or die(mysqli_error($conn))的分號前添加mysqli_query($conn, 'SELECT...')。
確保在推送到生產之前刪除它，因此不要向公眾輸出有關數據庫的信息。
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