如何显示MySQLi查询的错误？

Question

如何显示MySQLi查询的错误？

我使用以下脚本来处理表单以向我的网站添加信息。我遇到的问题是，当我提交表单时，没有任何内容被提交到数据库，并且没有错误。如何向查询添加错误报告？

<?php

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);

?>

Answer 1

只需or die(mysqli_error($db));在查询结束时添加，就会打印出mysqli错误。&nbsp;mysqli_query($db,"INSERT&nbsp;INTO&nbsp;stockdetails&nbsp;(`itemdescription`,`itemnumber`,`sellerid`,`purchasedate`,`otherinfo`,`numberofitems`, &nbsp;`isitdelivered`,`price`)&nbsp;VALUES&nbsp;('$itemdescription','$itemnumber','$sellerid','$purchasedate','$otherinfo','$numberofitems',' &nbsp;$numberofitemsused','$isitdelivered','$price')")&nbsp;or&nbsp;die(mysqli_error($db));作为旁注，我会说你有风险mysql injection，请点击这里如何在PHP中阻止SQL注入？。您应该使用准备好的语句来避免任何风险。

Answer 2

mysqli_error()如：$sql&nbsp;=&nbsp;"Your&nbsp;SQL&nbsp;statement&nbsp;here";$result&nbsp;=&nbsp;mysqli_query($sql)&nbsp;or&nbsp;trigger_error("Query&nbsp;Failed!&nbsp;SQL:&nbsp;$sql&nbsp;-&nbsp;Error:&nbsp;".mysqli_error(),&nbsp; E_USER_ERROR);触发错误比die更好，因为你可以将它用于开发和生产，它是永久的解决方案。

Answer 3

出于开发目的，您可以在行中or die(mysqli_error($conn))的分号前添加mysqli_query($conn, 'SELECT...')。确保在推送到生产之前删除它，因此不要向公众输出有关数据库的信息。