我已经在这方面工作了一段时间并整理了整个后端,我只是在与mySQL数据库挣扎,因为这是我第一次在项目中使用它! 我试图从数据库中回忆一些信息,但是从2个不同的表中回忆并显示它。 我正在尝试在循环中调用信息时收到T_VARIABLE错误。
<p><?php
$inttotalcredits=0;
$result = mysql_query("SELECT * FROM site_products");
$set = mysql_fetch_array($result)
$prod_id = $row[0]
echo "<table border='0' width='100%'>
<tr>
<th>Product ID</th>
<th>Part Name</th>
<th>Part Number</th>
<th>Description</th>
<th>Level</th>
</tr>";
while($row = mysql_fetch_array($result)
{
echo "<tr>";
echo "<td>" . "<center> $row[0] </center>" . "</td>";
echo "<td>" . "<center> $row[1] </center>" . "</td>";
echo "<td>" . "<center> $row[2] </center>" . "</td>";
echo "<td>" . "<center> $row[4] </center>" . "</td>";
}
$result1 = mysql_query("SELECT * FROM site_trans WHERE trans_product = $prod_id");
while($row1 = mysql_fetch_array($result1))
{
echo "<td>" . "<center> $row1[6] </center>" . "</td>";
$inttotalcredits += $row1[6];
echo "</tr>";
echo "</center>";
}
echo "<td>" . "" . "</td>";
echo "<td>" . "" . "</td>";
echo "<td>" . "<b>Total Stock Items</b>" . "</td>";
echo "<td>" . "<b><center> $inttotalcredits </center></b>" . "</td>";
echo "</table>";
?></p>
我不知道这是一个直接的错误我或者我只是没有正确构建循环。
奥利弗
答案 0 :(得分:4)
您遗失的;
:
$set = mysql_fetch_array($result)
$prod_id = $row[0]
应该是:
$set = mysql_fetch_array($result);
$prod_id = $row[0];
同样如Lyth所指出的,你在while语句中缺少parethesis:
while ($row = mysql_fetch_array($result)
应该是:
while ($row = mysql_fetch_array($result))
我建议重写代码,不要在PHP代码中回显HTML元素:
<p>
<table border='0' width='100%'>
<tr>
<th>Product ID</th>
<th>Part Name</th>
<th>Part Number</th>
<th>Description</th>
<th>Level</th>
</tr>
<?php
$inttotalcredits = 0;
$result = mysql_query("SELECT * FROM site_products");
$set = mysql_fetch_array($result)
$prod_id = $row[0]
while($row = mysql_fetch_array($result)) :
?>
<tr>
<td><center><?php echo $row[0] ?></center></td>
<td><center><?php echo $row[1] ?></center></td>;
<td><center><?php echo $row[2] ?></center></td>
<td><center><?php echo $row[3] ?></center></td>
<?php endwhile ?>
依此类推...它更干净,更容易调试。