javascript - PHP AJAX 和 MySQL

Question

我尝试遵循本教程 http://www.w3schools.com/php/php_ajax_database.asp
我想从 select(带有 onchange)发送值来更改 sql 查询。

function myfunctionTime(time) {
      if (time=="") {
        document.getElementById("txtHint").innerHTML="";
        return;
      } 
      if (window.XMLHttpRequest) {
        // code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
      } else { // code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
      }
      xmlhttp.onreadystatechange=function() {
        if (xmlhttp.readyState==4 && xmlhttp.status==200) {
          document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
        }
      }
      xmlhttp.open("GET","book.php?id=<?php echo $FILM_ID; ?>&q="+time,true);
      xmlhttp.send();
    }

这是带有 onchange 的选择:

<p for="session">Time:</p>
<select class="form-control" id="session" name="session"onchange="myfunctionTime(this.value)"> 
    <option selected="selected" value="12:00">12:00</option>
    <option value="16:00">16:00</option>
    <option value="20:00">20:00</option>
</select>

这是我想要获取变量的 PHP。

// Print all the get variables
    print_r($_GET);

    // Film session or time
    $session = "12:00"; 
    $session = $_GET['q'];
    echo "Session: ".$session;

    $query = "SELECT * FROM booking WHERE (FILM_ID = '$film_id' AND BOOKING_SESSION = '$session')";

这是我得到的错误:

Notice: Undefined index: q in C:\xampp\htdocs...\book.php on line 271

Answer 1

将您的 $session 值更改为此

 $session = $_GET['id'];