javascript - 使用ajax显示正在选择的内容不起作用，PHP，AJAX，JAVASCRIPT

Question

使用ajax，我试图显示正在选择的内容，但由于某种原因它根本没有显示任何内容。我知道 ajax 函数本身被调用了，通过在函数内部使用警报，我认为真正的问题实际上是在 test2.php 中，但我不确定我做错了什么。请看一下:

test1.php

<?php

include('ajax.php');

echo "<select name = 'select' onchange = 'ajax(\"test2.php\",\"output\")'>";

echo "<option value = '1'> 1 </option>";
echo "<option value = '2'> 2 </option>";
echo "<option value = '3'> 3 </option>";

echo "</select>";
echo "<div id = 'output'/>";

?>

test2

<?php

$select = $_POST['select'];
echo $select;

?>

ajax.php

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>

<script type = "text/javascript"> 

function ajax(url,id) {

      $.ajax({
           type: "POST",
           url: url,
           error: function(xhr,status,error){alert(error);},
           success:function(data) {
             document.getElementById( id ).innerHTML = data;
           }

      });

}

</script>

Answer 1

您尚未将数据发布到test2!!

<?php

include('ajax.php');

echo "<select id = 'select' onchange = 'ajax(\"test2.php\",\"output\")'>";

echo "<option value = '1'> 1 </option>";
echo "<option value = '2'> 2 </option>";
echo "<option value = '3'> 3 </option>";

echo "</select>";
echo "<div id = 'output'/>";

?>

function ajax(url,id) {

      $.ajax({
           type: "POST",
           url: url,
           data : {select:$('#select').find("option:selected").val()}, 
           error: function(xhr,status,error){alert(error);},
           success:function(data) {
             document.getElementById( id ).innerHTML = data;
           }

      });

}