如何将用户选择的选项的值从下拉选择框列表存储到mysql数据库中？ [英] How can I store the value of a user selected option from a drop down select box list into mysql database?

问题描述

具有下拉选择框列表的表单。如何将用户选择的选项的值从下拉选择框列表存储到mysql数据库中？谢谢。

FORM

  ; form action =xxx.phpclass =wellid =xxxnamexxxmethod =post> 
 
 
< select name =extrafield5> 
< option value =NOWselected =selected>立即提交订单< / option> 
< option value =REVIEW>提交我的订单以查看< / option> 
< button id =btn1type =submitvalue =提交>提交< / button> 
 
< / select> 
 
 
< / form> 
  

PHP FILE

 <？php 
 
 define（'DB_NAME'，'xxx'）; 
 define（'DB_USER'，'xxx'）; 
 define（'DB_PASSWORD'，'xxx'）; 
 define（'DB_HOST'，'xxx'）; 
 
 $ connection = mysqli_connect（DB_HOST，DB_USER，DB_PASSWORD）; 
 
 if（！$ connection）{
 die（'Database connection failed：'。mysqli_connect_error（））; 
} 
 
 $ db_selected = mysqli_select_db（$ connection，DB_NAME）; 
 
 if（！$ db_selected）{
 die（'Can not use'.DB_NAME''：'。mysqli_connect_error（））; 
} 
 
 echo'Connected successfully'; 
 
 
 if（isset（$ _ POST ['extrafield5']））{
 $ extrafield5 = $ _POST ['extrafield5']; 
} 
 
 else {$ extrafield5 ='';} 
 
 
 
 
 $ sql =INSERT INTO seguin_orders extrafield5）
 VALUES（'$ extrafield5'）; 
 
 if（！mysqli_query（$ connection，$ sql））{
 die（'Error：'。mysqli_connect_error（$ connection））; 
} 
  

DATABASE

http://oi60.tinypic.com/9ppc0i.jpg

解决方案

您需要提交表单。

 code>< form action =xxx.phpclass =wellid =xxxnamexxxmethod =post> 
< select name =extrafield5> 
< option value =NOWselected =selceted>立即提交订单< / option> 
< option value =REVIEW>提交我的订单以查看< / option> 
< / select> 
< input type =submitname =submitvalue =Submit/> 
< / form> 
  

并在PHP中获取：

<$ p $（$ _ code）if（isset（$ _ POST ['submit']））{
if（isset（$ _ POST ['extrafield5']））{
$ extrafield5 = $ _POST [ 'extrafield5'];
}
else {$ extrafield5 ='';}
}

Have a form with a dropdown select box list. How can I store the value of a user selected option from a drop down select box list into mysql database? Thanks.

FORM

<form action="xxx.php" class="well" id="xxx" name"xxx" method="post">


<select name="extrafield5">
 <option value="NOW" selected="selected">Submit order now</option>
 <option value="REVIEW">Submit my order for review</option>
<button id="btn1" type="submit" value="Submit">Submit</button>

</select>


</form>

PHP FILE

<?php

define('DB_NAME', 'xxx');
define('DB_USER', 'xxx');
define('DB_PASSWORD', 'xxx');
define('DB_HOST', 'xxx');

$connection = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);

if(!$connection){
die('Database connection failed: ' . mysqli_connect_error());
}

$db_selected = mysqli_select_db($connection, DB_NAME);

if(!$db_selected){
die('Can\'t use ' .DB_NAME . ' : ' . mysqli_connect_error());
}

echo 'Connected successfully';


if (isset($_POST['extrafield5'])){
    $extrafield5 = $_POST['extrafield5'];
}

else {$extrafield5 = '';}




$sql = "INSERT INTO seguin_orders (extrafield5) 
        VALUES ('$extrafield5')";

if (!mysqli_query($connection, $sql)){
die('Error: ' . mysqli_connect_error($connection));
}

DATABASE

http://oi60.tinypic.com/9ppc0i.jpg

解决方案

You need to submit the form.

<form action="xxx.php" class="well" id="xxx" name"xxx" method="post">
<select name="extrafield5">
 <option value="NOW" selected="selceted">Submit order now</option>
 <option value="REVIEW">Submit my order for review</option>
</select>
<input type="submit" name="submit" value="Submit"/>
</form>

And get it in PHP:

if (isset($_POST['submit'])){
  if (isset($_POST['extrafield5'])){
    $extrafield5 = $_POST['extrafield5'];
  }
  else {$extrafield5 = '';}
}

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