我有一个命令的问题不会给搜索中获得结果空字段
Notice: Undefined index: q in C:\xampp\htdocs\Capstone - Copy\index.php on line 118
即使该字段为空我为这是唯一的问题,该网站的代码在那里,如果我不在我的搜索中没有放置一个值,它会给出这个错误,我可以删除当我重新加载站点时得到的值,所以它只会给出默认输出,即整行和图片,我会把打印屏幕该网站的屏幕截图是该网站的默认视图。
> <?php
>
> $searchtext = $_GET['q'];
>
>
> $per_page =5;
> $pages_query = mysql_query("SELECT COUNT('PersonID') FROM persons");
> $pages = ceil(mysql_result($pages_query,0)/$per_page);
>
> $page = (isset($_GET['page'])) ? (int)$_GET['page'] : 1;
> $start = ($page - 1) * $per_page;
>
>
> $query=mysql_query("select * from persons where firstname like
> '%$searchtext' or lastname like '%$searchtext' order by date desc
> LIMIT $start,$per_page "); while($test = mysql_fetch_array($query))
> {
> $id = $test['PersonID'];
>
>
>
> echo"<div class = content />";
> echo"<img height=200 width=200 src='upload/". $test['Image'] ."'/>";
> echo"" .$test['LastName']." ";
> echo"". $test['MiddleName']. " ";
> echo"". $test['FirstName']. "";
> echo"<right> <a href ='view.php?PersonID=$id'>Edit</a></right>";
> echo"<right> <a href ='del.php?PersonID=$id'>Delete</a></right>";
> echo"</div>";
>
>
>
> }
> if ($pages >=1 && $page <= $pages) {
> for ($x=1; $x<=$pages; $x++) {
> echo ($x == $page) ? '<strong><a href="?page='.$x.'">'.$x.' </a></strong> ' : '<a
> href="?page='.$x.'">'.$x.' </a>';
> }
> } ?>
>
>
,这里是不会给一个错误的样本结果,因为有一个默认值
如果$ searchText为null,他希望运行该查询。他的查询将返回所有失踪人员! –