您好，我是 PHP 的新手。我试图从 PHP 中的单选按钮获取所选值，但我无法获取所选值。我通过连接到我的数据库 (MySQL) 填充了这些值，但我无法从单选按钮获取分配的值。它总是逃避 if 条件并显示“未选择值”，我无法分配值并将其保存到我的数据库
感谢您的帮助。
我的index.php如下`
<?php
session_start();
$_SESSION['timein']= time();
?>
<?php
include("config.php");
$conn = mysqli_connect($dbHost, $dbuser, $dbpassword, $dbDatabase);
$query_salutation_type='SELECT salutation_description FROM tbl_salutation;';
$select_salutation_type=mysqli_query($conn, $query_salutation_type);
while($row1 = mysqli_fetch_array($select_salutation_type))
{
echo '<input type="radio" name="salutation_description" value="'.$row1[0].'"/>'.$row1[0];
}
?>
<html>
<body>
<form method="post" action="capture_data.php" >
<input type="submit" name="submit" value="Submit"/>
</form>
</body>
</html>
`
我的capture_data.php如下
<?php
if(isset($_POST['submit']))
{
if(isset($_POST['salutation_description']))
{
$selected_val = $_POST['salutation_description'];
echo "You have selected :" .$selected_val;
}
else
{
echo 'No Value Selected';
}
}
?>
最佳答案
表单仅发送介于 <form> 之间的输入和 </form> .您在 <form> 之前回显单选按钮.
<?php
session_start();
$_SESSION['timein']= time();
?>
<html>
<body>
<form method="post" action="capture_data.php" >
<?php
include("config.php");
$conn = mysqli_connect($dbHost, $dbuser, $dbpassword, $dbDatabase);
$query_salutation_type='SELECT salutation_description FROM tbl_salutation;';
$select_salutation_type=mysqli_query($conn, $query_salutation_type);
while($row1 = mysqli_fetch_array($select_salutation_type))
{
echo '<input type="radio" name="salutation_description" value="'.$row1[0].'"/>'.$row1[0];
}
?>
<input type="submit" name="submit" value="Submit"/>
</form>
</body>
</html>
关于php - 动态创建单选按钮并分配选定的值 PHP，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/35053645/