如何使用jquery ajax方法访问php页面中动态变化的变量到其他html页面 - How can i access a dynamically changing variable in php page to other html page using jquery ajax method?

How can i access a dynamically changing variable in php page to other html page using jquery ajax method?

这是响应变量$correct的php页面。初始值为0。它是随着它的值的增量变化，但当我加载这个值后，用jquery ajax方法将它返回到其他页面，它只返回0而不是增量值。

<?php
$i=1;
while($i<=10) {
    $answer="answerswer_".$i;
    ${"answerswer_$i"}=$_POST[$answer];
    $i++;
}
$correct=0;
if($answer_1=="a")
$correct=$correct+10;
else {
    if($answer_1=="b" || $answer_1=="c" || $answer_1=="d")
        $correct=$correct-10;
}
if($answer_2=="a")
$correct=$correct+10;
else {
if($answer_2=="b" || $answer_2=="c" || $answer_2=="d")
            $correct=$correct-10;
}
if($answer_3=="a")
$correct=$correct+10;
else {
    if($answer_3=="b" || $answer_3=="c" || $answer_3=="d")
        $correct=$correct-10;
}
if($answer_4=="a")
$correct=$correct+10;
else
{   if($answer_4=="b" || $answer_4=="c" || $answer_4=="d")
        $correct=$correct-10;
}
if($answer_5=="a")
$correct=$correct+10;
else
{   if($answer_5=="b" || $answer_5=="c" || $answer_5=="d")
        $correct=$correct-10;
}
if($answer_6=="a")
$correct=$correct+10;
else
{   if($answer_6=="b" || $answer_6=="c" || $answer_6=="d")
        $correct=$correct-10;
}
if($answer_7=="a")
$correct=$correct+10;
else
{   if($answer_7=="b" || $answer_7=="c" || $answer_7=="d")
        $correct=$correct-10;
}
if($answer_8=="a")
$correct=$correct+10;
else {
    if($answer_8=="b" || $answer_8=="c" || $answer_8=="d")
        $correct=$correct-10;
}
if($answer_9=="a")
$correct=$correct+10;
else
{
    if($answer_9=="b" || $answer_9=="c" || $answer_9=="d")
        $correct=$correct-10;
}
if($answer_10=="a")
$correct=$correct+10;
else
{
    if($answer_10=="b" || $answer_10=="c" || $answer_10=="d")
        $correct=$correct-10;
}
echo "$correct";
?>

这是其他页面上的脚本访问此变量$correct，但它总是导致0，虽然变量值在php页面上发生变化?我怎样才能纠正这个问题?

<script>
$(function(){
$('form').submit(function(event){
event.preventDefault();
$.ajax({
url:"results.php",
type:"POST",
success:function(result){
console.log(result);
$('#results').html("<p>"+result+"</p>");
}
});
});
});
</script>

您没有将任何答案从AJAX传递到PHP。
因此PHP中的所有条件都不满足。
因此它返回0。

要与post请求一起传递数据，请参阅ajax文档(http://api.jquery.com/jQuery.ajax/)。您需要查看data属性。

你会想要这样的东西:

$.ajax({
    url:"results.php",
    type:"POST",
    data: {
        answer_1: 'blue'    // what colour is the sky?
    },
    success:function(result){
        console.log(result);
        $('#results').html("<p>"+result+"</p>");
    }
});

然后你的PHP需要收集数据但我想你已经设置好了:

$answer1 = $_POST['answer1'];

人力资源>

<<strong>更新:包含来自用户注释的示例信息。

<input type="radio" name="answerswer_1" value="a" id="question_1_answer_a"/> 
<label for="question_1_answer_a">anirudh</label> 
<input type="radio" name="answerswer_1" value="b" id="question_1_answer_b"/> 
<label for="question_1_answer_b">anirudh</label>

所以你想要这样写:

data: {
    answer_1: $('input[name="answerswer_1"]').val(),    // from HTML example
    answer_2: $('input[name="answerswer_2"]').val(),
    answer_3: $('input[name="answerswer_3"]').val()
    // etc
},

如果你不打算用jQuery修改任何表单数据，你可以将它序列化并发布:http://api.jquery.com/serialize/