我无法将选定的下拉列值检索到PHP变量
<!DOCTYPE html>
<html>
<body>
<form action="#" method="post">
<select name="Color">
<option value="Red">Red</option>
<option value="Green">Green</option>
<option value="Blue">Blue</option>
<option value="Pink">Pink</option>
<option value="Yellow">Yellow</option>
</select>
</form>
<?php
$selected_val = $_POST['Color']; // Storing Selected Value In Variable
echo "You have selected :" .$selected_val; // Displaying Selected Value
?>
</body>
</html>
请帮忙。
答案 0 :(得分:1)
我认为你应该使用javascript,我给你简单的脚本,它将为你获得所选选项的价值。用户在选择时更改值时的功能加载。
看看:
function run() {
document.getElementById("resultColorValue").innerHTML = document.getElementById("Color").value;
}
&#13;
<p>Choose your Color:</p>
<select id="Color" onchange="run()"> <!--Call run() function-->
<option value=""></option>
<option value="red">Red</option>
<option value="green">Green</option>
<option value="blue">Blue</option>
<option value="yellow">Yellow</option>
</select>
<p>Your color is: </p><p id="resultColorValue"></p>
&#13;
如果您正在寻找,请告知我们。)
答案 1 :(得分:0)
我建议使用多个
<form action="#" method="post">
<select name="Color[]">
<option value="Red">Red</option>
<option value="Green">Green</option>
<option value="Blue">Blue</option>
<option value="Pink">Pink</option>
<option value="Yellow">Yellow</option>
</select>
<input type="submit" name="submit" value="Submit" />
</form>
PHP:
<?php
if(isset($_POST['submit'])) {
foreach ($_POST['Color'] as $select)
{
echo $select;
}
}
?>