Question

我已经在“阅读更多”页面上工作了很长一段时间。我基本上想要的是当你点击阅读更多时，它将重定向到新闻项页面。

的index.php

$p=mysqli_query($con, "SELECT id, titel, tekst, datum FROM nieuws LIMIT $start, $limit");

      if(mysqli_num_rows($p) > 0) {
        while ($row = mysqli_fetch_array($p)){
          $nieuws = $row['0'];

          echo "<h2>".$row['1']."</h2>";
          echo "<div class='clr'></div>";
          echo "<p>".$row['2']."</p>";
          echo "<p> Geplaatst op: ".$row['3']."</p>";
          echo "<a href='pages/nieuws.php?nieuwsitem=$nieuws'>Lees meer..</a></br></br>";
        }
      } else {
        echo "Geen items gevonden.";
      }

nieuws.php

<?php
      $con = mysqli_connect("localhost", "root", "", "groenevingers");

      if (mysqli_connect_errno()) {
        die("Failed to connect to MySQL: " . mysqli_connect_error());
      }

      $p=mysqli_query($con, "SELECT id, titel, tekst, datum FROM nieuws WHERE id='$nieuws'");

      if(mysqli_num_rows($p) > 0) {
        while ($row = mysqli_fetch_array($p)){
          echo "<h2>".$row['1']."</h2>";
          echo "<div class='clr'></div>";
          echo "<p>".$row['2']."</p>";
          echo "<p> Geplaatst op: ".$row['3']."</p>";
        }
      } else {
        echo "Item niet gevonden.";
      }
    ?>

在'nieuws.php'中，它应该显示带有id的特定新闻项。但是，我收到一条错误消息“通知：未定义的变量：nieuws”。我可以理解为什么它说错误，因为我没有定义$ row ['0']这是我的新闻表中的新闻ID，但我无法以某种方式解决它。

无论如何你们可以帮帮我吗？非常感谢帮助。

亲切的问候， vaxzz

Answer 1

您可能需要更改新闻页面的代码，如下所示：

<?php
    $con = mysqli_connect("localhost", "root", "", "groenevingers");

    if (mysqli_connect_errno()) {
    die("Failed to connect to MySQL: " . mysqli_connect_error());
    }
    if(isset($_GET['nieuwsitem'])) {
        $nieuws = intval($_GET['nieuwsitem']);
        $p=mysqli_query($con, "SELECT id, titel, tekst, datum FROM nieuws WHERE id='$nieuws'");

        if(mysqli_num_rows($p) > 0) {
            if ($row = mysqli_fetch_array($p)){
                echo "<h2>".$row['1']."</h2>";
                echo "<div class='clr'></div>";
                echo "<p>".$row['2']."</p>";
                echo "<p> Geplaatst op: ".$row['3']."</p>";
            }
        } else {
            echo "Item niet gevonden.";
        }
    } else {
        echo "Item niet gevonden.";
    }  
?>

这将做几件事：

if(isset($_GET['nieuwsitem'])) {

检查您的网址中是否有ID

$nieuws = intval($_GET['nieuwsitem']);

检索该ID并将其转换为整数，因此它安全＆＃34;要在您的查询中使用，它还会设置您的$nieuws变量

Answer 2

您必须在使用之前定义变量$ nieuws： nieuws.php： -

 <?php
   $con = mysqli_connect("localhost", "root", "", "groenevingers");

 $nieuws = $_GET['nieuwsitem'];

  if (mysqli_connect_errno()) {
     die("Failed to connect to MySQL: " . mysqli_connect_error());
 }

 $p=mysqli_query($con, "SELECT id, title, text, datum FROM nieuws WHERE id='$nieuws'");

 if(mysqli_num_rows($p) > 0) {
  while ($row = mysqli_fetch_array($p)){
  echo "<h2>".$row['1']."</h2>";
  echo "<div class='clr'></div>";
  echo "<p>".$row['2']."</p>";
  echo "<p> Geplaatst op: ".$row['3']."</p>";
}
} else {
  echo "Item niet gevonden.";
 }
?>

PHP - 阅读更多页面返回'undefined variable'错误

2 个答案: