我使用mysql_*
方式进行了查询,select
的某些值为from
table
,where
行id
是1
。然后我将所有这些值放在变量中,这样我就可以将它们回显到我的表单中。如何使用Joomla的数据库查询方式做同样的事情?
以下是使用mysql_*
:
<?php // DATABASE QUERY
$query="SELECT countdown_module, hometeam_header
FROM jos_gm_nextmatch
WHERE id = 1";
$result=mysql_query($query);
// DATABASE VARIABLES
$countdown_module = mysql_result($result,$i,"countdown_module"); ?>
$hometeam_header = mysql_result($result,$i,"hometeam_header"); ?>
<form action="" method="post" name="form">
<input name="countdown_module" value="<?php echo $countdown_module ?>" type="text" />
<input name="hometeam_header" value="<?php echo $hometeam_header ?>" type="text" />
<input name="submit" type="submit" value="UPDATE" />
</form>
答案 0 :(得分:0)
好的我找到了!!!这是一个例子......
<?php // DATABASE QUERY
$db = JFactory::getDBO();
$query="SELECT countdown_module, hometeam_header
FROM jos_gm_nextmatch
WHERE id = 1";
$db->setQuery($query);
$rows = $db->loadObjectList();
$itemrow = $rows[0];
// DATABASE VARIABLES
$countdown_module = $itemrow->countdown_module;
$hometeam_header = $itemrow->hometeam_header; ?>
<form action="" method="post" name="form">
<input name="countdown_module" value="<?php echo $countdown_module ?>" type="text" />
<input name="hometeam_header" value="<?php echo $hometeam_header ?>" type="text" />
<input name="submit" type="submit" value="UPDATE" />
</form>