数据在数据库中输入错误 [英] Data entering error in databas

查看:96
本文介绍了数据在数据库中输入错误的处理方法,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

我创建了一个表单,现在我想在数据库中提交它的数据。我有以下代码



 <?php  

$ dbconn = new MySQLi( localhost root < span class =code-string> Tajweed);

{

$ pname = mysqli_real_escape_string($ dbconn,$ _POST [' pname']);
$ plink = mysqli_real_escape_string($ dbconn,$ _POST [' PLINK']);
$ sql = $ dbconn-> query( INSERT INTO main_page(pname,plink)VALUES(' $ PNAME, '$ PLINK'));
}


?>

< 表格 name = myForm action = <?php $ _PHP_SELF?> 方法 = 发布 >
页面名称:< 输入 类型 = text 占位符 = 页面名称: 名称 = pname / >
页面链接:< 输入 type = text 占位符 = 页面链接: 名称 = plink / >
< input type = 提交 value < span class =code-keyword> =
提交 >
< / form >




< pre lang = PHP > < <跨度class =code-attribute> lang = PHP >







当我运行此代码时,它会给出错误

un-deafind index:pname和plink





请帮我解决问题。



我尝试了什么:





$ dbconn =新的MySQLi(localhost,root,,Tajweed);



{



$ pname = mysqli_real_escape_string($ dbconn,$ _POST ['pname']);

$ plink = mysqli_real_escape_string($ dbconn,$ _POST ['plink']);

$ sql = $ dbconn-> query(INSERT INTO main_page(pname,plink)VALUES('$ pname' ,'$ plink'));

}





?>



< form name =myFormaction =method =post>

页面名称:<输入type =textplaceholder =page name:name =pname/>

页面链接:< input type =textplaceholder =Page Link:name =plink />

< input type =submitvalue =submit>

< / form>





i尝试此代码

解决方案

dbconn = MySQLi( localhost root Tajweed);

{


pname = mysqli_real_escape_string(


dbconn,

I created a Form and now i want to submit it's data in the Database. i have the following code

    <?php

$dbconn = new MySQLi("localhost","root","","Tajweed");

{
 
 $pname = mysqli_real_escape_string($dbconn, $_POST['pname']);
$plink = mysqli_real_escape_string($dbconn, $_POST['plink']);
 $sql=$dbconn->query("INSERT INTO main_page (pname,plink) VALUES('$pname','$plink')");
}


?>
    
<form name="myForm" action="<?php $_PHP_SELF ?>" method="post">
Page Name: <input type="text" placeholder="page name :" name="pname" />
Page Link: <input type="text" placeholder="Page Link :" name="plink" />
<input type="submit" value="submit">
</form>




<pre lang="PHP"><pre lang="PHP">




when i run this code it's give an error that
un-deafind index: "pname" and "plink"


please help me to solve the problem.

What I have tried:


$dbconn = new MySQLi("localhost","root","","Tajweed");

{

$pname = mysqli_real_escape_string($dbconn, $_POST['pname']);
$plink = mysqli_real_escape_string($dbconn, $_POST['plink']);
$sql=$dbconn->query("INSERT INTO main_page (pname,plink) VALUES('$pname','$plink')");
}


?>

<form name="myForm" action="" method="post">
Page Name: <input type="text" placeholder="page name :" name="pname" />
Page Link: <input type="text" placeholder="Page Link :" name="plink" />
<input type="submit" value="submit">
</form>


i tried this code

解决方案

dbconn = new MySQLi("localhost","root","","Tajweed"); {


pname = mysqli_real_escape_string(


dbconn,


这篇关于数据在数据库中输入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IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆