Question

懂PHP的进来一下帮忙解决个问题

网页显示Parse error: syntax error, unexpected T_DOUBLE_ARROW in /data/home/qxu0100220/htdocs/wap/category.php on line 31
是怎么回事
是category.php出错了吗？
我把这个文件内容贴出来，求大神帮忙回答一下怎么解决！
<?php
define('IN_PHPMPS', true);
require dirname(__FILE__) . '/include/common.inc.php';
if(isset($_REQUEST['id']))$catid = intval($_REQUEST['id']);

if($catid) {
$cat_info = get_cat_info($catid);
if(empty($cat_info)) {
header("Location: ./");
exit;
}
}

$cat_row = get_cat_children($catid, 'array');
if(!empty($cat_row)) {
$cat_arr = array();
foreach($cat_row as $val) {
$val['catname'] = $val['name'];
$val['url'] = url_rewrite('category',array('cid'=>$val['id'],'eid'=>$areaid));
$cat_arr[] = $val;
}
$cats = get_cat_children($catid);
}
if(empty($cats))$cats = $catid;

$cat_sql = " and i.catid in ($cats) ";
$page = empty($_REQUEST['page']) ? '1' : intval($_REQUEST['page']);
$sql = "SELECT COUNT(*) FROM {$table}info as i WHERE is_check=1 $cat_sql $area_sql";
$count = $db->getone($sql);
$size = '20';
$pager['search'] = array_merge('catid'=>$catid, 'areaid'=>$areaid);
$pager = get_pager('category.php',$pager['search'],$count,$page,$size);
$sql = "SELECT id,title,postdate,enddate,i.catid,c.catname,i.areaid,a.areaname,thumb,i.description FROM {$table}info AS i LEFT JOIN {$table}category AS c ON i.catid=c.catid LEFT JOIN {$table}area AS a ON a.areaid=i.areaid WHERE is_check=1 $cat_sql $area_sql ORDER BY postdate DESC limit $pager[start],$pager[size]";
$res = $db->query($sql);
$info = array();
while($row=$db->fetchRow($res)) {
$row['url'] = url_rewrite('view',array('vid'=>$row['id']));
$row['postdate'] = date('y年m月d日', $row['postdate']);
$row['lastdate'] = enddate($row['enddate']);
$row['intro'] = cut_str($row['description'], 50);
$info[$row['id']] = $row;
}

$seo['title'] = $cat_info['catname'] . '信息列表 - Powered by Phpmps';
include tpl('category');
?>

Answer 1

消除下缓存试试，也可能是模板问题，出现不对称的标签，所以需要修改或者看看问题，你进不去就做不到这些了。

解决方法：

1、将代码中的<?  ?>全部替换成<?php  ?>

2、打开 php.ini ，找到 short_open_tag = Off 这一行，将 Off 修改为 On，保存退出并重启 Apache 即可解决问题

parse error 意思是解析错误;语法错误;分析错误

syntax error 意思是：句法误差

unexpec 意思是：没有料到的，想不到的，意外的

Answer 2

array_merge 参数给错了嘛,应该给两个数组呀,没你那种写法的

Answer 3

$pager['search'] = array_merge('catid'=>$catid, 'areaid'=>$areaid); 这一行改成

$pager['search'] = array_merge(array('catid'=>$catid, 'areaid'=>$areaid));

Answer 4

刚学习还不太懂

Answer 5

就是简单的语法错误，这个还是要自己再检查检查,毕竟不是什么太难的事，自己也可以得到提升，相信你可以解决的,呵呵