【发布时间】：2017-10-06 22:11:05
【问题描述】：
我目前正在使用while 语句来获取meal_id 的每个关联记录，但我的foreach 没有按预期工作。我的目标是为找到的每餐生成餐食行，然后在其下方显示与该餐食相关的餐食项目，然后循环到下一餐并重复。
目前，结果是显示正确的一餐，但随后显示其下每餐的每餐项目。如：
Meal
Meal Item
Meal Item
Unrelated Meal Item
Unrelated Meal Item
Unrelated Meal Item
Unrelated Meal Item
Unrelated Meal Item
Unrelated Meal Item
Unrelated Meal Item
什么时候该做：
Meal 1
Meal Item
Meal Item
Meal 2
Meal Item
Meal 3
Meal Item
Meal Item
Meal Item
Meal Item
我还想获得那顿饭的项目数，并将其显示在表的餐行中。 到目前为止，这是我的代码，您可以提供的任何帮助都会很棒！ :-)
<?php
$con=mysqli_connect($DBServer,$DBUser,$DBPass,$DBName);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$view = 'empty';
if(isset($_GET["v"]))
$view = $_GET["v"];
switch($view)
{
default:
$result = mysqli_query($con,"SELECT * FROM tblmeals INNER JOIN tblmealitems ON tblmeals.meal_id = tblmealitems.meal_id WHERE DATE(meal_date) = CURDATE() ORDER BY meal_date DESC, meal_time DESC");
$viewstatus = "Today (default view)";
break;
case 'all':
$result = mysqli_query($con,"SELECT * FROM tblmeals INNER JOIN tblmealitems ON tblmeals.meal_id = tblmealitems.meal_id ORDER BY meal_date DESC, meal_time DESC");
$viewstatus = "All Records";
break;
case 'today':
$result = mysqli_query($con,"SELECT * FROM tblmeals INNER JOIN tblmealitems ON tblmeals.meal_id = tblmealitems.meal_id WHERE DATE(meal_date) = CURDATE() ORDER BY meal_date DESC, meal_time DESC");
$viewstatus = "Today";
break;
case 'yesterday':
$result = mysqli_query($con,"SELECT * FROM tblmeals INNER JOIN tblmealitems ON tblmeals.meal_id = tblmealitems.meal_id WHERE DATE(meal_date) = CURDATE() - INTERVAL 1 DAY ORDER BY meal_date DESC, meal_time DESC");
$viewstatus = "Yesterday";
break;
case 'week':
$result = mysqli_query($con,"SELECT * FROM tblmeals INNER JOIN tblmealitems ON tblmeals.meal_id = tblmealitems.meal_id WHERE meal_date BETWEEN CURDATE() - INTERVAL 7 DAY AND CURDATE() ORDER BY meal_date DESC, meal_time DESC");
$viewstatus = "Last 7 Days";
break;
case "fortnight":
$result = mysqli_query($con,"SELECT * FROM tblmeals INNER JOIN tblmealitems ON tblmeals.meal_id = tblmealitems.meal_id WHERE meal_date BETWEEN CURDATE() - INTERVAL 14 DAY AND CURDATE() ORDER BY meal_date DESC, meal_time DESC");
$viewstatus = "Last 14 Days";
break;
} ?><br>
<div class="viewdesc">Current View: <strong><?php echo($viewstatus); ?></strong></div><br>
<br>
<?php
// Start record fail error checking
if (!$result) {
printf("Error: %s\n", mysqli_error($con));
exit();
}
// Start record fail error checking */
if (mysqli_num_rows($result) === 0) {?>
<div id="resultserror">
No results could found for the selected view.</div>
<?php } else { ?>
<table id="results" width="100%" cellpadding="3" cellspacing="0">
<tr>
<th align="center">Meal #</th>
<th align="center">User #</th>
<th align="center">Meal Items</th>
<th align="center">Meal Date</th>
<th align="center">Meal Time</th>
<th align="center">Meal Notes</th>
</tr>
<?php
while($row = mysqli_fetch_assoc($result))
{
?>
<tr>
<td align='center'><?php echo($row['meal_id']);?></td>
<td align='center'><?php echo($row['user_id']);?></td>
<td align='center'><?php echo("Items: " . mysqli_num_rows($result) . "");?></td>
<td align='center'><?php echo(date("d/m/Y", strtotime($row['meal_date'])));?></td>
<td align='center'><?php echo(date("g:i A", strtotime($row['meal_time'])));?></td>
<td align='center'><?php echo($row['meal_notes']);?></td>
</tr>
<?php foreach ($result as $row)
{?>
<tr>
<td align='center' colspan='4'><?php echo($row['meal_item_name']);?></td>
<td align='center'><?php echo($row['meal_item_measure']);?></td>
<td align='center'><?php echo($row['meal_item_measurement']);?></td>
</tr>
<?php
}
}
?>
<tfoot>
<tr>
<td colspan="8" align="center">- end of report -</td></tr></tfoot>
</table>
<?php
mysqli_close($con);
}?>
【问题讨论】：
-
我建议不要使用
JOIN，而是在循环内执行第二组查询（每餐一个，具体获取该餐项）。使用 JOIN 执行此操作会导致您为每顿包含该餐的每餐获得 [该餐中的项目数] 行。在您的案例中循环时，您需要重组/处理它，而在使用额外查询路由时您不必这样做。 -
尽管有成千上万的用户使用数百种餐点，每个用户都有自己的数千种餐点，但这在性能方面如何公平？来回查询不同的表会不会归结为过度处理？
-
取决于您的索引，现代 DBMS 可以毫不费力地处理它。
-
好的，试一试！谢谢:-)
标签： php mysqli foreach while-loop