解决lists标签中，加上where后其他条件失效的问题



问题描述：

{pc:content action="lists" catid="$catid" where="posids`!='0'" thumb="1" num="10"}

发现不是调用当前栏目而是全部栏目的，并且不管有没有缩略图都显示出来，在PHPCMS官方论坛搜索了一下，发现这个问题存在很久了。今天我们说一下怎么修复吧！

 

打开 /phpcms/modules/content/classes/目录下的 content_tag.class.php这个文件，把下面的代码（大概第63行）

if(isset($data['where'])) {

$sql = $data['where'];

} else {

$thumb = intval($data['thumb']) ? " AND thumb != ''" : '';

if($this->category[$catid]['child']) {

$catids_str = $this->category[$catid]['arrchildid'];

$pos = strpos($catids_str,',')+1;

$catids_str = substr($catids_str, $pos);

$sql = "status=99 AND catid IN ($catids_str)".$thumb;

} else {

$sql = "status=99 AND catid='$catid'".$thumb;

}

}

替换为下面的代码即可。

/*以下为创想工作室修改开始*/

if(isset($data['where'])) {

$where = (isset($data['where'])&&(!empty($data['where'])))?' AND '.$data['where']:'';

$thumb = intval($data['thumb']) ? " AND thumb != ''" : '';

if($this->category[$catid]['child']) {

$catids_str = $this->category[$catid]['arrchildid'];

$pos = strpos($catids_str,',')+1;

$catids_str = substr($catids_str, $pos);

$sql = "status=99".$where." AND catid IN ($catids_str)".$thumb;

} else {

$sql = "status=99".$where." AND catid='$catid'".$thumb;

}

} else {

$thumb = intval($data['thumb']) ? " AND thumb != ''" : '';

if($this->category[$catid]['child']) {

$catids_str = $this->category[$catid]['arrchildid'];

$pos = strpos($catids_str,',')+1;

$catids_str = substr($catids_str, $pos);

$sql = "status=99 AND catid IN ($catids_str)".$thumb;

} else {

$sql = "status=99 AND catid='$catid'".$thumb;

}

}