Question

我是MySQLi和PHP的新手，我试图通过匹配postid从我的数据库中选择评论。它应该选择具有该id的所有行，并显示每个行的名称和值。但是没有工作。

      // Create connection
  $conn = mysqli_connect($hostname, $username, $password, $dbname);
  // Check connection
  if (!$conn) {
      die("Connection failed: " . mysqli_connect_error());
      echo "<script>alert('dead!')</script>";
  }

    $get_comments = "SELECT * FROM 'articlecomments' WHERE 'commentpost' = 'post1' LIMIT 0 , 30";
  $check_comments = mysqli_query($conn, $get_comments);

    while ( $row = mysqli_fetch_assoc($check_comments) )  {
        echo "<script>alert('we have comments!');</script>

      <div class='comment-single'>
        <div class='row'>
          <div class='col-md-3'>
            " . $check_comments['commentname'] . "
          </div>
          <div class='col-md-9'>
            " . $check_comments['commentvalue'] . "
         </div>
       </div>
      </div>
    ";

}

连接建立正常，因为我可以在同一个脚本中访问数据库的其他部分。我怀疑我的查询存在问题。

由于

Answer 1

使用tick来转义列而不是单引号：

SELECT *
FROM `articlecomments`
WHERE `commentpost` = 'post1'
LIMIT 30

Answer 2

您必须使用$row['commentname']和$row['commentvalue']从数据库中获取数据。

echo "<script>alert('we have comments!');</script>

  <div class='comment-single'>
    <div class='row'>
      <div class='col-md-3'>
        " . $row['commentname'] . "
      </div>
      <div class='col-md-9'>
        " . $row['commentvalue'] . "
     </div>
   </div>
  </div>
";

Answer 3

关于表格名称的单引号错误：

$get_comments = "SELECT * FROM 'articlecomments' WHERE 'commentpost' = 'post1' LIMIT 0 , 30";

正确的是：

$get_comments = "SELECT * FROM articlecomments WHERE commentpost = 'post1' LIMIT 0, 30";

必须写$ row：

 " . $row['commentname'] . "
      </div>
      <div class='col-md-9'>
        " . $row['commentvalue'] . "

PHP MySQLi：选择多行并为每个行显示所选的列值

3 个答案: