Question

我是AJAX的新手，我不知道如何获取阵列，我希望通过AJAX实时自动更新通知栏。如果我只是从气球更新数字，但我已经完成了工作，但我也无法通过PHP / AJAX从MySQL获取消息

$sql=mysql_query("select * from $tableName where is_read=0");                   
$comment_count=mysql_num_rows($sql);
$comm_array = array('count' => $comment_count);

//first output need for get number of notications
echo json_encode($comm_array);

$listsql=mysql_query("select * from notification order by execute_at desc");
while($rowsmall=mysql_fetch_array($listsql))
{ 
   $n_id=$rowsmall['notification_id'];
   $date=date("Y-m-d H:i:s", strtotime($rowsmall["execute_at"]));
   $command=$rowsmall['command'];
   $server=$rowsmall['hostname'];
   $status=$rowsmall['status'];
   $variable[] = array( 'data' => "$date", 
                    'command' => "$command",
                     'server' => "$server",
                     'status' => "$status" );
   if($rowsmall['is_read'] == 0)
   {
      //this was just a try for get with different colors the unread notification
      //second output - get notification from mysql
      echo json_encode($variable);
   }
   else 
   {
      //this was just a try for get with different colors the read notification
      //second output - get notification from mysql
      echo json_encode($variable);
   }
}

这是AJAX，如果我只得到计数，但是我不知道如何使用AJAX这个PHP循环以及如何获得两个输出？

    <script>
$(document).ready(function() {
    setInterval("ajaxcall()",2000);
});

function ajaxcall() {
//-----------------------------------------------------------------------
// 2) Send a http request with AJAX http://api.jquery.com/jQuery.ajax/
//-----------------------------------------------------------------------
    $.ajax({     
    type: "GET",
    url: '/notifications/api.php',                  //the script to call to get data          
    success: function(response){          //on recieve of reply
        json_object = JSON.parse(response)
        var count = json_object.count;
        var date = json_object.notifications.data;
        var server = json_object.notifications.server;
        var command = json_object.notifications.command;
        var status = json_object.notifications.status;
        //--------------------------------------------------------------------
        // 3) Update html content
        //--------------------------------------------------------------------
        $('#mes').html(count); //Set output element html
        $('#not_read').html("Date: "+date+" Host: "+server+" Command: "+command+" Status: "+status);
        //recommend reading up on jquery selectors they are awesome 
        // http://api.jquery.com/category/selectors/
    } 
});
}
</script>

这就是我想要做的。红色是警报的数量，但我不知道如何使两者都工作 IMAGE HOW LOOK IN WEB

Answer 1

出于安全考虑，请使用mysqli。

你的问题在于你需要只放置1个json响应，这里至少提供2个。然后，只需1个查询和一个简单的循环就可以简化所有这些。像（未经测试）的东西：

    $objMysql = new mysqli('localhost', 'my_user', 'my_password', 'my_db');

    $r = array();
    $comment_count = 0;
    if(!$objMysql->connect_error){
      $strQuery = "select a.num as comment_count, b.* from
      (select count(*) as num from {$tableName} where is_read=0) as a,
      (select * from notification order by execute_at desc) as b";

      $res = $objMysql->query($strQuery);

      while($row = mysql_fetch_assoc($res)){
        $comment_count = $row['comment_count'];

        $r[] = array(
          'data' => date("Y-m-d H:i:s", strtotime($rowsmall["execute_at"])),
          'command' => $row['command'],
          'server' => $row['hostname'],
          'status' => $row['status'],
          'color' => ($row['is_read'] == 0)? 'red' : 'blue'
        );
      }

    }else{
      $r[] = 'mysql connect error: '.$objMysql->connect_error;
    }
    echo json_encode(array('comment_count' => $comment_count, 'results' => $r));

之后，您的所有值都将出现在javascript / jquery的response变量中。

AJAX从多个数组中获取PHP输出

1 个答案: