javascript - 使用 AJAX 和 PHP 将值插入 MYSQL db OnChange 的下拉框

Question

当下拉框发生变化时，我在通过 AJAX 执行 SQL 时遇到问题，如果可能，我希望获得一些帮助。

背景信息我的任务是创建一个每日日历，显示在健身房运行的所有类(class)，最多是每小时 5 节课，每节课 6 (30) 人，持续 14 小时。我不是专业人士，我可能已经创建了一个关于这个问题的复杂方法，如果我有，请告诉我。

我已经设法创建了一个 View ，该 View 由 14 列 30 个下拉框组成(每小时 5 节课，每节 6 节课，持续 14 小时)。每个下拉框都会轮询数据库，如果有条目，它将使用预订用户的名称填充该框。如果没有找到任何预订，它将创建一个下拉框，用于轮询成员(member)表并显示健身房的所有成员(member)，当更改时，希望可以预订该人。-这是我当前的问题!

每个下拉框的名称对应于我打算传递给 javascript 函数并最终传递给 SQL 语句的时间、组和人数。每个选项的值都与 memberid 相对应，该 memberid 也将传递给构造 SQL 所需的所有信息。

我目前的代码

HTML - 从 php 循环生成的片段

    <div id="results">
<div id="07" class="column">07:00<br/>
<div id="group1">
<select name="07:00-1-0" onchange="getda(this.value,this)">
    <option value="none">---------------</option>
    <option value="2">John Doe</option>
    <option value="1">Joe Bloggs</option>
</select>
<select name="07:00-1-1" onchange="getda(this.value,this)">
    <option value="none">---------------</option>
    <option value="2">John Doe</option>
    <option value="1">Joe Bloggs</option>
</select>

PHP

<?php
$mysqli = new mysqli("localhost", "root", "", "gym");

/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
function hyphenate($str) {
return implode("-", str_split($str, 2));
}
function getmembers($time,$group,$iteration)
{
$date=$_GET["date"];

$date=hyphenate($date);
$date = explode('-', $date);
$new_date = $date[2].'-'.$date[1].'-'.$date[0];

$mysqli = new mysqli("localhost", "root", "", "gym");
if ($iteration == 0){
$result = $mysqli->query("select members.memberid, members.firstname, members.lastname from bookings inner join members on bookings.memberid = members.memberid where bookings.date = '$new_date' and time = '$time' and bookings.groupnumber = '$group' order by bookings.bookingid ASC limit 1");
}
else {$result = $mysqli->query("select members.memberid, members.firstname, members.lastname from bookings inner join members on bookings.memberid = members.memberid where bookings.date = '$new_date' and time = '$time' and bookings.groupnumber = '$group' order by bookings.bookingid ASC limit 1,$iteration");
}
$rowcount=mysqli_num_rows($result);
if ($rowcount==$iteration && $iteration == 0)
{
 $result = $mysqli->query("select firstname, lastname,memberid from members order by firstname ASC");


echo '<select name="'.$time.'-'.$group.'-'.$iteration.'" onchange="getda(this.value,this)"><option value="---------------">---------------</option>';

while ($row = $result->fetch_assoc()) {

              unset($firstname, $lastname);
              $firstname = $row['firstname'];
              $lastname = $row['lastname']; 
              $memberid = $row['memberid'];
              echo '<option value="'.$memberid.'">'.$firstname . ' ' . $lastname .'</option>';

}

echo "</select>";
}
else if ($rowcount>=$iteration){
echo '<select name="'.$time.'-'.$group.'-'.$iteration.'" onchange="getda(this.value,this)">';

while ($row = $result->fetch_assoc()) {

              unset($firstname, $lastname);
              $firstname = $row['firstname'];
              $lastname = $row['lastname']; 
              $memberid = $row['memberid'];
              echo '<option value="'.$memberid.'">'.$firstname . ' ' . $lastname .'</option><option value="cancel">Cancel</option>';
}
echo "</select>";
}
else{
     $result = $mysqli->query("select firstname, lastname, memberid from members order by firstname ASC");


echo '<select name="'.$time.'-'.$group.'-'.$iteration.'" onchange="getda(this.value,this)"><option value="---------------">---------------</option>';

while ($row = $result->fetch_assoc()) {

              unset($firstname, $lastname);
              $firstname = $row['firstname'];
              $lastname = $row['lastname']; 
              $memberid = $row['memberid'];
              echo '<option value="'.$memberid.'">'.$firstname . ' ' . $lastname .'</option>';

}

echo "</select>";

 }
}

 ?> 

JS

function getda(id,booking){
 $.ajax({
 type: 'post',
 url: 'samefile.php',
 data: {
   get_option:id
    },
 success: function (response) {
   document.getElementById("result").innerHTML=response; 
 }
   });
}

相同文件.php

<?php 
if(isset($_POST['get_option']))
   {
inlude 'config/config.php';

$name=$_POST["get_option"];
echo "<SCRIPT>
alert('$name');
</SCRIPT>";

$sql = "insert into bookings (memberid,date,time,groupnumber) values (1,'2016-04-14','09:00',3)";
$query = mysqli_query($sql);



$mysqli->close();


  ?>

chrome 中的控制台看起来不错(下图)，但没有插入任何记录，也没有显示 php 警报。我没有将任何变量传递给 SQL，因为我第一次测试查询是否正确执行

jquery.min.js:4 XHR 完成加载:POST "http://localhost/gym/samefile.php ".send @ jquery.min.js:4n.extend.ajax @ jquery.min.js:4getda @ cal.php?date=140416 :42onchange @ cal.php?date=140416:36ListPicker._handleMouseUp @ about:blank:535

Answer 1

可能需要查看 jQuery 的 .change() .我认为它可以与下面的代码一起使用。你也可以让它调用你的函数，其中也有 ajax

    $( ".class" ).change(function() { //can use #id here too
         $.ajax({
         type: 'post',
         url: 'samefile.php',
         data: {
             get_option:this.value
         },
         success: function (response) {
             document.getElementById("result").innerHTML=response; 
         }
         });
   });