使用ajax显示所选内容不起作用,PHP,AJAX,JAVASCRIPT [英] Using ajax to display what is being selected not working, PHP, AJAX, JAVASCRIPT
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问题描述
我使用ajax尝试显示正在选择的内容,但是由于某种原因它根本不显示任何内容.我知道通过使用函数内部的alert可以调用ajax函数本身,并且我认为真正的问题实际上在test2.php中,但是我不确定自己做错了什么.请看一下:
Using ajax, I'm trying to display what is being selected, but it's not displaying anything at all for some reason. I know the ajax function itself got called, by using alert inside the function, and I think the real problem is actually in test2.php, but I'm not sure what I did wrong. Please take a look:
test1.php
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
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>
推荐答案
您尚未将数据发布到 test2
!!
you have not post data to 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;
}
});
}
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