如何从html表单的下拉列表中选择选项的值到MySQL数据库？ [英] How to get value of selected option from a html form's dropdown list into mysql database?
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问题描述
制作一个带有下拉列表的表单，其中第一个选项由defualt选择。我怎样才能获得用户在我的数据库中选择的价值？提前感谢您的帮助？
HTML表格下降列表：
< select name =extrafield5>
< option value =NOWselected =selected>立即提交订单< / option>
< option value =审核>提交我的订单以供审核< / option>
< / select>
在PHP文件中
if（isset（$ _ POST ['extrafield5']））{
$ extrafield5 = $ _POST ['extrafield5'];
}
else {$ extrafield5 ='';}
解决方案
以下是一个具有类似要求的示例段（PHP-MySQL）：
filename.php
<？php
$ servername =localhost;
$ username =root;
$ password =;
$ tablename =name_of_table;
$ db_name =db_name;
//创建连接
$ conn = new mysqli（$ servername，$ username，$ password）; （$ conn-> connect_error）{
die（Connection failed：。$ conn-> connect_error）;
}
if（isset（$ _ POST ['extrafield5']））{
$ extrafield5 = $ _POST ['extrafield5'];
else {$ extrafield5 ='';}
$ sql =INSERT INTO $ tablename（fieldname）VALUES（'$ extrafield5'）;;
mysql_select_db（$ db_name）;
$ retval = mysql_query（$ sql，$ conn）;
if（！$ retval）
{
die（'Could not enter data：'。mysql_error（））;
}
回显输入数据成功\;
mysql_close（$ conn）;
？>
<！DOCTYPE html>
< html>
< head>
< title> PhpFiddle初始密码< / title>
< script type =text / javascript>
/ *您的脚本在这里* /
< / script>
< style type =text / css>
/ *您的css在这里* /
< / style>
< / head>
< body>
< div style =margin：30px 10％;>
< h3>我的表单< / h3>
< form action =method =postid =myformname =myform>
< select name =extrafield5>
< option value =NOWselected =selected>立即提交订单< / option>
< option value =审核>提交我的订单以供审核< / option>
< / select>
< input type =submit>
< / form>
< / div>
< / body>
< / html>
根据您的游览需求进行适当的更改。
这里PHP代码写在同一个文件中（否则用.php文件名指定表单操作）。
Have a form with a dropdown list with the first option selected by defualt. How can I get the value the user selects into my data base? Thanks in advance for your help?
HTML FORMS DROP DOWN LIST:
<select name="extrafield5">
<option value="NOW" selected="selected">Submit order now</option>
<option value="REVIEW">Submit my order for review</option>
</select>
IN PHP FILE
if (isset($_POST['extrafield5'])){
$extrafield5 = $_POST['extrafield5'];
}
else {$extrafield5 = '';}
解决方案
Here is a sample segment with a similar requirement(PHP-MySQL):
filename.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$tablename = "name_of_table";
$db_name = "db_name";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if (isset($_POST['extrafield5'])){
$extrafield5 = $_POST['extrafield5'];
}
else {$extrafield5 = '';}
$sql = "INSERT INTO $tablename (fieldname) VALUES('$extrafield5');";
mysql_select_db($db_name);
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
mysql_close($conn);
?>
<!DOCTYPE html>
<html>
<head>
<title>PhpFiddle Initial Code</title>
<script type="text/javascript">
/* Your scrips here */
</script>
<style type="text/css">
/* Your css here */
</style>
</head>
<body>
<div style="margin: 30px 10%;">
<h3>My form</h3>
<form action="" method="post" id="myform" name="myform">
<select name="extrafield5">
<option value="NOW" selected="selected">Submit order now</option>
<option value="REVIEW">Submit my order for review</option>
</select>
<input type="submit">
</form>
</div>
</body>
</html>
Make appropriate changes according yo tour requirement. Here the PHP code is written in the same file(Otherwise specify the form action with the .php file name).
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