Question

我看过其他答案，但他们没有帮助。

$postid = $_GET['p'];

$stmt = $conn->prepare('SELECT comment_id,comment_submitter_id,comment_text,comment_lava,comment_icy,comment_date,comment_parent,post_id FROM comments WHERE post_id = ? AND comment_parent = 0');
$stmt->bind_param('s', $postid);

当我在phpmyadmin中运行相同的SQL时，它会正确地获取行（？替换为1）。我的查询字符串是？p = 1。这怎么可能失败？

  $postid = $_GET['p'];

  $stmt = $conn->prepare('SELECT * FROM posts WHERE post_id = ?');
  $stmt->bind_param('s', $postid);

  $stmt->execute();

  $stmt->bind_result($post_id, $post_submitter_id, $post_title, $post_text, $post_lava, $post_icy, $post_url, $post_nsfw, $post_views, $post_date, $post_heat, $post_sub);

  $stmt->fetch();

一些HTML，然后是更多PHP

      $query = $conn->prepare('SELECT comment_id,comment_submitter_id,comment_text,comment_lava,comment_icy,comment_date,comment_parent,post_id FROM comments WHERE post_id = ? AND comment_parent = 0');
      $query->bind_param('s', $postid);
      $query->execute();
      $query->bind_result($comment_id, $comment_submitter_id, $comment_text, $comment_lava, $comment_icy, $comment_date, $comment_parent, $post_id);

      while($query->fetch()){

        $posts[] = [
             'comment_id' => $comment_id,
             'comment_submitter_id' => $post_submitter_id,
             'comment_text' => $comment_text,
             'comment_lava' => $comment_lava,
             'comment_icy' => $comment_icy,
             'comment_date' => $comment_date,
             'comment_parent' => $comment_parent,
             'comment_post_id' => $post_id
        ];

      }

Answer 1

基于标题，我假设是你得到的错误，你准备好的语句未能创建。问题出在你甚至调用bind_param之前，它正在“准备”（或更早）。

这可能有很多原因，所以我们真的需要看到更多代码。

在布尔PHP上调用成员函数bind_param（）

1 个答案: