Question

注意：我希望这个Show data from database on load完全正确

我正在使用ajax进行实时搜索一切正常，但是仅需要解决一个问题，因此问题是ajax无法在页面加载中工作，我的意思是我正在从这样的URL获取数据。即http://localhost/basic_hackings/search.php?q=usman，这里的“üsman”是要搜索的数据。因此，一切工作正常，但刷新后消失了，我也在jQuery中使用onload，但徒劳无功。详细了解代码。

ajax_search.php是：

<?php 
$host = "localhost";
$dbUser = "root";
$password = "";
$database = "basic_hackings";

$dbConn = new mysqli($host,$dbUser,$password,$database);

if($dbConn->connect_error)
{
    die("Database Connection Error, Error No.: ".$dbConn->connect_errno." | ".$dbConn->connect_error);
}

$search_data = $_GET['q'];
$qry = "select first_name, last_name, email, phone, city, country from customers where first_namelike '%$search_data%'";

$rs = $dbConn->query($qry);

$fetchAllData = $rs->fetch_all(MYSQLI_ASSOC);

$createTable = '<table>';

$createTable .= '<tr>';
$createTable .= '<th>First Name</th>';
$createTable .= '<th>Last Name</th>';
$createTable .= '<th>Email</th>';
$createTable .= '<th>Phone</th>';
$createTable .= '<th>City</th>';
$createTable .= '<th>Country</th>';
$createTable .= '</tr>';


foreach($fetchAllData as $customerData)
{
    $createTable .= '<tr>';
    $createTable .= '<td>'.$customerData['first_name'].'</td>';
    $createTable .= '<td>'.$customerData['last_name'].'</td>';
    $createTable .= '<td>'.$customerData['email'].'</td>';
    $createTable .= '<td>'.$customerData['phone'].'</td>';
    $createTable .= '<td>'.$customerData['city'].'</td>';
    $createTable .= '<td>'.$customerData['country'].'</td>';
    $createTable .= '</tr>';    
}

$createTable .= '</table>';

echo $createTable;

$rs->close();

$dbConn->close();

?>

和ajax_search.js是：

$(document).ready(function() {
    $('#form1').submit(function(e) {
        e.preventDefault();
        var search_data = $('#search_data').val();
            $.ajax({
                url: "ajax_search.php/?q=" + search_data,
                type: "GET",
                data: new FormData($("#form1")[0]),
                contentType: false,
                processData: false,
                beforeSend: function(data) {
                    $(".main_search_content").html("");
                    $(".main_search_content").addClass("searching");
                },
                success: function(data) {
                    setTimeout(function() {
                        if (!(data == "")) {
                            $(".main_search_content").html(data);
                        }
                        else {
                            $(".main_search_content").html("Nothing Found!");
                        }
                        $(".main_search_content").removeClass("searching");
                    }, 500)
                }
            });
            history.pushState(null, null, "?q=" + search_data);
        }
    });
});

这里是site.js，我正在使用onload，请参见：

$(window).on("load", function() {
    $.ajax({
        url: 'ajax_search.php',
        success: function(data){
            $(".main_search_content").html(data);
        }
    });
});

加载时未从数据库加载数据（php + ajax）

0 个答案: