Question

我试图为我的网站制作一个评论部分，到目前为止我已经完成了显示名称和头像，但没有显示评论文本和日期。评论表是： id，member_id，comment，date，product_id 问题是我无法在循环中获取结果。我尝试了另一个查询，但它一直给我相同的评论。

<?php
include("includes/db.php");
{
$sql_query="SELECT * FROM product_comm WHERE product_id='$id'";
    $result_set=mysqli_query($conn,$sql_query) or die('error');
    $get_comm=mysqli_fetch_array($result_set);
    $comm_date=$get_comm['date'];
    $comm_mem_id=$get_comm['mem_id'];
    $comm_text=$get_comm['comment'];
}
{
$sql_query="SELECT * FROM products WHERE str='$str'";
    $result_set=mysqli_query($conn,$sql_query) or die('error');
    $get_products=mysqli_fetch_array($result_set);
    $id=$get_products['id'];
      }
{
$sql_query="SELECT * FROM member WHERE id='$comm_mem_id'";
    $result_set=mysqli_query($conn,$sql_query) or die('error');
    $get_member=mysqli_fetch_array($result_set);
    $member_name_first=$get_member['first_name'];
     $member_name_last=$get_member['last_name'];
     $member_avatar=$get_member['avatar'];
     }

$result = mysqli_query($conn,"SELECT id  FROM products WHERE str='$str'");

$result = mysqli_query($conn,"SELECT * FROM product_comm WHERE product_id=$id");

$result = mysqli_query($conn,"SELECT * FROM member  WHERE id='$comm_mem_id'");
if ($result) { 

    if($result->num_rows === 0)
    {
        echo "<center>No comments</center>";
    }
    else
    {

while($row = mysqli_fetch_array($result)) {

     echo "

<div class='media'>
  <div class='media-left'>
    <img src='http://maikatideeba.byethost32.com/images/profile_pics/".$row['avatar']."' class='img-circle' style='width:60px'>
  </div>
  &nbsp;<div class='media-body'>
    <h4 class='media-heading'>&nbsp; ".$row['first_name']." ".$row['last_name']." <small><i> on ".$row['date']." </i></small></h4>
  <p> ".$row['comment']." </p> 
<hr>  
</div>

</div>";
}
}
}


mysqli_close($conn);
?>

正确地将所有内容正确地插入到数据库中，但是当我尝试查看结果时（注释如此）它什么都没有。Result image

Answer 1

从不必要的查询中清理您的来源。这是一团糟，很难理解你要做的事情。

学习将查询分配给变量，例如：

$sql_members = "SELECT * FROM member WHERE id= ".$comm_mem_id;
$result = mysqli_query( $conn , $sql_members );

以后调试和回显查询要容易得多。例如，输出是什么：

echo $sql_members;

避免在HTML输出中使用单引号

非常 好奇的 子域名选择...“maikatideeba.byethost32.com”。

PHP无法从DB获取结果（同时）

1 个答案: