Question

我在使用表单中的ajax调用连接到我的函数和数据库时遇到了一些问题。

<?php require_once("../includes/session.php"); ?>
<?php require_once("../includes/db_connection.php"); ?>
<?php require_once("../includes/functions.php"); ?>
<?php require_once("../includes/validation_functions.php"); ?>
<?php include("../includes/layouts/header.php"); ?>


<form action="Assets/ajax/userdetails.php" id="ajax-user" method="post">
    <div class="well" style="border-color: #b81e1e; display: none;" id="details1">
        <div>
            <label for="FirstName">First Name:<br /></label>
            <input name="FirstName" type="text" />
        </div>
        <div>
            <label for="LastName">Last Name:<br /></label>
            <input name="LastName" type="text"  />
        </div>
        <div>
            <select name="gender">
                <option value="male">Male</option>
                <option value="female">Female</option>
            </select>
        </div>

        <input class="btn btn-hero right" type="submit">

    </div>
</form>

<?php include("../includes/layouts/footer.php"); ?>

Javascript看起来像这样;

$(document).on("submit", "form#ajax-user", function() {

    var that = $(this),
        url = that.attr("action"),
        type = that.attr("method"),
        data = {};


    that.find('[name]').each(function(index, value) {
        var that = $(this),
            name = that.attr('name'),
            value = that.val();

        data[name] = value;

    });

    $.ajax({
        url: url,
        type: type,
        data: data,
        success: function(response) {
            console.log(response);
        }

    });

    return false;
});

我已经在提交上运行了ajax，这很好。然后是运行php，但我一直收到错误;

警告：mysqli_query（）要求参数1为mysqli，在 /home/public/Assets/ajax/userdetails.php 行中给出null

最后，php调用看起来如下，我认为它与范围有关，但我仍然在学习并且在这几天内一直在努力，如果有人能给我一些指示它会很大赞赏。

由于

<?php 

    if (isset($_POST['FirstName'], $_POST['LastName'], $_POST['gender'])) {

    $current_user = $_SESSION["admin_id"];
    $firstname = $_POST['FirstName'];
    $lastname = $_POST['LastName'];
    $gender = $_POST['gender'];

    $query = "UPDATE users SET FirstName = '$firstname', LastName = '$lastname', Gender = '$gender' WHERE id = $current_user";
    $result = mysqli_query($connection, $query) or die(mysqli_error($connection));
    }
?>

修改

$ connection在db_connection;

中定义

<?php
    define("DB_SERVER", "localhost");
    define("DB_USER", "****");
    define("DB_PASS", "****");
    define("DB_NAME", "****");
  // 1. Create a database connection
  $connection = mysqli_connect(DB_SERVER, DB_USER, DB_PASS, DB_NAME);
  // Test if connection succeeded
  if(mysqli_connect_errno()) {
    die("Database connection failed: " . 
         mysqli_connect_error() . 
         " (" . mysqli_connect_errno() . ")"
    );
  }
?>

Answer 1

根据mysqli，你有enlid /没有连接语法。

使用它像：

$connection=mysqli_connect("localhost","Uname","PWD","DBName");
if (mysqli_connect_errno()) {
    throw new Exception(mysqli_connect_error(), mysqli_connect_errno());
}

<强>更新

根据您编辑的问题：

使用<?php require_once("../includes/db_connection.php"); ?> 在你的userdetails.php里面。因为你需要$ connection参数。

Answer 2

正如其他人所指出的，您需要先设置连接。此外，您的查询字符串在$ current_user变量周围缺少单引号：

$query = "UPDATE users SET FirstName = '$firstname', LastName = '$lastname', Gender = '$gender' WHERE id = '$current_user'";

Answer 3

您忘了定义$ connection

<?php 

if (isset($_POST['FirstName'], $_POST['LastName'], $_POST['gender'])) {

$current_user = $_SESSION["admin_id"];
$firstname = $_POST['FirstName'];
$lastname = $_POST['LastName'];
$gender = $_POST['gender'];
$connection = mysqli_connect("myhost","myuser","mypassw","mybd") or die("Error " . mysqli_error($connection));
$query = "UPDATE users SET FirstName = '$firstname', LastName = '$lastname', Gender = '$gender' WHERE id = $current_user";
$result = mysqli_query($connection, $query) or die(mysqli_error($connection));
}

＆GT;

Ajax PHP Scope问题

3 个答案: