php - PHP 将 MySQL 数据库提取到表中时出现问题

Question

我正在尝试生成代码以使用嵌入 html 的 php 将信息从数据库提取到表中。

<div class="entry">
 <?php
  $host="localhost";
  $user="";
  $password="";
  $dbname = "";
  mysql_connect($host,$user,$password);
  mysql_select_db($dbname) or die( "Unable to select database");
  $query = "SELECT * FROM QA";
  $result = mysql_query($query);
  $info = mysql_fetch_array($result);
  print '<table="1">';
  print   '<tr>';
  print        '<th>...</th>';
  print        '<th>Questions and Answers</th>';
  print    '</tr>';
  while ($row = mysql_fetch_array($result))
  {
    print '<tr>';
    print '<td>Question</td>'
    print '<td>'.$row['question'].'</a></td>'
    print '</tr>';
    print '<tr>';
    print '<td>Answer:</td>';
    print '<td>'.$row['answer'].'</a></td>'
    print '</tr>';
  }
  print '</table>';  
  mysql_close();
  ?>
  </div>

输出如下所示。

'; print ''; print '...'; print 'Questions and Answers'; print ''; while ($row = mysql_fetch_array($result)) { print ''; print 'Question' print ''.$row['question'].'' print ''; print ''; print 'Answer:'; print ''.$row['answer'].'' print ''; } print ''; mysql_close(); ?>

我做错了什么？

Answer 1

<div class="entry">
 <?php
  $host="localhost";
  $user="";
  $password="";
  $dbname = "";
  mysql_connect($host,$user,$password);
  mysql_select_db($dbname) or die( "Unable to select database");
  $query = "SELECT * FROM QA";
  $result = mysql_query($query);
  $info = mysql_fetch_array($result);
  echo '<table="1">';
  echo   '<tr>';
  echo        '<th>...</th>';
  echo        '<th>Questions and Answers</th>';
  echo    '</tr>';
  while ($row = mysql_fetch_array($result))
  {
    echo '<tr>';
    echo '<td>Question</td>'
    echo '<td>'.$row['question'].'</a></td>'
    echo '</tr>';
    echo '<tr>';
    echo '<td>Answer:</td>';
    echo '<td>'.$row['answer'].'</a></td>'
    echo '</tr>';
  }
  echo '</table>';  
  mysql_close();
  ?>
  </div>

试试这个，我相信这是由于打印