嘿所以我的代码存在问题,我试图过滤来自数据库的数据并将其显示在表格中。我正在使用AJAX将请求发送到PHP页面。我没有找到解决方案的运气。 (它类似于您的常见房地产网站或零售等,用户可以在搜索框中输入位置,搜索它,然后使用2个下拉菜单过滤显示的数据)。
我的index.php页面有3个输入(一个文本框和2个下拉列表)
<form action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="text" class="searchForm" id="search" placeholder="Stuff" autocomplete="off">
<div id="here"></div>
<select class="orderType" name="type" id="orderByType" data-toggle="dropdown" onchange="displaySelection(this.value)">
<option value="" selected>--------</option>
<option value="dropdown1" selected>Dropdown1</option>
<option value="dropdown1" selected>Dropdown1</option>
</select>
<select class="order" name="order" id="orderBy" data-toggle="dropdown">
<option value="" selected>--------</option>
<option value="lowest">Lowest</option>
<option value="highest">Highest</option>
</select>
</form>
<div id="searchTable">
然后我的ajax调用了index.php页面(之后的AJAX将是另一个问题,因为我确信有比我更好的方式来发送数据)
function fill(Value)
{
$('#search').val(Value);
$('#here').hide();
}
$(document).ready(function(){
$("#search").keyup(function(){
var x = $('#search').val();
if(x==""){
$("#here").html("");
$('#searchTable').html("");
}
else{
$.ajax({
type:'POST',
url:'test.php',
data:'q='+x,
success:function(html){
$("#here").html(html).show();
}
});
}
});
$('.searchForm').change(function(){
var type = $('#search').val();
var city = $('#city').text();
$.ajax({
type: 'POST',
url: 'test.php',
data: { search : type, city : city },
success: function(response){
$("#searchTable").html(response);
$('#search').live("keypress",function(e){
var code = (e.keyCode ? e.keyCode : e.which);
if(code == 13){
e.preventDefault();
e.stopPropagation();
$('#searchTable').show();
}
});
}
});
});
$('.orderClass').change(function(){
var order = $('#orderBy').val();
var city = $('#city').text();
$.ajax({
type: 'POST',
url: 'test.php',
data: { orderBy : order, city : city },
success: function(response){
$("#searchTable").html(response);
}
});
});
$('.orderType').change(function(){
var type = $('#orderByType').val();
var city = $('#city').text();
$.ajax({
type: 'POST',
url: 'test.php',
data: { orderByType : type, city : city},
success: function(response){
$("#searchTable").html(response);
}
});
});
});
然后在test.php上 (我可以使用2个下拉菜单过滤数据,这样可以正常工作,但我不知道如何过滤搜索输入框中显示的数据。)
$stmt = "SELECT * FROM places";
if(isset($_POST['search'])){
$search = htmlspecialchars($_POST['search']);
$stmt .= " WHERE name = :search";
}
if(isset($_POST['orderByType'])){
$selection = $_POST['orderByType'];
$stmt .= " AND type = :selection";
}
if(isset($_POST['orderBy'])){
$order = $_POST['orderBy'];
$selection = $_SESSION['id'];
$stmt .= " ORDER BY".$order;
}
$stmt = $conn->prepare($stmt);
$search = "%".$search."%";
$stmt->bindValue(':search', $search, PDO::PARAM_STR);
$stmt->bindParam(":selection", $selection);
if($stmt->rowCount() > 0){
$result = $stmt->fetchAll();
foreach($result as $row){
echo $row['data'];
}
}
//Search input live search
if(!empty($_POST['q'])){
$name = $_POST['q'];
$name = htmlspecialchars($name);
$liveSearch = $conn->prepare("SELECT name, city FROM places WHERE name LIKE :name OR city LIKE :name");
$name = "%".$name."%";
$liveSearch->bindValue(':name', $name, PDO::PARAM_STR);
$result = $liveSearch->fetchAll();
if($liveSearch->rowCount() > 0){
foreach($result as $row){
echo $row['name'];
}
}
else{
echo "No results found";
}
}
(如果有一个很棒的系统可以使用用户输入进行搜索,然后使用下拉菜单进行过滤,那么请告诉我)
提前致谢。