Question

求php大神帮忙看下代码

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("skkd", $con);
$result = mysql_query("SELECT * FROM r_order");
echo "<table border='1'>
<tr>
<th>订单编号</th>
<th>顾客姓名</th>
<th>顾客电话</th>
<th>顾客人数</th>
<th>就餐时间</th>
<th>备注</th>
<th>菜品信息</th>
<th>总价</th>
</tr>";

while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['orderID'] . "</td>";
echo "<td>" . $row['user_name'] . "</td>";
echo "<td>" . $row['tel'] . "</td>";
echo "<td>" . $row['people'] . "</td>";
echo "<td>" . $row['time'] . "</td>";
echo "<td>" . $row['contract'] . "</td>";
echo "<td>" . $row['dinner_id'] . "</td>";
echo "<td>" . $row['total'] . "</td>";
echo "</tr>";
}
echo "</table>";

mysql_close($con);
?>
数据库包括
r_order
r_food(food_id,food_name)
我打印出r_order中的数据，里面的dinner_id就是food里面的food_id，怎么把打印表里的dinner_id换成打印food_name

Answer 1

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("skkd", $con);
$result = mysql_query("SELECT a.*,b.food_name FROM r_order a INNER JOIN r_food b on a.dinner_id = b.food_id");
echo "<table border='1'>
<tr>
<th>订单编号</th>
<th>顾客姓名</th>
<th>顾客电话</th>
<th>顾客人数</th>
<th>就餐时间</th>
<th>备注</th>
<th>菜品信息</th>
<th>总价</th>
</tr>";

while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['orderID'] . "</td>";
echo "<td>" . $row['user_name'] . "</td>";
echo "<td>" . $row['tel'] . "</td>";
echo "<td>" . $row['people'] . "</td>";
echo "<td>" . $row['time'] . "</td>";
echo "<td>" . $row['contract'] . "</td>";
echo "<td>" . $row['food_name'] . "</td>";
echo "<td>" . $row['total'] . "</td>";
echo "</tr>";
}
echo "</table>";

mysql_close($con);
?>

追问

菜名输出是 ？？？

追答

echo "" . $row['food_name'] . "";

追问

我的意思是 他没输出菜品名，只输出了问号

Answer 2

你要先将你的 $row 变成一个二维数组 之后再将所编辑的二维数组的键值变成以id开头的拿数据的时候可以用遍历拿出

Answer 3

直接把$row['dinner_id']改成$row['food_name']不就可以了吗？