【发布时间】：2014-06-10 08:18:58
【问题描述】：
您好，我有一个小项目，它将通过使用 html 中的选择选项为您提供过去 4、6、10 周的新闻。我收到语法错误，选择选项标记中出现意外的 T_STRING。我理解错误，因为我无法在打开的 php 标记中使用 php。但我不知道如何解决这个错误。任何帮助。谢谢。 这是我的代码：
<?php
$page['doctype'] = true;
$param = array_merge($_GET, $_POST);
$return = array();
if($param['aktion'] == 'edit-news')
{
$page['register-edit-news'] = array(
1 => array( 'Edit-News','aktiv',$page['script'],'',''),
);
if(isset($_POST['BtnSubmit']))
{
if(($_POST['news'])==4){
$sql=" SELECT DISTINCT ad_news_texte.headline, ad_news.datum_archiv
FROM ad_news_texte
INNER JOIN ad_news_oe ON ad_news_texte.news_id = ad_news_oe.id_ad_news
INNER JOIN ad_news ON ad_news_oe.id_ad_news = ad_news.id
WHERE ad_news.datum_archiv
BETWEEN curdate( ) - INTERVAL DAYOFWEEK( curdate( ) ) +28
DAY AND curdate( )
";
$sql_select=mysql_query($sql);
while($row = mysql_fetch_array($sql_select)) {
echo $row['headline'] . " " .$row['datum_archiv'] ;
echo "<br>";
}
}
if(($_POST['news'])==10){
$sql=" SELECT DISTINCT ad_news_texte.headline, ad_news.datum_archiv
FROM ad_news_texte
INNER JOIN ad_news_oe ON ad_news_texte.news_id = ad_news_oe.id_ad_news
INNER JOIN ad_news ON ad_news_oe.id_ad_news = ad_news.id
WHERE ad_news.datum_archiv
BETWEEN curdate( ) - INTERVAL DAYOFWEEK( curdate( ) ) +70
DAY AND curdate( )
";
$sql_select=mysql_query($sql);
while($row = mysql_fetch_array($sql_select)) {
echo $row['headline'] . " " .$row['datum_archiv'] ;
echo "<br>";
}
}
$html =
'
<body bgcolor="#FFFFFF">
<table width="100%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td class="welcome-page-hint">
<table width="538" cellpadding="0" cellspacing="0" border="0">
<tr>
<td>
<h3>News</h3>
<form name="UserInformationForm" method="POST" action="#">
<select name="news">
<option value="4" <?php if($_POST['news']=="4") echo "selected=selected"; ?>>Show news from the last 4 weeks</option>
<option value="6" <?php if($_POST['news']=="6") echo "selected=selected"; ?>>Show news from the last 6 weeks</option>
</select>
<br/><br/>
<input name="BtnSubmit" type="submit" value="Submit">
</form>
</td>
</tr>
</table>
</td>
</tr>
</table>';
$return = array(
'status' => 1,
'html' => $html
);
echo(json_encode($return)) ;
}
?>
【问题讨论】：
-
你需要转义你的引号。
-
看看你代码的语法高亮，我就可以清楚地看到问题了。