我有两个从数据库上的SQL查询生成的下拉列表。它们如下:
<?php
$conn = new mysqli('localhost', 'root', '', 'Rosters')
or die ('Cannot connect to db');
$result = $conn->query("SELECT City, Name FROM Teams");
echo "<select name='Teams'>";
while ($row = $result->fetch_assoc()) {
unset($city, $team);
$city = $row['City'];
$name = $row['Name'];
$fullname = $city." ".$name;
echo '<option value="'.$fullname.'">'.$fullname.'</option>';
}
echo "</select>";
?>
和
<?php
$conn = new mysqli('localhost', 'root', '', 'Rosters')
or die ('Cannot connect to db');
$team = "Chicago Blackhawks";
$result = $conn->query("SELECT Number, First, Last FROM `$team`");
echo "<select name='Players'>";
while ($row = $result->fetch_assoc()) {
unset($number, $first, $last);
$number = $row['Number'];
$first = $row['First'];
$last = $row['Last'];
$fullname = $first." ".$last;
echo '<option value="'.$fullname.'">'.$number." - ".$fullname.'</option>';
}
echo "</select>";
?>
第一个是NHL的球队名单。第二个是来自该团队的球员名单。我试图使第二个更新第一个时更新(基于“选项”的“值”)。为此,需要更新第二段代码中的$ team变量。由于PHP是服务器端的,无法动态更新,我该怎么做?即使使用AJAX,答案似乎也不明显。我是否完全使用了一种有缺陷的方法?
答案 0 :(得分:0)
你必须在更改第一个下拉列表时使用ajax并从ajax调用php文件并将数据从php文件返回到ajax并在第二个下拉列表中显示它。
答案 1 :(得分:0)
首先,写一个onchange事件处理程序,它使用ajax将“team”选项发送到服务器端,然后编写一个php来接收客户端的“团队”选项,从客户端检索来自DB的播放器信息,之后以XML或json格式重新格式化数据,发送到客户端。
最后,编写一个javascript函数来解析服务器端响应并更新网页。
这是解决问题的合理流程。您可以去谷歌搜索上面的关键字以获取示例代码。
这是一个简单的示例代码: HTML文件内容:
import numpy as np
import matplotlib.pyplot as plt
data = np.random.random((10,10))
# To make a standalone example, I'm skipping initializing the
# `Figure` and `FigureCanvas` and using `plt.figure()` instead...
# `plt.draw()` would work for this figure, but the rest is identical.
fig, ax = plt.subplots()
ax.set(title='Click to update the data')
im = ax.imshow(data)
def update(event):
im.set_data(np.random.random((10,10)))
fig.canvas.draw()
fig.canvas.mpl_connect('button_press_event', update)
plt.show()
php文件(文件名:getResult.php)
<html>
<head>
<title>PHP/Ajax update 2nd drop down box base on 1st drop down box value</title>
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script language=javascript>
function updateData(v)
{
var value=v.options[v.selectedIndex].value;
$("#number").empty(); //empty "digit" drop down box
if (value!="") //Ensure no empty value is submitted to server side
{
jQuery.post("getResult.php","type="+value,updateNumber);
}
}
function updateNumber(data)
{
var numberData=jQuery.trim(data).split("\n");//split server side response by "\n"
var number=document.getElementById("number");
for (i=0;i<numberData.length;i++)
{
value=numberData[i].split("-")[0];//get value from server response
text=numberData[i].split("-")[1]; //get text from server response
option=new Option(text,value); //for better IE compatibility
number.options[i]=option;
}
}
</script>
</head>
<body>
<h1>PHP/Ajax update 2nd drop down box base on 1st drop down box value Demo</h1>
No number type
<select id=type name="type" onchange="updateData(this)">
<option value="">Please select</option>
<option value="1">Odd No.</option>
<option value="0">Even No.</option>
</select>
Number
<select id="number" name="number">
</select>
</body>
</html>