我在这里遇到了一个问题:我试图创建一个jQuery/AJAX/PHP实时搜索栏。我调用search.php很好，但是每当我在控制台上输出响应时，我都会得到master.php文件的内容(这只是站点范围的布局)以及json编码的结果。我想不出是什么原因导致这种情况的发生。
$(function() {
$("#search-text").keyup(function() {
var $res = $(".search-results");
$.ajax({
type: "POST",
url: "search.php",
data: { query: $(this).val() },
cache: false,
success: function(html) {
$res.show();
$res.append(html);
console.log(html);
},
error: function(xhr, status, error) {
console.log("XHR: " + xhr);
console.log("Status: " + status);
console.log("Error: " + error);
}
});
return false;
});
});
和search.php:
$key = $_POST["query"];
$db = new Database();
$db->query("SELECT * FROM users WHERE firstname LIKE :key OR lastname LIKE :key OR firstname AND lastname LIKE :key");
$db->bind(":key", '%' . $key . '%');
$rows = $db->resultset();
echo json_encode($rows);
谢谢!
在search.php的回显语句后写入exit()。
这样的:
$key = $_POST["query"];
$db = new Database();
$db->query("SELECT * FROM users WHERE firstname LIKE :key OR lastname LIKE :key OR firstname AND lastname LIKE :key");
$db->bind(":key", '%' . $key . '%');
$rows = $db->resultset();
echo json_encode($rows);
exit();
它应该防止显示整个页面。
In search.php before sending $rows in json_encode() give one condition that will check $row is empty or not.like :
if(!empty($res))
echo json_encode(array('status'=>'success','result'=>$rows));
else
echo json_encode(array('status'=>'failed','result'=>[]));
In ajax check
success: function(html)
{
if(html.status=='success')
{
$res.show();
$res.append(result.html);
console.log(result.html);
}
}
may be it will work..