php - 显示来自 3 个 SQL 表的表数据并使用 php 条件显示 - 使用完全连接

Question

我是 PHP 新手，所以请耐心等待。我正在工作中从事一个数据库项目，并在那里实习。我正在尝试显示一个表格，因为数据已从数据库中获取。我有3张 table

表 vege_plant

• uuid
• 姓名

表 vege_location

• uuid
• 地点

表 vege_data

• uuid
• vege_plants_uuid
• vege_location_uuid
• 横断面编号
• 象限编号
• 还活着
• 死了

我的 PHP 代码如下所示:

<?php

       $sql = mysql_query('SELECT vege_data.*, vege_location.*, vege_plants.*
       FROM vege_data
       JOIN vege_location ON vege_location.uuid =vege_data.vege_location_uuid
       JOIN vege_plants ON vege_plants.uuid = vege_data.vege_plants_uuid;);
            echo '<table class = "table table-hover">
<thead>
    <tr>
        <th>Location</th>
        <th>Transect Number</th>
        <th>Quadrat Number</th>
        <th>Plant Name</th>
        <th>Alive</th>
        <th>Dead</th>
        <th>Action</th>
    </tr>
</thead>
<tbody>';
            $count = 1;    

while ($row = mysql_fetch_assoc($result)) {
            echo '<tr>';
            echo '<td contenteditable="true">';
            echo '<td>';
            echo $row['location'];
            echo '</td>';
            echo '</td>';
            echo '<td contenteditable="true">';
            echo $row['transect_num'];
            echo '</td>';
            echo '<td contenteditable="true">';
            echo $row['quadrant_num'];
            echo '</td>';
            echo '<td contenteditable="true">';
            echo $row['name'];
            echo '</td>';
            echo '<td contenteditable="true">';
            echo $row['alive'];
            echo '</td>';
            echo '<td>';
            echo $row['dead'];
            echo '</td>';
            echo '</tr>';
        }

但是，它无法显示，也没有警告任何错误。我不知道我哪里错了。

Answer 1

尝试一下，第一行有一个语法错误，缺少单个 qoute

   $sql = mysql_query('SELECT vege_data.*, vege_location.*, vege_plants.*
   FROM vege_data
   JOIN vege_location ON vege_location.uuid =vege_data.vege_location_uuid
   JOIN vege_plants ON vege_plants.uuid = vege_data.vege_plants_uuid;');


while ($row = mysql_fetch_array($result))