1
這是與變量$correct
相呼應的php頁面。其初始值爲0
。它隨着它的值增加而改變,但是當我使用jquery ajax方法將該值加載到其他頁面後,它只返回0
而不是增量值。如何使用jquery ajax方法在php頁面訪問其他html頁面的動態變量變量?
<?php
$i=1;
while($i<=10) {
$answer="answer_".$i;
${"answer_$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 "$correct";
?>
這是訪問這個變量$correct
其他網頁上的腳本,但它總是導致0
,雖然該變量的值是PHP頁面上的變化?我怎樣才能糾正這一點?
<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>
我知道這是我的形式,然後我通過它傳遞什麼數據。假設它由10個問題組成,每個4個選項都是這樣的。這是我第一個問題的兩個選項 \t \t \t \t \t <標籤= 「question_1_answer_b」> anirudh – TeamA1