php 如何为我的 MySQLi 查询显示错误？
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How to display errors for my MySQLi query?
提问by Iain Simpson
I am using the following script to process a form to add info to my website. The problem I am having is when I submit the form nothing gets submitted to the database, and there are no errors. How can I add error reporting to my query?
我正在使用以下脚本来处理表单以将信息添加到我的网站。我遇到的问题是当我提交表单时，没有任何内容提交到数据库，并且没有错误。如何将错误报告添加到我的查询中？
<?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='.$key);
?>
回答by Fabio
Just simply add or die(mysqli_error($db));at the end of your query, this will print the mysqli error.
只需or die(mysqli_error($db));在查询的末尾添加，这将打印 mysqli 错误。
mysqli_query($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));
As a side note I'd say you are at risk of mysql injection, check here How can I prevent SQL injection in PHP?. You should really use prepared statements to avoid any risk.
作为旁注，我会说您有风险mysql injection，请在此处查看如何防止 PHP 中的 SQL 注入？. 你真的应该使用准备好的语句来避免任何风险。
回答by Jessica
mysqli_error()
As in:
如：
$sql = "Your SQL statement here";
$result = mysqli_query($conn, $sql) or trigger_error("Query Failed! SQL: $sql - Error: ".mysqli_error($conn), E_USER_ERROR);
Trigger error is better than die because you can use it for development AND production, it's the permanent solution.
触发错误比死好因为您可以将其用于开发和生产，这是永久解决方案。
