我有3个选择下拉菜单,当我选择第一个,第二个显示,第二个选择时,第三个显示,使用if(isset($_post[first_one]))
,第三个使用{ {1}}
SQL:
if(isset($_post[second_one]))
PHP / HTML:
$conn = new mysqli($servername, $username, $password, $dbname);
$sql = "SELECT * FROM sp_meeting_log ";
$result1 = $conn->query($sql);
答案 0 :(得分:1)
基本上这就是这个想法。您希望第一页只从第二页获取并将结果发送回第一页到正确的位置:
<强> page1.php中强>
<script type="text/javascript" src="http://code.jquery.com/jquery-1.9.1.js"></script>
<form>
<label>Org
<select id="org" name="org">
<option value="1">One</option>
<option value="2">Two</option>
</select>
</label>
<!-- This is where the forum html will drop into after ajax runs -->
<div id="forum"></div>
<!-- This is where the user html will drop into after ajax runs -->
<div id="user"></div>
</form>
<script type="text/javascript">
$(document).ready(function() {
// On change of a select menu
$(this).on('change','select',function(e){
// Assign selection
var thisSelect = $(this);
// Get the id name, this will tell page two
// what it's receiving
var sendType = thisSelect.attr('id');
// Get the actual value of the selection
var sendVal = thisSelect.val();
// Create essentially a POST
var sendData = { field: sendType, value: sendVal };
$.ajax({
// Send to page 2
url : '/page2.php',
// Use post method
type: "POST",
// Use post data from above
data : sendData,
// This is what will run on success
success:function(response){
// Parse the json coming back for placement
var jSon = JSON.parse(response);
// Save the correct html into the correct drop spot
$('#'+jSon.type).html(jSon.html);
},
error: function(response){
console.log(response);
}
});
});
});
</script>
<强>使page2.php 强>
if(!empty($_POST)) {
$data = '';
ob_start();
if(isset($_POST['field'])) {
if($_POST['field'] == 'org') {
$type = 'forum';
?>
<label>Forum
<select id="forum" name="forum">
<option value="1">One</option>
<option value="2">Two</option>
</select>
</label>
<?php
}
elseif($_POST['field'] == 'forum') {
$type = 'user';
?>
<label>user
<select id="user" name="user">
<option value="1">One</option>
<option value="2">Two</option>
</select>
</label>
<?php }
$data = ob_get_contents();
ob_end_clean();
die(json_encode(array('type'=>$type,'html'=>$data)));
}
die(json_encode(array('type'=>'error','html'=>false)));
}