本文介绍了如何显示我的MySQLi查询错误?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

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.如何在查询中添加错误报告?

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?

<?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);
?>

推荐答案

只需在查询末尾添加or die(mysqli_error($db));,这将显示mysqli错误.

Just simply add or die(mysqli_error($db)); at the end of your query, this will print the mysqli error.

 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));

作为旁注,我想说您有患mysql injection的风险,请在此处检查.您应该真正使用准备好的语句来避免任何风险.

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.

这篇关于如何显示我的MySQLi查询错误?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-19 02:07