javascript - 如何停止 json_encode 返回未定义

Question

我从一个页面发送了许多动态帖子 ID，并且 PHP 服务器端页面(server.php)使用这些 id 进行查询以查找 mysql 中的新添加的数据。

如果在 mysql 中未找到任何新添加的数据，则返回一个未定义值。因此，根据我的脚本，它会每隔一段时间一个又一个地附加一个 undefined 。

如何停止这个未定义的值？

我的 JavaScript:

var CID = []; // Get all dynamic ids of posts (works well)
$('div[data-post-id]').each(function(i){
CID[i] = $(this).data('post-id');
});

function addrep(type, msg){
CID.forEach(function(id){
    $("#newreply"+id).append("<div class='"+ type +""+ msg.id +"'><ul><div class='cdomment_text'>"+ msg.detail +"</ul></div>");
});
}

function waitForRep(){
    $.ajax({
        type: "GET",
        url: "server.php",
        cache: false,
        contentType: "application/json; charset=utf-8",
        data: {
        // this way array containing all ID's can be sent:
        CID : CID
    },
        timeout:15000, 
        success: function(data){ 
            addrep("postreply", data);
            setTimeout(
                waitForRep, 
                15000 
            );
        },
        error: function(XMLHttpRequest, textStatus, errorThrown){
            setTimeout(
                waitForRep, 
                15000); 
        }
    });
}

$(document).ready(function(){
    waitForRep();
});

服务器.php

while (true) {
    if($_REQUEST['CID']){  //cid got all dynamic post id as: 1,2,3,4 etc.
      foreach($_REQUEST['CID'] as $key => $value){

        $datetime = date('Y-m-d H:i:s', strtotime('-15 second'));
        $res = mysqli_query($dbh,"SELECT * FROM reply WHERE qazi_id=".$_REQUEST['tutid']."  AND date >= '$datetime' ORDER BY id DESC LIMIT 1") or die(mysqli_error($dbh));
    $data = array();
        while($rows =  mysqli_fetch_assoc($res)){

          $data[]=$rows;

          $data['id'] = $rows['id']; 
          $data['qazi_id'] = $rows['qazi_id'];
          $data['username'] = $rows['username'];
          $data['description'] = $rows['description'];
          $data['date'] = $rows['date'];
          //etc. all
             $id = $rows['id'];
             $qazi_id = $rows['qazi_id'];
             $username = $rows['username'];
             $description = $rows['description'];
             //etc. all
          } //foreach close
      } //foreach close

          if ($description=="") {$detail .= '';}
            else {$detail .=''.$description.'';}
          $data['detail'] = $detail;
          // do somethig

           if (!empty($data)) {
              echo json_encode($data);
              flush();
              exit(0);
           }

    } //request close
    sleep(5);
} //while close

Answer 1

编辑您的 addrep 函数，使其在未定义时不插入:

function addrep(type, msg){
CID.forEach(function(id){
    if(msg.detail != "undefined") {
       $("#newreply"+id).append("+ msg.detail +");
    }
});
}

或者如果未定义则中断它:

function addrep(type, msg){
    CID.forEach(function(id){
        if(msg.detail == "undefined")
           break;
        $("#newreply"+id).append("+ msg.detail +");
    });
}