Php 将变量值传递给URL
我使用下面的代码从URL获取参数Php 将变量值传递给URL,php,Php,我使用下面的代码从URL获取参数 <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post" role="search"> <select name="seme" id="seme"> <option value="sem1">Semester-1</option> <option value="sem2">Semes
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post" role="search">
<select name="seme" id="seme">
<option value="sem1">Semester-1</option>
<option value="sem2">Semester-2</option>
<option value="sem3">Semester-3</option>
<option value="sem4">Semester-4</option>
</select>
<input type="text" name="find" id="find" placeholder="Enter worksheet / file name..." />
<button type="submit" class="btn btn-theme">Search</button>
</form>
<?php
if ($_SERVER["REQUEST_METHOD"] == "POST") {
// collect value of input field
$name = $_REQUEST['find'];
$sem = $_REQUEST['seme'];
if (empty($name)) {
echo "Name is empty";
} else {
echo $name; echo $seme;
}
}
?>
但问题是我只得到文本框的值。我也想得到选择框的值。我该怎么办?或者我缺少什么
谢谢 为什么不使用$\u POST[]方法而不是$\u REQUEST[]来获取值?请尝试$\u POST['seme']您正在为selectbox值打印不同的变量名,您将在
$sem
和打印$seme
使用以下命令:
if ($_SERVER ["REQUEST_METHOD"] == "POST") {
// collect value of input field
$name = $_REQUEST ['find'];
$seme = $_REQUEST ['seme'];
if (empty ( $name )) {
echo "Name is empty";
} else {
echo $name . "<br/>";
echo $seme;
}
}
if($\u服务器[“请求\u方法”]=“发布”){
//收集输入字段的值
$name=$_请求['find'];
$seme=$_请求['seme'];
if(空($name)){
echo“名称为空”;
}否则{
echo$name。“
”;
echo$seme;
}
}
尝试以下方法:
<?php
if (isset($_POST["find"])&&isset($_POST["seme"])) {
// collect value of input field
$name = $_POST["find"];
$sem = $_POST["seme"];
echo $name; echo $seme;
}else{
if (!isset($_POST["find"])){
echo "Name is empty";
}
if (!isset($_POST["seme"])){
echo "Semester type is empty";
}
}
?>
您如何知道没有获取selectbox值?你回应了吗?谢谢你的提问!!已解决:)