我正在展示两张桌子上的记录,但我正在重复。
表章节
chapterID,ID,cTitle,cDescription
表页
pageId,chapterID,pageName,Icon我得到的当前输出:

期望的输出:

<?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;
}
?>发布于 2015-12-25 05:57:30
而不是使用
SELECT DISTINCT .... 试着使用
SELECT ....
GROUP BY chapter.cTitle,chapter.cDescription;您是否在这两个表之间维护了外键关系?
https://stackoverflow.com/questions/34460305
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