Question

我有一个页面可以从数据库中检索两个值（Atender和Posto）。

我想刷新textbox1 Atender值和textbox2 Posto值。目前，我只能这样做，所以这个功能同时刷新1个值或2个值，但两个值都出现在两个文本框中。

编辑：忘记添加，你看到的PHP就是PainelUtilizador.php。 HTML也是如此，所有代码都在PainelUtilizador.php

中

这是我的代码，我确定您在看完之后能够知道我的意思：

＆＃13;

<?php
include '../Login/db_login.php';

if(isset($_GET['ajax'])) {
$sql1 = "SELECT Atender,Posto FROM atendercliente WHERE ID=1";
$sql2 = "SELECT Atender,Posto FROM atendercliente WHERE ID=2";
$resultA = $conn->query($sql1);
   while($row = $resultA->fetch_assoc()) {
		$returnArrayA = array('AtenderA' => $row["Atender"], 'PostoA' => $row["Posto"]);
   };
$resultB = $conn->query($sql2);
   while($row = $resultB->fetch_assoc()) {
		$returnArrayB = array('AtenderB' => $row["Atender"], 'PostoB' => $row["Posto"]);
   };
   echo json_encode($returnArrayA);
   echo json_encode($returnArrayB);exit;
}

?>

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0" />
<title>Painel do utilizador</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
        <script>
            setInterval(function(){
                $.ajax('PainelUtilizador.php?ajax=1').done(function(data) {
                    data = JSON.parse(data);
                    $("#refreshASenha").val(data['Atender']);
                    $("#refreshAPosto").val(data['Posto']);
				          	$("#refreshBSenha").val(data['Atender']);
                    $("#refreshBPosto").val(data['Posto']);
                })
            }, 1000);
        </script>
</head>
<body>
<table border="1">
<tr>
  <td> <input type="text" id="refreshASenha" disabled> </td>
  <td> <input type="text" id="refreshAPosto" disabled> </td>
</tr>
</table>
<table border="1">
<tr>
  <td> <input type="text" id="refreshBSenha" disabled> </td>
  <td> <input type="text" id="refreshBPosto" disabled> </td>
</tr>
</table>
</body>
</html>

＆＃13;

我还是初学者，我自己还在学习AJAX和JQuery

Answer 1

您需要为ajax＆amp;添加支票。然后只调用DB方法。你可以简单地在一个ajax中完成，而不是调用两个

<强>更新

<?php
include '../Login/db_login.php';

if(isset($_GET['ajax'])) {
$sql1 = "SELECT Atender,Posto FROM atendercliente WHERE ID=1";
$sql2 = "SELECT Atender,Posto FROM atendercliente WHERE ID=2";
$resultA = $conn->query($sql1);
   while($row = $resultA->fetch_assoc()) {
     $returnArrayA = array('Atender' => $row["Atender"], 'Posto' => $row["Posto"]);
   };
$resultB = $conn->query($sql2);
   while($row = $resultB->fetch_assoc()) {
     $returnArrayB = array('Atender' => $row["Atender"], 'Posto' => $row["Posto"]);
   };
   echo json_encode(array('A' => $returnArrayA, 'B' => $returnArrayB));exit;
}

?>

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0" />
<title>Painel do utilizador</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
        <script>
            setInterval(function(){
                $.ajax('PainelUtilizador.php?ajax=1').done(function(data) {
                    data = JSON.parse(data);
                    $("#refreshASenha").val(data['A']['Atender']);
                    $("#refreshAPosto").val(data['A']['Posto']);
		    $("#refreshBSenha").val(data['B']['Atender']);
                    $("#refreshBPosto").val(data['B']['Posto']);
                })
            }, 1000);
        </script>
</head>
<body>
<table border="1">
<tr>
  <td> <input type="text" id="refreshASenha" disabled> </td>
  <td> <input type="text" id="refreshAPosto" disabled> </td>
</tr>
</table>
<table border="1">
<tr>
  <td> <input type="text" id="refreshBSenha" disabled> </td>
  <td> <input type="text" id="refreshBPosto" disabled> </td>
</tr>
</table>
</body>
</html>

使用AJAX刷新两个不同的文本框

1 个答案: