使用 jquery ajax 检索和显示多个不同的 mysql 查询 - Retrieve and display multiple and different mysql queries using jquery ajax

Retrieve and display multiple and different mysql queries using jquery ajax

我有几个div标签，我想用几个不同的mysql查询来更新。

我

下面的代码只更新了其中一个div，我怎样才能将其扩展到 1-6，并为将来的扩展空间留出更多div 的空间？

我还需要能够使用特定表的特定查询单独更新每个div。

<html>
<div id="1"> </div>
<div id="2"> </div>
<div id="3"> </div>
<div id="4"> </div>
<div id="5"> </div>
<div id="6"> </div>
//...and so on
</html>

<script type="text/javascript">
$(document).ready(function () {
function display($id) {
    $.ajax({
        async: false,
        type: "POST",
        url: "apifile.php",
        data: {id:$id},
        dataType: "json",
        success: function(msg) {
            if(msg.success) {
                $($id).next(".one").html(msg);
            } else {
                alert("error");
            }
        }
    });
}
</script>

<?php
mysql_connect(host, user, pass);
mysql_select_db(name);
mysql_query("select round((count(*)*100)/(select count(*) from test),1) as percent from test group by field1 order by percent desc");
$reply['success'] = "Success"; 
if($return = display($id)) {
    $reply['success'] = "Success";
} else {
    $reply['error'] = "Error";
}
echo json_encode($reply);
?>

您可以将每个查询的数据/消息/其他信息存储在$reply数组的不同部分中

query1 goes here ...
   $reply[1]['success'] = "Success"; 
if($return = display($id)) {
    $reply[1]['success'] = "Success";
} else {
    $reply[1]['error'] = "Error";
}
query2 goes here ...
$reply[2]['success'] = "Success"; 
$reply[2]['data'] = "3,4,5"; 
query3 goes here ...
$reply[3]['success'] = "Success"; 
$reply[3]['data'] = "beans,soup,guac"; 
echo json_encode($reply);

你可以向div 添加一个类并在类上运行 ajax

$( ".div" ).each(function() {
    var div_id = $(this).attr('id');
    display(div_id);
});