[英]How can i access a dynamically changing variable in php page to other html page using jquery ajax method?
這是php頁面,它回顯變量$correct
。 其初始值為0
。 它隨着其值的增量而變化,但是當我使用jquery ajax方法將其回顯到其他頁面后加載此值時,它僅返回0
而不是增量值。
<?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 "$correct";
?>
這是其他頁面上的腳本,訪問此變量$correct
但始終會導致0
,盡管變量值在php頁面上正在更改? 我該如何糾正?
<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>
您沒有將任何答案從AJAX傳遞到PHP。
因此,您的PHP條件都不滿足。
因此它返回0。
要通過發布請求傳遞數據,請查閱ajax文檔(http://api.jquery.com/jQuery.ajax/)。 您將要查看data
屬性。
您將需要這樣的東西:
$.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,您的PHP需要收集數據,但我想您已經有了該設置:
$answer1 = $_POST['answer1'];
<input type="radio" name="answer_1" value="a" id="question_1_answer_a"/> <label for="question_1_answer_a">anirudh</label> <input type="radio" name="answer_1" value="b" id="question_1_answer_b"/> <label for="question_1_answer_b">anirudh</label>
所以你想要這樣的東西:
data: { answer_1: $('input[name="answer_1"]').val(), // from HTML example answer_2: $('input[name="answer_2"]').val(), answer_3: $('input[name="answer_3"]').val() // etc },
如果您不打算使用jQuery修改任何表單數據,則始終可以序列化並將其發布: http : //api.jquery.com/serialize/
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