Question

我有一个库存表，可以使用每个事务进行更新。目前有4种产品，但在某些时候可能会有更多。以下是我用来拉动所有内容的表格。

    <?php
$db = new mysqli('', '', '', 'inventory');

if($db->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
}

$sql = "SELECT * FROM inventory";

if(!$result = $db->query($sql)){
    die('There was an error running the query [' . $db->error . ']');
}

echo "<table style='border: 2px;font-family: tahoma;'><caption><b>Entire Database Contents</b></caption><tr><td>ID</td><td>Time Stamp</td><td>Staff</td><td>Client</td><td>Needed</td><td>Product</td><td>Amount</td><td>Totals</td><td style='width: 200px;'>Comments</td></tr>";
    while($row = $result->fetch_assoc()){
        echo '<td>' . $row['id'] . '</td>';
        echo '<td>' . $row['timeStamp'] . '</td>';
        echo '<td>' . $row['staff'] . '</td>';
        echo '<td>' . $row['client'] . '</td>';
        echo '<td>' . $row['dateNeeded'] . '</td>';
        echo '<td>' . $row['product'] . '</td>';
        echo '<td>' . $row['amt'] . '</td>';
        echo '<td>' . $row['tot'] . '</td>';
        echo '<td>' . $row['comments'] . '</td></tr>';
}

echo "</table>";

?>

我想做的是仅提取每种产品的最新条目。我想也许DISTINCT和LAST可能在这里发挥作用，但不知道如何设置它。有什么指针吗？

Answer 1

您应该创建date_added字段并按其排序：

$sql = "SELECT * FROM inventory ORDER BY date_added DESC";

Answer 2

您只需选择@TargetApi(Build.VERSION_CODES.KITKAT) private void enableRemoteDebugging() { try { WebView.setWebContentsDebuggingEnabled(true); } catch (IllegalArgumentException e) { LOG.d(TAG, "You have one job! To turn on Remote Web Debugging! YOU HAVE FAILED! "); e.printStackTrace(); } }时间戳：

的产品

MAX

Answer 3

这应该有效：

select distinct product, id, timeStamp, staff, client, dateNeeded,  amt, tot, comments from inventory order by timestamp desc;

如何提取每件产品的最后一个条目

3 个答案: