我正在使用的项目中使用了很多jQuery。
我有一个javascript函数,它向一个以JSON返回数据的控制器发出ajax请求。
我想显示一条用户友好的消息,通知用户他/她尚未存储任何信息。但是我对如何在JSON中发送响应感到困惑,所以我的javascript函数可以确定用户是否有要显示的信息。
这是我的javascript函数:
function latest_pheeds() {
var action = url+"pheeds/latest_pheeds";
$('#pheed-stream').html('<div class="loading"></div>');
$('.loading').append('<img src="'+pheed_loader_src+'" />');
$.ajax({
url:action,
type:'GET',
dataType:'json',
error: function () {
},
success:function(data) {
$('.loading').fadeOut('slow');
$.each(data,function(index,item) {
$('#pheed-stream').append
(
'<div class="pheed" id="'+item.pheed_id+'">'+
'<p><a class="user_trigger" href="users/info/'+item.user_id+'">'
+item.user_id+'</a></p>'+
'<p>'+item.pheed+'</p>'+
'<div class="pheed_meta">'+
'<span>'+item.datetime+' Ago</span>'+
'<span class="cm">'+item.comments+
'<img class="comment_trigger" src="/pheedbak/assets/img/comment.png" title="Click to comment on pheed" onclick="retrieve_comments('+item.pheed_id+')">'+
'</span>'+
'<span>'+item.repheeds+
' Repheeds'+
'<img class="repheed_trigger" src="/pheedbak/assets/img/communication.png" title="Click to repheed" onclick="repheed('+item.pheed_id+')">'+
'</span>'+
'<span>'+
'Favourite'+
'<img class="favourite_trigger" src="/pheedbak/assets/img/star.png" title="Click to make this a favourite" onclick="favourite_pheed('+item.pheed_id+')" />'+
'</span>'+
'</div>'+
'</div>'
);
});
}
});
}
继续控制器功能ajax请求
function latest_pheeds() {
//Confirm if a user is logged before allowing access
if($this->isLogged() == true) {
//load the pheed model for database interaction
$this->load->model('pheed_model');
//load user model
$this->load->model('user_model');
//load comment model
$this->load->model('comment_model');
//store the pheeds to a the $data variable
$data = $this->pheed_model->get_latest_pheeds();
//Load the date helper to calculate time difference between post time and current time
$this->load->helper('date');
//Current time(unix timetamp)
$time = time();
//pheeds
$pheeds = array();
if(count($data) > 0 ) {
foreach($data as $pheed) {
$row['pheed_id'] = $pheed->pheed_id;
$row['user_id'] = $this->user_model->return_username($pheed->user_id);
$row['pheed'] = $pheed->pheed;
$row['datetime'] = timespan($pheed->datetime,$time);
$row['comments'] = $this->comment_model->count_comments($pheed->pheed_id);
$row['repheeds'] = $pheed->repheeds;
$pheeds[] = $row;
}
echo json_encode($pheeds);
$response['response'] = "Ok";
$res[] = $response;
echo json_encode($res)."\n";
}
} else {
}
它生成JSON输出,但语法被破坏所以我无法用javascript读取它, 但是一旦我从上面的方法中删除了以下代码,它就能正常工作
$response['response'] = "Ok";
$res[] = $response;
echo json_encode($res)."\n";
答案 0 :(得分:14)
你必须在响应中只使用一次json_encode($ data),如果你想输出更多东西,你需要将你的数据合并到一个数组中并发送它。
编辑:要明确,你能做到的一种方法就是这样:
echo json_encode(array('pheeds' => $pheeds, 'res' => $res));
然后在JS中你会得到一个带有“pheeds”和“res”键的数组。
虽然它可能没什么实际意义,但我建议你在回显json编码的字符串之前这样做:
header('Content-Type: application/json');