内部联接表并显示在php中

Question

I want to select data from more tables with Inner join.

These are my tables.

teams (id, team_name)
badges (id, badgename, badgeimage, badgedescription)
teambadges (id, team_id, badge_id)

I want to write a statement that shows the team name with all the badges they have. I also want to display this in a table

This is my statement.

$sql = mysqli_query($connection, 'SELECT teams.team_name,badges.badgename
FROM teambadges
INNER JOIN teams ON teams.id = teambadges.team_id
INNER JOIN badges ON badges.id = teambadges.badge_id;');

Php:

<table class="table table-condensed table-striped table-bordered table-hover">
            <thead>
                <tr>
                    <th width="5%"><center>No</center></th>
                    <th>team id</th>
                    <th>badge id</th>


                </tr>
            </thead>
            <tbody id="data">
            <?php $no=1; while ($row = mysqli_fetch_array($sql)) { ?>
                <tr>
                    <td align="center"><?php echo $no; ?></td>
                    <td><?php echo $row['team_name']; ?></td>
                    <td><?php echo $row['badgename']; ?></td>

                </tr>
            <?php $no++; } ?>   
            </tbody>
        </table>

This is executed inside the php page but i keep getting this error : Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result,

Answer 1

使用mysqli_query（）或die（mysqli_error（$ connection））来获取错误并检查查询是否在mysql上成功运行

Answer 2

i think you error is here

<td><?php echo $row['teams.name']; ?></td>
<td><?php echo $row['badges.name']; ?></td>

just chech if you have the right name column . if you have there table with the same name

teams (id, team_name)
badges (id, badgename, badgeimage, badgedescription)
teambadges (id, team_id, badge_id)

you must do this :

<td><?php echo $row['team_name']; ?></td>
    <td><?php echo $row['badgename']; ?></td>