首先,我在HTML文档中有一个选择选项
<form action="Journey.php" method="post">
<select name = "Startpoint">
<optgroup label = "Start point">
<option value = "GrimesDyke">GrimesDyke</option>
<option value = "SeacroftRingRoad">SeacroftRingRoad</option>
<option value = "WykeBeck">WykeBeck</option>
<option value = "FfordeGrene">FfordeGrene</option>
<option value = "St.JamesHospital">St.JamesHospital</option>
.........
我确定操作是使用 post 方法将数据传递给 Journey.php 文件以执行某些算法,但是当我单击浏览器中的提交按钮时,它向我显示了我的整个 PHP 代码....所以我决定运行一些这样的测试:
Journey.php
<?php
if(isset($_POST['submit'])){
$selected = $_POST['Startpoint']; // Storing Selected Value In Variable
echo "You have selected :" .$selected; // Displaying Selected Value
}
?>
这一次,它什么也没显示,我想做的是,将传递给 PHP 文件的 Startpoint 的值存储在一个$selected变量中,并在屏幕上回显它,但它仍然不起作用
我在网上检查了很多例子,但老实说,我看不出我做错了什么,请指出我的错误并告诉我我如何才能做到正确,非常感谢。
我很久以前就写过php,但我记得$_POST
表中没有'submit'
键。尝试改为检查'Startpoint'
密钥。
这对
我有用:
<?php
if(isset($_POST['submit'])){
$selected = $_POST['startpoint']; // Storing Selected Value In Variable
echo "You have selected: " . $selected; // Displaying Selected Value
};
?>
<form action="" method="post">
<select name = "startpoint">
<optgroup label = "Start point">
<option value = "GrimesDyke">GrimesDyke</option>
<option value = "SeacroftRingRoad">SeacroftRingRoad</option>
<option value = "WykeBeck">WykeBeck</option>
<option value = "FfordeGrene">FfordeGrene</option>
<option value = "St.JamesHospital">St.JamesHospital</option>
<input type="submit" name="submit" value="Submit">
</form>