Question

我有一个php发布脚本，我需要它来从数据库中获取数据。这是脚本：

    <?php
error_reporting(E_ALL);
  session_start();

  // If the session vars aren't set, try to set them with a cookie
  if (!isset($_SESSION['user_id'])) {
    }
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
  <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
  <title>Cheesecake Productions - Post Topic</title>
  <link rel="stylesheet" type="text/css" href="include/style/content.css" />
</head>
<body>

<?php

include ("include/header.html");

include ("include/sidebar.html");

?>
<div class="container">
<?php

  require_once('appvars.php');
  require_once('connectvars.php');

  // Make sure the user is logged in before going any further.
  if (!isset($_SESSION['user_id'])) {
    echo '<p class="login">Please <a href="login.php">log in</a> to access this page.</p>';
    exit();
  }
  else {
    echo('<p class="login">You are logged in as ' . $_SESSION['username'] . '. <a href="logout.php">Log out</a>.</p>');
  }

  // Connect to the database
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) or die('could not connect to mysql '.mysqli_connect_error());

// Grab the profile data from the database
$query = "SELECT first_name FROM ccp2_user WHERE first_name = '" . $_SESSION['user_id'] . "'";
    $data = mysqli_query($dbc, $query);

    ///////////////////////////
   ///What must I do after////
  //getting the data from////
 //database. I am new to////
//PHP//////////////////////
//////////////////////////



  $row = mysqli_fetch_array($data);
   $first_name = mysqli_real_escape_string($dbc, trim($_POST['first_name']));



  if (isset($_POST['submit'])) {
    // Grab the profile data from the POST
     $post1 = mysqli_real_escape_string($dbc, trim($_POST['post1']));

    // Update the profile data in the database
    if (!$error) {
      if (!empty($post1)) {
        // Only set the picture column if there is a new picture
    $query = "INSERT INTO `ccp2_posts` (`first_name`, `post_date`, `post`) VALUES ('$first_name', NOW(), '$post1')";
        mysqli_query($dbc, $query);

        // Confirm success with the user
        echo '<p>Your post has been successfully added. Would you like to <a href="viewpost.php">view all of the posts</a>?</p>';

        mysqli_close($dbc);
        exit();
      }
      else {
        echo '<p class="error">You must enter information into all of the fields.</p>';
      }
    }
  } // End of check for form submission
    else {
    echo '<p>Grr</p>';
    }

  mysqli_close($dbc);
?>

  <form enctype="multipart/form-data" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
    <input type="hidden" name="MAX_FILE_SIZE" value="<?php echo MM_MAXFILESIZE; ?>" />
    <fieldset>
      <legend>Post Here:</legend>     
      <label type="hidden" for="post1">Post Content:</label><br />
      <textarea rows="4"  name="post1" id="post1" cols="50">Post Here...</textarea><br />
    </fieldset>
    <input type="submit" value="Save Post" name="submit" />     
  </form>
   </div>
  <?php

include ("include/footer.html");

?>

</body> 
</html>

这个脚本应该从数据库中获取first_name而不是。帮助

编辑：这是完整的代码。

Answer 1

你的代码很多东西都很奇怪

我认为它是空白的，因为其中一个if / else搞砸了：

  if (isset($_POST['submit'])) {
  ....
  } 
  else {//here
    else {
      echo '<p class="error">There was a problem accessing your profile.</p>';
    }
  }

然后你有$ error变量没有意义

$error = false;

然后你的形式是：

  <input type="text" id="first_name" name="first_name" value="" /><br />

但你不想从那里抓住它，但数据库：

$query = "SELECT first_name FROM ccp2_user 
          WHERE user_id = '" . $_SESSION['user_id'] . "'";

然后你想从帖子中获取$ last_name

$ last_name = mysqli_real_escape_string（$ dbc，trim（$ _ POST ['last_name']））;

但你的表格中没有它

这一部分：

if (!empty($first_name) && !empty($post1)) {
    // Only set the picture column if there is a new picture
    if (!empty($new_picture)) {
        $query = "INSERT INTO `ccp2_posts` (`first_name`, `post_date`, `post`) 
                      VALUES ('$first_name', NOW(), '$post1')";
    }
    else {
        $query = "INSERT INTO `ccp2_posts` (`first_name`, `post_date`, `post`) 
                      VALUES ('$first_name', NOW(), '$post1')";
    }
}

你有关于new_picture的条件你在哪里初始化它。为什么再次使用相同的插入查询？

你不需要引用它吗？

你这里有很多问题，我建议你一步一步地麻烦拍摄。并重新设计整个事情。

Answer 2

我把一些快速的东西放在我的系统上。

这是一个基本方法，我的意思是基本的，所以你需要做其余的工作。

只需将数据库凭据更改为您自己的凭据，并将the_user_id分配给$_SESSION['user_id']

这是我能做的最好的帮助。

<?php
$DB_HOST = "xxx";
$DB_USER = "xxx";
$DB_PASS = "xxx";
$DB_NAME = "xxx";

$dbc = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($dbc->connect_errno > 0) {
  die('Connection failed [' . $dbc->connect_error . ']');
}

session_start();
$_SESSION['user_id'] = "the_user_id"; // change this to the user's id

// You can use * also as the line from below
// $sql = $dbc->query("SELECT * FROM `ccp2_user` WHERE `user_id` = '" . $_SESSION['user_id'] . "'");
$sql = $dbc->query("SELECT `first_name` FROM `ccp2_user` WHERE `user_id` = '" . $_SESSION['user_id'] . "'");

while($row= mysqli_fetch_array($sql))
{
echo $row['user_id'];
}

// for testing purposes
// var_dump($_SESSION['user_id']);
// var_dump($_SESSION);

mysqli_close($dbc);

Answer 3

它在这里，

require_once('appvars.php');
require_once('connectvars.php');

其中一个文件不能设置或php无法找到这些文件。所以它说“要求”意味着直到我们没有得到这个文件它不会继续。所以它停止了那里的执行。

尝试：

include('appvars.php');
include('connectvars.php');

你看到页面然后问题就在这里。

Php发布脚本无法从数据库获取数据

3 个答案: