php - 将表二中的数据添加到表一中

Question

我需要显示一个表格，显示将笑话输入数据库的用户。我有一个笑话名称“joke”的表和一个名为“author”的用户的表 笑话表有 id、joketext、jokedate、authorid 作者表有 id、姓名、电子邮件 所以'authorid'与作者表中的'id'相同。

<?php
$con=mysqli_connect("localhost","user","password","ijdb");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$result = mysql_query("SELECT joke.id, joketext, name, email FROM joke INNER JOIN author
ON authorid = author.id");

echo "<form action='delete1.php' method='post'>
<table border='1'>
<tr>
<th>Joke</th>
<th>Date</th>
<th>Name</th>
<th>Email</th>
<th>Delete</th>
</tr>";

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['joketext'] . "</td>";
  echo "<td>" . $row['jokedate'] . "</td>";
  echo "<td>" . $row['name'] . "</td>";
  echo "<td>" . $row['email'] . "</td>";
  echo "<td><input type='submit' name='deleteItem' value='".$row['id']."' /></td>";
  echo "</tr>";
  }
  echo "</table>";

  echo "</form></br>";

mysqli_close($con);
?>

我收到这个错误 警告:mysqli_fetch_array() 期望参数 1 为 mysqli_result，第 22 行 C:\xampp\htdocs\kimmy\jokes\joke1.php 中给出的 bool 值 和一张空 table

谢谢

Answer 1

应该是:

$result = mysqli_query($con, "SELECT joke.id, joketext, name, email FROM joke INNER JOIN author
                              ON authorid = author.id");

当您使用 mysqli 扩展时，您不能使用 mysql_query()。