Question

我的test.php页面如下：

<!DOCTYPE html>
<html lang="en">
<head>
</head>
<body>
    <h3>Setting</h3>
    <form>
        <p>Name:<input type="text" id="updatetherone1"/></p>
        <p>Email:<input type="text" id="updateotherone2"/></p>
        <p><input id="updateone" type="button" value="Save"/></p>
    </form>
    <span id="updateotherotherone"></span>
    <script src="js/jquery.js"></script>
    <script src="js/ajax.js"></script> 
</body>
</html>

我的ini.php页面如下：

<?php
  session_start();
  $_SESSION['state'] ='2';
  $conn=new mysqli('localhost','root','','people');
?>

我的test1.php页面如下：

<?php
  include 'ini.php';

  if (isset($_POST['name'], $POST['email'])){
    $name = mysqli_real_escape_string(htmlentities($POST['name']));
    $email = mysqli_real_escape_string(htmlentities($POST['email']));

    $update = mysqli_query("UPDATE state SET Name='$name', email='$email' WHERE Id=".$_SESSION['state']);

    if($update === true){
        echo 'Setting have been updated.';
    }else if ($update === false){
        echo 'There was an error updating your setting.';
    }
  }
?>

我的ajax.js页面如下：

$('[id=updateone]').click(function(){
  var name=$('[id=updateotherone1]').val();
  var email=$('[id=updateotherone2]').val();
  $('[id=updateotherotherone]').text('Loading...');

  $.post('test1.php',{name:name,email:email},function(data){
    $('[id=updateotherotherone]').text(data);
  });
});

最终代码不起作用也不显示任何错误，我怀疑test1.php页面有问题，请任何人指导：

Answer 1

请注意mysqli_query()的程序界面，第一个参数需要连接。

$update = mysqli_query($conn, "UPDATE state SET Name='$name', email='$email' WHERE Id=".$_SESSION['state']);

如果这些是拼写错误$POST，那么应该在代码中修复它。它应该是$_POST。（除非是问题的拼写错误。）这是一个superglobal。

我建议你只使用面向对象的界面。因此，您不需要每次都添加它：

<?php
include 'ini.php';

if (isset($_POST['name'], $_POST['email'])){
    $name = $conn->real_escape_string(htmlentities($_POST['name']));
    $email = $conn->real_escape_string(htmlentities($_POST['email']));

    $update = $conn->query("UPDATE state SET Name='$name', email='$email' WHERE Id = " . $_SESSION['state']);

    if($conn->affected_rows > 0) {
        echo 'Setting have been updated.';
    } else {
        echo 'There was an error updating your setting.';
    }
}
?>

由于mysqli支持它，所以也可以使用预备语句：

if (isset($_POST['name'], $_POST['email'])){
    $name = htmlentities($_POST['name']);
    $email = htmlentities($_POST['email']);

    $sql = 'UPDATE state SET Name = ?, email = ? WHERE Id = ?';
    $update = $conn->prepare($sql);
    $update->bind_param('ssi', $name, $email, $_SESSION['state']);
    $update->execute();

    if($update->affected_rows > 0) {
        echo 'Setting have been updated.';
    } else {
        echo 'There was an error updating your setting.';
    }
}

在JS部分：

您在表单id和JS：

上也有拼写错误

<p>Name:<input type="text" id="updatetherone1"/></p> <!-- missing o -->
              var name=$('[id=updateotherone1]').val();

应该是：<p>Name:<input type="text" id="updatetherone1"/></p>

旁注：

如果你想避免那种愚蠢的拼写错误，只需正确识别并标记它们即可。例如：

<p>Name:<input type="text" id="name_field"/></p>
<p>Email:<input type="text" id="email_field"/></p>

数据库表值不通过AJAX＆amp;更新jQuery的

1 个答案: