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如何将PHP文件中的多个值回显到Javascript Ajax调用?_Javascript_Php_Ajax - Fatal编程技术网

如何将PHP文件中的多个值回显到Javascript Ajax调用?

如何将PHP文件中的多个值回显到Javascript Ajax调用?,javascript,php,ajax,Javascript,Php,Ajax,参考下面附带的代码,我希望在index.php文件中引用在processUsersAnswers.php中初始化的变量 index.php中的代码片段 <script type="text/javascript"> function checkAnswers(){ var xhr = new XMLHttpRequest(); /* code */ xhr.open("POST", "processUsersAnswers.php", true);

参考下面附带的代码,我希望在index.php文件中引用在processUsersAnswers.php中初始化的变量

index.php中的代码片段

<script type="text/javascript"> 
function checkAnswers(){
    var xhr = new XMLHttpRequest();

    /* code */

    xhr.open("POST", "processUsersAnswers.php", true);
    xhr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xhr.onreadystatechange = function() {
        if(xhr.readyState == 4 && xhr.status == 200) {
            /* code */
        }
    }
    xhr.send(vars); // 
}
<?php 
$score = 403;
$userAnswers = array();
$userAnswers[0] = "bla bla";
$userAnswers[1] = "asfasfaeg";
$userAnswers[2] = "erhehdfh";

/*How can I retrieve $userAnswers in my index.php file? */
?>

函数checkAnswers(){
var xhr=new XMLHttpRequest();
/*代码*/
open(“POST”,“processUsersAnswers.php”,true);
setRequestHeader(“内容类型”,“应用程序/x-www-form-urlencoded”);
xhr.onreadystatechange=函数(){
如果(xhr.readyState==4&&xhr.status==200){
/*代码*/
}
}
xhr.send(变量);//
}

processUsersAnswers.php中的代码片段

<script type="text/javascript"> 
function checkAnswers(){
    var xhr = new XMLHttpRequest();

    /* code */

    xhr.open("POST", "processUsersAnswers.php", true);
    xhr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xhr.onreadystatechange = function() {
        if(xhr.readyState == 4 && xhr.status == 200) {
            /* code */
        }
    }
    xhr.send(vars); // 
}
<?php 
$score = 403;
$userAnswers = array();
$userAnswers[0] = "bla bla";
$userAnswers[1] = "asfasfaeg";
$userAnswers[2] = "erhehdfh";

/*How can I retrieve $userAnswers in my index.php file? */
?>

例如,如何在index.php文件中检索$score的值?

试试以下方法:

新的
processUsersAnswers.php
文件

<?php 
$score = 403;
$userAnswers = array();
$userAnswers[0] = "bla bla";
$userAnswers[1] = "asfasfaeg";
$userAnswers[2] = "erhehdfh";

echo json_encode('score' => $score);
exit;
试试这个:

新的
processUsersAnswers.php
文件

<?php 
$score = 403;
$userAnswers = array();
$userAnswers[0] = "bla bla";
$userAnswers[1] = "asfasfaeg";
$userAnswers[2] = "erhehdfh";

echo json_encode('score' => $score);
exit;

使用simple
echo
尝试:
echo$score
但变量很多。你能说清楚一点吗。你解决问题了吗?是的,我用逗号分隔了元素,然后创建了一个函数,将字符串解释为数组。谢谢。使用simple
echo
试试:
echo$score
但是变量很多。你能说清楚一点吗。你解决问题了吗?是的,我用逗号分隔了元素,然后创建了一个函数,将字符串解释为数组。谢谢