将PHP表单提交给XAMPP MySQL数据库 [英] Submission of PHP Form to XAMPP MySQL database

问题描述

我为餐饮服务创建了一个php表单，我打算将其插入到MySQL数据库中：php表单称为index.php：

 <！DOCTYPE html> 
< html lang =en> 
< head> 
< meta charset =utf8> 
< meta http-equiv =X-UA-Compatiblecontent =IE = edge> 
< meta name =viewportcontent =width = device-width，initial-scale = 1> 
< link rel =stylesheethref =css / bootstrap.min.css> 
< link rel =stylesheethref =css / font-awesome.min.css> 
< link rel =stylesheethref =css / style.css> 
< title> BOOTSTRAP< / title> 
< / head> 
< div class =container-fluid> 
< div class =row> 
< div class =inquiry col-md-6> 
< h1>立即查询< / h1> 
< form method =postaction =inquire.phpname =inquireformid =inquireForm> 
< div class =form-group> 
< label for =InputName>名称*< / label> 
< input class =form-controltype =textid =inputNameplaceholder =Namename =Cust`Name> 
< / div> 
< div class =form-group> 
< label for =InputLocation>位置*< / label> 
< input class =form-controltype =textid =inputLocationplaceholder =Locationname =Location> 
< / div> 
< / form> 
< form class =form-inline> 
< div class =form-group> 
< label for =SelectDate>事件日期*< / label> 
 < option> Jan< / option> 
< option> Feb< / option> 
< option> Mar< / option> 
< option> Apr< / option> 
< / option> May< / option> 
< option> Jun< / option> 
< option> Jul< / option> 
< option> Aug< / option> 
< option> Sept< / option> 
< option> Oct< / option> 
< option> Nov< / option> 
< option> Dec< / option> 
< / select> 
< select class =form-controlid =SelectDayname =Day> 
< option> 1< / option> 
< option> 2< / option> 
< option> 3< / option> 
< option> 4< / option> 
< option> 5< / option> 
< option> 6< / option> 
< option> 7< / option> 
< option> 8< / option> 
< option> 9< / option> 
< option> 10< / option> 
< option> 11< / option> 
< option> 12< / option> 
< option> 13< / option> 
< option> 14< / option> 
< option> 15< / option> 
< option> 16< / option> 
< option> 17< / option> 
< option> 18< / option> 
< option> 19< / option> 
< option> 20< / option> 
< option> 21< / option> 
< option> 22< / option> 
< option> 23< / option> 
< option> 24< / option> 
< option> 25< / option> 
< option> 26< / option> 
< option> 27< / option> 
< option> 28< / option> 
< option> 29< / option> 
< option> 30< / option> 
< option> 31< / option> 
< / select> 
< select class =form-controlid =selectYearplaceholder =Yearname =Year> 
< option> 2017< / option> 
< option> 2018< / option> 
< option> 2019< / option> 
< option> 2020< / option> 
< option> 2021< / option> 
< option> 2022< / option> 
< / select> 
< / div> 
< / form> 
< form> 
< div class =form-group> 
< label for =InputNumber>客人人数*< / label> 
< input class =form-controltype =Numberid =inputNumberplaceholder =Numbername =Guests> 
< / div> 
< div class =form-group> 
< label for =InputContact>联络电话号码*< / label> 
< input class =form-controltype =textid =inputContactplaceholder =Contact Numbername =ContNum> 
< / div> 
< input class =btn btn-defaulttype =submitvalue =submit> 
< / form> 
< / div> 
< / div> 
< / div> 
< / body> 
< / html> 
  
并且inquire.php是： 
 
 
 <？php 
 $ mysql_host =localhost; 
 $ mysql_username =root; 
 $ mysql_password =password; 
 $ mysql_database =catering; 
 
 $ mysqli = new Mysqli（$ mysql_host，$ mysql_username，$ mysql_password，$ mysql_database）; 
 $ prepare = $ mysqli-> prepare（INSERT INTO`inquiry`（`CustName`，`Location`，`Month`，`Day`，`Year`，`Guests`，`ContNum`）VALUES ）（，，，，，，？？？？？？？）; 
 $ prepare-> bind_param（sssssss，print_r（$ POST）['CustName']，print_r（$ POST）['Location']，print_r（$ POST）['Month']，print_r $ POST）['Day']，print_r（$ POST）['Year']，print_r（$ POST）['Guests']，print_r（$ POST）['ContNum']）; 
 $ prepare-> execute（）; 
 $ mysqli-> close（）; 
？> 
  
我从其他问题中得到：如何将信息从html表单写入MySQL数据库 
 
 
 
但是，我的工作不正常。 php url只是改变，没有数据被插入到我的数据库中。我想可能是我的设置或安装有问题，也许它是下列之一？
 
 
我的电脑没有安装PHP， 
 
 
我将index.php和inquire.php放入xampp> htdocs文件夹中。   我以管理员模式运行XAMPP。 / li> 
 
 print_r（$ _ POST）可能不正确。
 
 
我真的需要帮助。我确保数据库名称，表名和列名匹配。我在这里做错了什么？
 
 
 
数据库图片：
 在这里输入图片描述 
 
 
 
更新;  
 
 
 
 
尚未插入任何数据，网址只会从file：//index.php更改为
 file：//index.php？Guests = 2& ContNum = 123 
 
 
 
 
解决方案
使用localhost / your_project_name.html在URL栏中访问您的项目。基本上你不使用xampp localhost，你只需双击我所得到的html文件即可。
 
 
 
好的，我的代码有一些更多的问题，我斑。  
 
 
首先在查询中将 print_r（$ POST ['something']）改为 $ _ POST [ '东西'] 。向前发展，你的表单中有错误的值，例如
中的 $ _ POST ['Cust Name'] 设置为name =Cust`Name。那么你有太多的形式，你必须缩小它。对于1个HTML文件，你不能有这么多的表单。 PHP不能读取元素它是服务器端，所以如果你保持这种结构，也许你应该考虑Ajax调用PHP。
 
 
 
 < ; form method =postaction =inquire.phpname =inquireformid =inquireForm> 
< div class =form-group> 
< label for =InputName>名称*< / label> 
< input class =form-controltype =textid =inputNameplaceholder =Namename =Cust`Name> 
< / div> 
< div class =form-group> 
< label for =InputLocation>位置*< / label> 
< input class =form-controltype =textid =inputLocationplaceholder =Locationname =Location> 
< / div> 
< / form> 
  
这部分在这里将向php文件发出请求，但这就是其余部分不会通过请求发送因为它们被包含在另一个表单中，没有动作，并且php文件不包含在那里。
 
I created a php form for a catering service that I'm planning to insert into a MySQL database: The php form is called index.php:
<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="utf8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1">
    <link rel="stylesheet" href="css/bootstrap.min.css">
    <link rel="stylesheet" href="css/font-awesome.min.css">
    <link rel="stylesheet" href="css/style.css">
    <title>BOOTSTRAP</title>
</head>
<div class="container-fluid">
<div class="row">
<div class="inquiry col-md-6">
        <h1>Inquire Now</h1>
            <form method="post" action="inquire.php" name="inquireform" id="inquireForm">
                <div class="form-group">
                    <label for="InputName">Name*</label>
                    <input class="form-control" type="text" id="inputName" placeholder="Name" name="Cust`Name">
                </div>
                <div class="form-group">    
                    <label for="InputLocation">Location*</label>
                    <input class="form-control" type="text" id="inputLocation" placeholder="Location" name="Location">
                </div>
            </form>
            <form class="form-inline">
                <div class="form-group">
                    <label for="SelectDate">Date of Event*</label>
                    <select class="form-control" id="SelectMonth" name="Month">
                        <option>Jan</option>
                        <option>Feb</option>
                        <option>Mar</option>
                        <option>Apr</option>
                        <option>May</option>
                        <option>Jun</option>
                        <option>Jul</option>
                        <option>Aug</option>
                        <option>Sept</option>
                        <option>Oct</option>
                        <option>Nov</option>
                        <option>Dec</option>
                    </select>
                    <select class="form-control" id="SelectDay" name="Day">
                        <option>1</option>
                        <option>2</option>
                        <option>3</option>
                        <option>4</option>
                        <option>5</option>
                        <option>6</option>
                        <option>7</option>
                        <option>8</option>
                        <option>9</option>
                        <option>10</option>
                        <option>11</option>
                        <option>12</option>
                        <option>13</option>
                        <option>14</option>
                        <option>15</option>
                        <option>16</option>
                        <option>17</option>
                        <option>18</option>
                        <option>19</option>
                        <option>20</option>
                        <option>21</option>
                        <option>22</option>
                        <option>23</option>
                        <option>24</option>
                        <option>25</option>
                        <option>26</option>
                        <option>27</option>
                        <option>28</option>
                        <option>29</option>
                        <option>30</option>
                        <option>31</option>
                    </select>
                    <select class="form-control" id="selectYear" placeholder="Year" name="Year">
                        <option>2017</option>
                        <option>2018</option>
                        <option>2019</option>
                        <option>2020</option>
                        <option>2021</option>
                        <option>2022</option>
                    </select>
                </div>
                </form>
                <form>
                <div class="form-group">    
                    <label for="InputNumber">Number of Guests*</label>
                    <input class="form-control" type="Number" id="inputNumber" placeholder="Number" name="Guests">
                </div>
                <div class="form-group">    
                    <label for="InputContact">Contact Number*</label>
                    <input class="form-control" type="text" id="inputContact" placeholder="Contact Number" name="ContNum">
                </div>
                <input class="btn btn-default" type="submit" value="submit">
            </form>
        </div>
    </div>
    </div>
</body>
</html>
And the inquire.php is: 
<?php
$mysql_host     = "localhost";
$mysql_username = "root";
$mysql_password = "password";
$mysql_database = "catering";

$mysqli  = new Mysqli($mysql_host, $mysql_username, $mysql_password, $mysql_database);
$prepare = $mysqli->prepare("INSERT INTO `inquiry`(`CustName`,`Location`,`Month`,`Day`,`Year`,`Guests`,`ContNum`) VALUES (?,?,?,?,?,?,?)");
$prepare->bind_param("sssssss", print_r($POST)['CustName'], print_r($POST)['Location'], print_r($POST)['Month'], print_r($POST)['Day'], print_r($POST)['Year'], print_r($POST)['Guests'], print_r($POST)['ContNum']);
$prepare->execute();
$mysqli->close();
?>
Which I got from this other question: How to write information from html form to MySQL Database

However, mine doesn't work. The php url just changes, no data is inserted into my database. I'm thinking maybe something's wrong with my setup or installation maybe its one of the following?

I have no PHP installed in my PC, just XAMPP.
I put the index.php and inquire.php in a folder in xampp >htdocs
I run XAMPP in Admin mode.
the print_r($_POST) might be incorrect.
Guys i really need help. I made sure the database name, table name, and column names matched. What could I be doing wrong here?

DATABASE PIC: 
enter image description here

Update ;

  
No data still inserted, the url only changes from file://index.php to
  file://index.php?Guests=2&ContNum=123

解决方案
Use localhost/your_project_name.html to access your project in the URL bar. Basically you don't use "xampp localhost" you just double-click on the html file from what i get.

Ok there are some MORE problems with the code that I spotted. 

First of all inside the print_r($POST['something']) in your query change it to $_POST['something']. Moving forward you have wrong values inside post for example $_POST['Cust Name'] in your form is set name="Cust`Name". Then you have too many forms you must narrow it down. For 1 html file you can't have that many form. Php can not "read elements" it's server side so if you keep that structure maybe you should consider ajax call to php.
<form method="post" action="inquire.php" name="inquireform" id="inquireForm">
    <div class="form-group">
        <label for="InputName">Name*</label>
        <input class="form-control" type="text" id="inputName" placeholder="Name" name="Cust`Name">
    </div>
    <div class="form-group">    
        <label for="InputLocation">Location*</label>
        <input class="form-control" type="text" id="inputLocation" placeholder="Location" name="Location">
    </div>
</form>
This part here will make the request to php file but that's it the rest will not be send through request because they are included in another form , without action and php file is not included there.

                        
这篇关于将PHP表单提交给XAMPP MySQL数据库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！
                        
                    
                    
                
				

                            
                        
                

            
        
        

            



        
        
        

            



        
    
    


    

    

        

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