Question

我有一些PHP将MySQL条目拉入HTML表中。当查询未提供结果时，表标题仍然可见，因此我想使用IF函数在前端显示替代消息。

到目前为止，我对PHP的了解有限，这意味着我正在努力寻找任何方法来检查结果。我在网上遇到的每种方法都死定了。

有人可以看到我所缺少的吗？

sort.Slice(mySuperSlice, ascX0DescX1(mySuperSlice))

Answer 1

$ row ['Location'] $ row 在while循环之前不可用

$result = mysqli_query($con,"SELECT * FROM stock WHERE (SKU='$SKUsearch1' AND Location='London Store') OR (SKU='$SKUsearch2' AND Location='London Store') OR (SKU='$SKUsearch3' AND Location='London Store') OR (SKU='$SKUsearch4' AND Location='London Store') OR (SKU='$SKUsearch5' AND Location='London Store') OR (SKU='$SKUsearch6' AND Location='London Store') OR (SKU='$SKUsearch7' AND Location='London Store') OR (SKU='$SKUsearch8' AND Location='London Store') OR (SKU='$SKUsearch9' AND Location='London Store') OR (SKU='$SKUsearch10' AND Location='London Store') OR (SKU='$SKUsearch11' AND Location='London Store') OR (SKU='$SKUsearch12' AND Location='London Store') ");


    if (mysqli_num_rows($result)>0) {

                    echo "<table class='availableTable' border='1'>
                    <tr>
                    <th>Location</th>
                    <th>Item</th>
                    <th>Availability</th>
                    </tr>";

                    while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
                    {
                    echo "<tr>";
                    echo "<td>" . $row['Location'] . "</td>";
                    echo "<td>" . $row['Name'] . "</td>";

                    if(($row['Available']) > 0) {
                    echo "<td> In Stock </td>";
                    }
                    else {
                    echo "<td> Out of Stock </td>";
                    }

                    echo "</tr>";
                    }
                    echo "</table>";




        }

        else {

                echo "<div class='availableText'>No instore stock information is currently available for this product.</div>";
    }

尝试 mysqli_num_rows（$ result） http://php.net/manual/en/mysqli-result.num-rows.php

PHP隐藏数据（如果mysql字符串为空）

1 个答案: