Question

我从一个页面发送了许多动态post id，并且php服务器端页面（server.php）使用这些id进行查询，以在mysql中找出新添加的数据。

如果在mysql中未找到任何新添加的数据，则返回undefined值。所以我添加了这个if (msg.id !== undefined && msg.detail !== undefined && msg.name !== undefined) { //do here }来隐藏undefined。

但是在添加上面一行之后，我的脚本隐藏了undefined值，但仅返回第一个CID的新增值。

这意味着如果CID向php发送了id（100,101,102,103等），它只返回100个id的新增值并附加它。

请问这里有什么问题？

N.B。如果没有上面一行，它会很好地返回所有CID的值，但如果没有找到新数据，也会返回undefined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){
if (msg.id !== undefined && msg.detail !== undefined && msg.name !== undefined) {
    $("#newreply"+id).append("<div class='"+ type +""+ msg.id +"'><ul><div class='newkochi'>"+ msg.name +"</div><div class='cdomment_text'>"+ msg.detail +"</ul></div>");
}
});
}

function waitForRep(){
    $.ajax({
        type: "GET",
        url: "server.php",
        cache: false,
        data: {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();
});

server.php

while (true) {
    if($_GET['CID']){  //cid got all dynamic post id as: 1,2,3,4 etc.
      foreach($_GET['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=".$_GET['tutid']."  AND date >= '$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
          } //while close
      } //foreach close

          $name .='<p class="name">'.$username.' Says:</p>';
          $detail .=''.$description.'';

          $data['name'] = $name;
          $data['detail'] = $detail;
          // do others something more like as above

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

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

Answer 1

javascript中的发现问题：

修复1：

var CID = [$('div[data-post-id]').length]; 

$('div[data-post-id]').each(function(i){
CID[i] = $(this).data('post-id');
});

修复2：

var CID = []; 
$('div[data-post-id]').each(function(i){
CID.push($(this).data('post-id'));
});

插入问题：

function addrep(type, msg){
CID.forEach(function(id){
//You are inserting the comment in all html tags havin id ='newreplay' + id
//Here you need to correct you logic like: if (msg.id == id) //do something

if (msg.id !== undefined && msg.detail !== undefined && msg.name !== undefined) {
    $("#newreply"+id).append("<div class='"+ type +""+ msg.id +"'><ul><div class='newkochi'>"+ msg.name +"</div><div class='cdomment_text'>"+ msg.detail +"</ul></div>");
}
});
}

隐藏未定义值后，JS仅从数组返回第一个id的值

1 个答案: