从下拉菜单获取输入到php
从下拉菜单获取输入到php
提问于 2018-05-23 12:30:53
我正在尝试制作一个下拉列表，并从用户那里获取三个参数，它们将存储在一个php文件中供以后使用。我当前的代码是：
<body>
<header>
<form action="parameter.php" method="post">
<label class="heading">First</label>
<select name="First">
<option value="First-1">First-1</option>
<option value="First-2">First-2</option>
</select>
<input type="submit">
</form>
<form action="parameter.php" method="post">
<label class="heading">Second</label>
<select name="Second">
<option value="Second-1">Second-1</option>
<option value="Second-2">Second-2</option>
</select>
<input type="submit">
</form>
<form action="parameter.php" method="post">
<label class="heading">Third</label>
<select name="Third">
<option value="Third-1">Third-1</option>
<option value="Third-2">Third-2</option>
<input type="submit">
</select>
</form>
</header>
</body>
这里的问题是，我不希望每个参数都有提交按钮，而是所有参数都应该只有一个按钮。我尝试了几种方法，但都不起作用。
我的php文件
<?php
echo $_POST['First'];
echo $_POST['Second'];
echo $_POST['Third'];
?> 我哪里错了？
编辑:更正php代码
回答 3
Stack Overflow用户
发布于 2018-05-23 12:34:38
只需将所有选择封装在一个表单中，而不是三个不同的表单中。确保input提交不在select标记内。
另外，请注意您的php代码正在尝试回显在您共享的html中不存在的参数。您选择的name将是参数名。
<form action="parameter.php" method="post">
<label class="heading">First</label>
<select name="First" >
<option value="First-1">First-1</option>
<option value="First-2">First-2</option>
</select>
<label class="heading">Second</label>
<select name="Second" >
<option value="Second-1">Second-1</option>
<option value="Second-2">Second-2</option>
</select>
<label class="heading">Third</label>
<select name="Third" >
<option value="Third-1">Third-1</option>
<option value="Third-2">Third-2</option>
</select>
<input type="submit">
</form>PHP
<?php
echo $_POST['First'];
echo $_POST['Second'];
echo $_POST['Third'];
?> Stack Overflow用户
发布于 2018-05-23 12:41:31
您可以在一个表单中添加所有内容
<form action="parameter.php" method="post">
<label class="heading">First</label>
<select name="First" >
<option value="First-1">First-1</option>
<option value="First-2">First-2</option>
</select>
<label class="heading">Second</label>
<select name="Second" >
<option value="Second-1">Second-1</option>
<option value="Second-2">Second-2</option>
</select>
<label class="heading">Third</label>
<select name="Third" >
<option value="Third-1">Third-1</option>
<option value="Third-2">Third-2</option>
<input type="submit">
</select>
<input type="submit" name="submit" value="submit">
</form> 在你的代码中，你遗漏了"name“按钮。parameter.php代码应该是这样的。
if(isset($_POST['submit'])){
echo $First = $_POST['First'];
echo $Second = $_POST['Second'];
echo $Third = $_POST['Third'];
}Stack Overflow用户
发布于 2018-05-23 14:22:59
<form action="parameter.php" method="post" id="frm">
<label class="heading">First</label>
<select name="First" >
<option value="First-1">First-1</option>
<option value="First-2">First-2</option>
</select>
<label class="heading">Second</label>
<select name="Second" >
<option value="Second-1">Second-1</option>
<option value="Second-2">Second-2</option>
</select>
<label class="heading">Third</label>
<select name="Third" >
<option value="Third-1">Third-1</option>
<option value="Third-2">Third-2</option>
</select>
<input type="submit">
</form>JQuery AJAX
<script>
$('#frm').on('submit', function(e){
e.preventDefault();
$.ajax({
url : $(this).attr('action'),
type: "POST",
dataType: "json",
data : $(this).serialize(),
success: function(data)
{
console.log(data)
}
});
});
</script>PHP
<?php
echo $_POST['First'];
echo $_POST['Second'];
echo $_POST['Third'];
?>页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50479826
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