选择 OnChange 更改 PHP 变量
Selection OnChange change PHP Variable
你好 Whosebug 人。
我需要你的帮助。我有这个代码:
<select name="type" onChange="idk?" class="form-control">
<option value="H">Week 51</option>
<option value="V">Week 52</option>
</select>
如果用户更改周,我想更改文件获取内容 :s
$Week = "";
echo file_get_contents('http://rooster.farelcollege.nl/'.$Week.'/c/c00050.htm');
谢谢!
来自荷兰的问候=)
您可以向 "file_get_contents" url 发送 Ajax 请求。
这就是我正在使用的并且工作正常:)
<select name="type" class="form-control">
<option value="H">Week 51</option>
<option value="V">Week 52</option>
</select>
$(document).ready(function() {
$( ".form-control" ).change(function() {
$.ajax({
data: {
// You are not sending any post variables right? Then leave this empty :-) otherwise use it as array 'var1': 'value', 'var2': 'value2' ...
},
url: 'http://rooster.farelcollege.nl/'+$(this).val()+'/c/c00050.htm',
method: 'POST',
success: function(msg) {
alert(msg); // Optional, show text that was given by URL, can be removed
}
});
});
});
你好 Whosebug 人。
我需要你的帮助。我有这个代码:
<select name="type" onChange="idk?" class="form-control">
<option value="H">Week 51</option>
<option value="V">Week 52</option>
</select>
如果用户更改周,我想更改文件获取内容 :s
$Week = "";
echo file_get_contents('http://rooster.farelcollege.nl/'.$Week.'/c/c00050.htm');
谢谢!
来自荷兰的问候=)
您可以向 "file_get_contents" url 发送 Ajax 请求。 这就是我正在使用的并且工作正常:)
<select name="type" class="form-control">
<option value="H">Week 51</option>
<option value="V">Week 52</option>
</select>
$(document).ready(function() {
$( ".form-control" ).change(function() {
$.ajax({
data: {
// You are not sending any post variables right? Then leave this empty :-) otherwise use it as array 'var1': 'value', 'var2': 'value2' ...
},
url: 'http://rooster.farelcollege.nl/'+$(this).val()+'/c/c00050.htm',
method: 'POST',
success: function(msg) {
alert(msg); // Optional, show text that was given by URL, can be removed
}
});
});
});