基于输入类型选择的表单操作 [英] Form action based on Input Type selection
问题描述
我在php页面index.php中创建一个表单。当页面被加载时,会创建以下表单:
I am creating a form in php on the page "index.php". When the page is loaded the following form is created:
if($_SERVER['REQUEST_METHOD']!='POST')
{
echo '
<form action="index.php" method="POST">
<fieldset style="width: 700px;">
<legend>Enter your search below</legend>
<textarea rows="1" cols="80" name="query">
</textarea>
</fieldset>
<p>
<input type="radio" value="Non-Aggregated"> Non-Aggregated
<input type="radio" value="Aggregated"> Aggregated
<input type="submit" value="Search">
</p>
</form>';
}
当用户点击提交按钮时,会显示相应的内容: p>
When the user clicks the submit button, the appropriate content is displayed:
else
{
if ($_POST['query'])
{
//content displayed after form submission
}
}
回到
Going back to the form, note the radio options:
<input type="radio" value="Non-Aggregated"> Non-Aggregated
<input type="radio" value="Aggregated"> Aggregated
是否有条件可以放在if语句中以执行基于不同的操作关于是否从单选按钮中选择 非聚合 或 聚合 ;如果是的话,我会如何去做这件事?
Is there a condition that I can place in the if-statement to carry out a different action based on whether Non-Aggregated or Aggregated is selected from the radio buttons; and if so how would I go about doing this?
感谢您的帮助。
推荐答案
<input type="radio" name="aggr" value="Non-Aggregated"> Non-Aggregated
<input type="radio" name="aggr" value="Aggregated"> Aggregated
执行POST方法后,您可以使用PHP检查值:
After POST method is executed you can check values with PHP:
if($_POST['aggr']=='Aggregated'){
//DO STUFF
}
if($_POST['aggr']=='Non-Aggregated'){
//DO OTHER STUFF
}
换句话说,您可以设置名称,如
In other way, you can set names, like
<input type="radio" name="Non-Aggregated" value="Non-Aggregated"> Non-Aggregated
<input type="radio" name="Aggregated" value="Aggregated"> Aggregated
检查 isset
if(isset($_POST['Aggregated'])){
//DO STUFF
}
if(isset($_POST['Non-Aggregated'])){
//DO OTHER STUFF
}
这篇关于基于输入类型选择的表单操作的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!