Question

我有两个表，它们是连接的，每个表的ID和下面的元素是相似的。

parentID | objectName | subID            ID| className| subName |
_____________________________            ________________________
    84   |   Test     |   14             14|    BOM        |   Test
    84   |   More     |   16             14|    PDF        |   Test
    84   |   Sub      |   15             15|    Schematics |   Test2

我想列出相关元素的categoryname和subID。几个ObjectNames将有几个相关的类。

PHP代码：

$objects = mysqli_query($con,"SELECT * from subobject");


$join = mysqli_query($con, "SELECT * FROM subrelation AS subrelation INNER JOIN subobject AS subobject ON subobject.subId = subrelation.ID;");

echo "<ul>";
while($obj = mysqli_fetch_array($objects) and $row = mysqli_fetch_array($join))
{

    echo "<li>". $obj['objectName'];

    echo "<ul>";
    //ITERATION GOES HERE
    if($obj['objectName'] == $row['subName'])
        echo "<li>". "$row[className]" . "</li>";

    //END OF ITTERATION
    echo "</ul>";
    echo "</li>";
}   

echo "</ul>";

?>

和输出列表：

-Test
   -BOM
-Sub
  -Schematics
-More

在每个字段下面应该有更多列出的值。

Answer 1

看起来您需要稍微简化一下代码。我的猜测是你的问题正在发生，因为你在每个结果集中有不同的行数。这使得while循环在完成较小的结果集（可能是$objects）时退出，即使较大的集合中还有更多元素。

解决方案是对查询结果进行排序，在while循环中只使用一个条件，并使用字符串objectName跟踪您当前使用的$curr_objectName个$join = mysqli_query($con, 'SELECT * FROM subrelation AS subrelation INNER JOIN subobject AS subobject ON subobject.subId = subrelation.ID ORDER BY subobject.objectName;'); $curr_objectName = ''; echo '<ul>'; while($row = mysqli_fetch_array($join)) { $subName = $row['subName']; if($subName != $curr_objectName)) { if($curr_objectName != '') { #close the previous list #will be skipped on the first loop iteration echo '</ul>'; echo '</li>'; } #start a new list $curr_objectName = $subName; echo '<li>'. $obj['objectName']; echo '<ul>'; } else { echo '<li>'. $row['className'] . '</li>'; } } echo '</ul>'; ：

{{1}}

从表中迭代列表值

1 个答案: