这是什么意思?我的页面一直工作正常,直到我评论出一些代码后突然出现。这是我正在使用的代码。它不会让我发布这个,因为没有足够的描述所以我只是填补一些空间。
Parse error: syntax error, unexpected $end in C:\xampp\htdocs\test4\mainpage.php on line 387
<?php
if($note['type'] == 'list')
{
$note_type='list';
print "<table>";
while($note_info = mysql_fetch_array( $note_content ))
{
print "<tr><td>";
if($note_info['complete'])
echo "<strike>";
echo $note_info['body'];
print "</td>";
if($note_info['complete']){
echo "</strike>";
echo "<td><input type='checkbox' name='complete_goal' value='".$note_info['note_id']."'></input></td>";
}
print "</tr>";
}
/*
print "<dl style=\"float:left\">";
while($note_info = mysql_fetch_array( $note_content ))
{
if($note_info['complete'])
echo "<strike>";
echo "<dt>".$note_info['body']."</dt>";
if($note_info['complete'])
echo "</strike>";
}
print "</dl>";
mysql_data_seek($note_content, 0);
print "<dl style=\"float:right\">";
while($note_info = mysql_fetch_array( $note_content ))
{
// if(!$note['complete'])
echo "<dt><input type='checkbox' name='complete_goal' value='".$note_info['note_id']."'></input></dt>";
}
print "</dl>";
}
else
{
echo $note['body'];
}
*/
?>
答案 0 :(得分:0)
在评论您的代码时,您还评论了最顶层}
条件的右括号if($note['type'] == 'list')
答案 1 :(得分:0)
试试这个:
if($note['type'] == 'list')
{
$note_type='list';
echo "<table>";
while($note_info = mysql_fetch_array( $note_content ))
{
print "<tr><td>";
if($note_info['complete'])
{
echo "<strike>";
echo $note_info['body'];
echo "</td>";
}
if($note_info['complete'])
{
echo "</strike>";
echo "<td><input type='checkbox' name='complete_goal' value='".$note_info['note_id']."'></input></td>";
}
echo "</tr>";
}
echo "</table>";
}
您还应该使用可以为您检测这些错误的IDE。给Netbeans一个机会。