javascript - 如何从 PHP 向 Javascript 发送变量？

Question

如何使用 Ajax 将变量值从 PHP 发送到 Javascript？下面是JS代码和PHP代码

PHP 代码:

<?php
    $username = "trainerapp";
    $password = "password";
    $hostname = "localhost";
    $link = @mysql_connect($hostname, $username, $password);

    if(@mysql_select_db("trainer_registration"))
    {
        echo "Connected Successfully";
    }
    else
    {
        echo "Connection Error";
    }



    $select_query_num = @mysql_query("select id,program_name,company,date_prog from program_details");
    $num_rows = @mysql_num_rows($select_query_num);
    for ($i = 0; $i <= $num_rows; $i++){
            $id = $_POST["idjs"];
            $pgmname = $_POST["pgmnamejs"];
            $comp = $_POST["compjs"];
            $datephp = $_POST["datephpjs"];
        $select_query = @mysql_query("select id,program_name,company,date_prog from program_details where id = $i");
        $fetch_query = @mysql_fetch_assoc($select_query);
        $id =  $fetch_query['id'];
        echo $id;
        $pgmname = $fetch_query['program_name'];
        echo $pgmname;
        $comp =  $fetch_query['company'];
        echo $comp;
        $datephp = $fetch_query['date_prog'];
        echo $datephp;  

        }
?>

JS 代码:

window.onload = function createdivs() {
        var id;
        var pgmname;
        var comp;
        var datephp;
        var i = 1;
        for (;i < 10;i++)
        {
            div = "<div>.display";
            var list = document.createElement("div");
            document.getElementById('fulldisplay').appendChild(list);
            list.className = "container content-rows";
        }
        $.ajax({
        url:'displaycontent.php',
        data:{idjs:id, pgmnamejs:pgmname, compjs:comp, datephpjs:datephp},
        type:'POST',
        success:function(retval){
        alert(retval);
        }
    });

    }

问题:

在 JS 代码中，对于 var i 的每次增量，我需要对 PHP 文件进行 Ajax 调用，该文件应返回第一个值数组和第二个值数组，依此类推。我实际上很困惑如何做到这一点。有解释会更好。通过上面的代码，我只得到最后一个数组值，并出现未识别的索引错误。

Answer 1

如果这不是正式作品，那么只需在 php 脚本中回显 HTML 即可。

$select_query = @mysql_query("select id,program_name,company,date_prog from program_details where id = $i");
$fetch_query = @mysql_fetch_array($select_query. MYSQL_NUM);
list($id, $pgname, $comp, $datephp) = $fetch_query;
echo "
  <div>
    $id $pgname $comp $datephp
  </div>
";