Question

我需要显示一个表格，显示将笑话输入数据库的用户。我有一个关于笑话名称'笑话'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，名称，电子邮件因此'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);
?>

我收到此错误警告：mysqli_fetch_array（）期望参数1为mysqli_result，第22行的C：\ xampp \ htdocs \ kimmy \ jokes \ joke1.php中给出布尔值和一张空表

由于

Answer 1

您正在使用错误的mysql_query 代替 mysqli_query

所以代码变成了

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

$result = mysqli_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);
?>

Answer 2

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mysql_query()扩展名时，您无法使用mysqli。

Answer 3

尝试更改以下内容

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

ON authorid = author.id“）;

到

$result = mysqli_query("SELECT joke.id, joketext, name, email FROM joke INNER JOIN author

ON authorid = author.id“）;

您正在使用mysql_query而不是mysqli_query

相对于表1，从表2中添加数据

3 个答案: