！php+ajax+json有关问题

求助！！php+ajax+ json 问题
本帖最后由 zl2727 于 2013-12-10 18:08:23 编辑

function gtou($s) { return iconv('gbk', 'utf-8', $s); }

$proquery ="select * from pro where typeid='".$id."' order by corank asc";

 $dsql->SetQuery($proquery); 

$dsql->Execute(); 

$a=array();

 while($row = $dsql->GetArray())

 {

 $a[]= array_map('gtou', $row); 

} 

$res = array(

 "status" => "1",

 "result" => $a,

 "moreUrl" => $moreurl, );

 echo json_encode($res);

 

现在前台 返回值是{"status":"1","result":[],"moreUrl":"http:\/\/wcn"} result 是空的
如果
echo json_encode($res);
换成下面的
//echo '{"status":"1","result":[{"id":"1","typeid":"111","goodstitle":"\u8865\u6c34\u6da6\u80a4\u818f \u771f\u7684\u5f88\u597d \u5f88\u8865\u6c34 \u76ae\u80a4\u6709\u5f39\u6027","goodsurl":"http:\/\/www.\/uploads\/allimg\/c131205\/13V23912150-114L_lit.jpg","brandtitle":"\u7f8e\u5b9d\u83b2","brandurl":"http:\/\/www.\/dsadasdas\/","imgurl":"http:\/\/www.\/sadas213\/","price":"59.9","corank":"1"},{"id":"2","typeid":"111","goodstitle":"\u8865\u6c34\u6da6\u80a4222","goodsurl":"http:\/\/www.\/uploads\/allimg\/c131205\/13V23912150-114L_lit.jpg","brandtitle":"\u7f8e222","brandurl":"http:\/\/www.\/dsadasdas\/","imgurl":"http:\/\/www.\/sadas213\/","price":"59.9","corank":"1"}],"moreUrl":"http:\/\/jianfei.\/shuiguo\/"}'; 就正确了 请大神们帮个忙 版主也麻烦您再次指导以下

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------解决方案--------------------
用 var_dump 输出
------解决方案--------------------
print_r($a); 看看