如何在PHP的mysql中只显示一次注释？ [英] How display the comment only once from mysql in PHP?

问题描述

我有两张桌子：multiple_image和帖子



数据库结构===>

 multiple_image：
 
 | id |图片| imagesid | 
 -------------------------- 
 | 1 |名字| 5557 | 
 | 2 |名字| 5557 | 
 | 3 |名字| 5585 | 
 -------------------------- 

帖子：
 
 | commentid |评论| iamgesid | 
 ------------------------------------ 
 | 1 |傻瓜| 5557 | 
 | 2 |傻瓜2 5585 | 

我想要实现的输出是：

我想要显示所有图像对于具有相同imagesid的评论，但我的问题是，评论正在重复。



我尝试了什么： < br $>

 $ sql =SELECT image，posts.imagesid，multiple_image.imagesid，comment 
 FROM multiple_image JOIN post ON（multiple_image.imagesid = posts.imagesid）; 
 $ result = $ conn-> query（$ sql）; 
 
 if（！$ result）{
 trigger_error（'无效查询：'。$ conn->错误）; 
} 
 
 if（$ result-> num_rows> 0）{
 
 //每行输出数据
 while（$ row = $ result-> fetch_assoc（））{
 
 echo $ row ['comment']; 
 $ imgs =< div id ='img_div'>< img width =''src ='upload /\".$ row ['image']。'>< / div> ; 
 echo $ imgs; 
} 
} 

解决方案

sql =SELECT image，posts.imagesid，multiple_image.imagesid，comment
FROM multiple_image JOIN帖子ON（multiple_image.imagesid = posts.imagesid）;

result =

conn-> query（

I have two tables: multiple_image and posts

The db structures ===>

multiple_image:

| id |  image | imagesid |
--------------------------
| 1  |  name  |    5557  |
| 2  |  name  |    5557  |
| 3  |  name  |    5585  |
--------------------------

posts:

| commentid |  comment  | iamgesid |
------------------------------------
|     1     |   fool    |   5557   |
|     2     |  fool2    |   5585   |

The output that I am trying to achieve is:
I want to display all the images for a comment which has the same imagesid, but my problem is that, the comment is repeating itself.

What I have tried:

$sql = "SELECT image, posts.imagesid, multiple_image.imagesid, comment
        FROM multiple_image JOIN posts ON (multiple_image.imagesid=posts.imagesid)";
$result = $conn->query($sql);

if (!$result) {
    trigger_error('Invalid query: ' . $conn->error);
}

if ($result->num_rows > 0) {

    // output data of each row
while($row = $result->fetch_assoc()) {

echo $row['comment'];
$imgs= "<div id='img_div'><img width='' src='upload/".$row['image']."' ></div>";
echo $imgs;
}
}

解决方案

sql = "SELECT image, posts.imagesid, multiple_image.imagesid, comment FROM multiple_image JOIN posts ON (multiple_image.imagesid=posts.imagesid)";

result =

conn->query(

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