I'm new to PHP, so please bear with me. I am working on a database project at work, where I'm interning. I'm trying to display a table given that the data has been fetched from the database. I have 3 tables
Table vege_plant
Table vege_location
Table vege_data
My PHP code looks like this:
<?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>';
}
However, it's failing to display, not alerting any errors. I don't know where I'm going wrong.
Try This there is a syntax error at line 1st single qoute missing
$sql = mysql_query('SELECT vege_data.*, vege_location.*, vege_plants.*
FROM vege_data
JOIN vege_location ON vege_location.uuid =vege_data.vege_location_uuid
JOIN vege_plants ON vege_plants.uuid = vege_data.vege_plants_uuid;');
while ($row = mysql_fetch_array($result))
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