我正在显示两张桌子的记录,但我正在重复。
表章
chapterID,ID,cTitle,cDescription
表格页
pageId,chapterID,pageName,Icon
所需的输出:
<?php
$con=mysql_connect( 'localhost', 'root', '') or die ( 'could not connect to db:'. mysql_error());
mysql_select_db( 'database') or die( 'No db found');
$count = 1;
$query = mysql_query( "SELECT DISTINCT chapter.cTitle,chapter.cDescription,page.pageName,page.pageIcon FROM chapter INNER JOIN page ON chapter.chapterID = page.chapterID") or die( 'Query failed');
while ($result=mysql_fetch_array($query)) {
echo $result[ 'cTitle']; echo $result[ 'cDescription'];
echo $result[ 'pageName']; $count +=1;
}
?>
答案 0 :(得分:-1)
而不是使用
SELECT DISTINCT ....
尝试使用
SELECT ....
GROUP BY chapter.cTitle,chapter.cDescription;
您是否在这两个表中保持了外键关系?