形式没有使用MySQL [英] form didnt working with mysql

问题描述

这是html

 < form action =addadd.phpmethod =postenctype =multipart / form-dataname =form1id =form1> 
< p> 
名字< br> 
< label for =firstname>< / label> 
< input type =textname =firstnameid =firstname/> 
< / p> 
< p> 
姓氏< br> 
< label for =lastname>< / label> 
< input type =textname =lastnameid =lastname/> 
< / p> 
< p> 
手机< br> 
< label for =mobile>< / label> 
< input type =textname =mobileid =mobile/> 
< / p> 
< p> 
电邮< br> 
< label for =email>< / label> 
< input type =textname =emailid =email/> 
< / p> 
< p> 
< input type =submitname =buttonid =buttonvalue =Submit/> 
< input type =resetname =button3id =button3value =Reset/> 
< / p> 
< / form> 
  

这是php

数据库连接

 <？php 
 $ con =的mysql_connect（ 本地主机， 根， ）; 
 if（！$ con）
 {
 die（connection to database failed.mysql_error（））; 
} 
 $ dataselect = mysql_select_db（qoot，$ con）; 
 if（！$ dataselect）
 {
 die（Database namelist not selected.mysql_error（））; 
} 
？> 
 
 
 <？php 
 $ unm = $ _SESSION ['name']; 
 $ fname = $ _ POST ['firstname']; 
 $ lname = $ _ POST ['lastname']; 
 $ ema = $ _ POST ['email']; 
 $ mob = $ _ POST ['mobile']; 
？> 
 
 
 $ b $ lt; $ php 
 $ qry = mysql_query（INSERT INTO address（firstname，lastname，mobile，email）VALUES（'$ fname'，$ lname'，'$ ema'，'$ mob'），$ con）; 
？> 
  

现在问题在于这不会在数据库中插入任何东西。

我还可以尝试什么来检查出现问题的地方？
已更新数据库连接详细信息

解决方案

将您的表名地址使用 mysqli _

使用




> <？php
/ start session
session_start（）;
//建立连接
$ con = mysqli_connect（localhost，root，，qoot）;
if（！$ con）
{
die（connection to database failed.mysqli_error（$ con））;
}
//从表单
$ unm = $ _SESSION ['name'];
$ fname = $ _ POST ['firstname'];
$ lname = $ _ POST ['lastname'];
$ ema = $ _ POST ['email'];
$ mob = $ _ POST ['mobile'];
？>



<？php
//插入到数据库
$ qry = mysqli_query（$ con，INSERT INTO`address`（firstname， （'$ fname'，$ lname'，'$ ema'，'$ mob'））或die（mysqli_error（$ con））;
？>


PS我会说你冒着 mysql注入风险，请点击此处如何防止PHP中的SQL注入？。您应该使用准备好的语句来避免任何风险。

this is the html

<form action="addadd.php" method="post" enctype="multipart/form-data" name="form1" id="form1">
    <p>
        First Name<br>
        <label for="firstname"></label>
        <input type="text" name="firstname" id="firstname" />
    </p>
    <p>
        Last Name<br>
        <label for="lastname"></label>
        <input type="text" name="lastname" id="lastname" />
    </p>
    <p>
        Mobile<br>
        <label for="mobile"></label>
        <input type="text" name="mobile" id="mobile" />
    </p>
    <p>
        Email<br>
        <label for="email"></label>
        <input type="text" name="email" id="email" />
    </p>
    <p>
        <input type="submit" name="button" id="button" value="Submit" /> 
        <input type="reset" name="button3" id="button3" value="Reset" />
    </p>
</form>

This is the php

database connection

<?php
$con = mysql_connect("localhost","root","");
if(!$con)
{
die("connection to database failed".mysql_error());
}
$dataselect = mysql_select_db("qoot",$con);
if(!$dataselect)
{
die("Database namelist not selected".mysql_error());
}
?>


<?php 
    $unm = $_SESSION['name']; 
    $fname=$_POST['firstname']; 
    $lname=$_POST['lastname']; 
    $ema=$_POST['email'];
    $mob=$_POST['mobile'];
?>



<?php 
    $qry=mysql_query("INSERT INTO address( firstname, lastname, mobile, email)VALUES('$fname',$lname','$ema','$mob')", $con);
?>

Now the problem is that this is inserting nothing in to my database.
What else can i try in order to check where things go wrong? updated database connection details

解决方案

Place your tablename address in backticks

Using mysqli_

  <?php
/start session
session_start();
//establish connection
$con = mysqli_connect("localhost","root","","qoot");
if(!$con)
{
die("connection to database failed".mysqli_error($con));
}
     //read the values from form
    $unm = $_SESSION['name']; 
    $fname=$_POST['firstname']; 
    $lname=$_POST['lastname']; 
    $ema=$_POST['email'];
    $mob=$_POST['mobile'];
?>



<?php 
   //insert to database
    $qry=mysqli_query($con,"INSERT INTO `address` ( firstname, lastname, mobile, email)VALUES('$fname',$lname','$ema','$mob')") or die(mysqli_error($con));
?>

P.S 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.

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