使用Jquery将php字符串数据转换为Json对象

Question

我需要从PHP获取一些Json数据到jquery。

我需要JavaScript方法中的以下格式。

function returnJson() 
{ 
    return { 
      events: [ 
      { 
       "id": 1, 
       "start": new Date(2013, 4, 26, 12), 
       "end": new Date(2013, 4, 26, 13, 30), 
       "title": "Lunch with Mike" 
      }, 
      { 
       "id": 2, 
       "start": new Date(2013, 4, 27, 14), 
       "end": new Date(2013,4, 27, 14, 45), 
       "title": "Dev Meeting" 
      }] 
     }; 
} 

对于这个我做下面的事情在javascript：

在PHP

function returnJson() 
{ 
    var eventResult = $.getJSON("../PHP/PhpAction.php?f=fetchCalendarEvent"); 
    return eventResult; 
} 

：

function fetchCalendarEvent() 
      { 
       $tablename = "tb_calendar"; 

       $sql = "SELECT eventId,userId,enentName,eventText,EXTRACT(YEAR FROM startTime) AS startyear,EXTRACT(MONTH FROM startTime) AS startMonth,EXTRACT(DAY FROM startTime) AS startDay,EXTRACT(HOUR FROM startTime) AS startHour,EXTRACT(MINUTE FROM startTime) AS startMin,EXTRACT(YEAR FROM endTime) AS endyear,EXTRACT(MONTH FROM endTime) AS endMonth,EXTRACT(DAY FROM endTime) AS endDay,EXTRACT(HOUR FROM endTime) AS e`enter code here`ndHour,EXTRACT(MINUTE FROM endTime) AS endMin FROM ".$tablename." WHERE userId='".$_SESSION['userid']."' AND isActive=1"; 

         $q = mysql_query($sql); 
         $i=1;   
         $eventData="{events: ["; 

         if (!mysql_num_rows($q)) { 
          echo 'No records found'; 
         } 
         else 
         {           
          while ($row = mysql_fetch_assoc($q)) { 

           $eventData.="{'id':".$row['eventId'].","; 


           $eventData.="'end': new Date(".$row['startyear'].",".$row['startMonth'].",".$row['startDay']."," .$row['startHour'].",".$row['startMin']."),"; 

           $eventData.="'start': new Date(".$row['endyear'].",".$row['endMonth'].",".$row['endDay'].",".$row['endHour'].",".$row['endMin']."),"; 

           $eventData.="'title':'".$row['enentName']."'},"; 

           $i++; 
          }     
         } 
         $eventData= rtrim($eventData, ","); 

         $eventData.="]}"; 
         echo json_decode($eventData); 
      } 

我检查的萤火php的方法返回的数据，如：

{ events: [ { 
     'id': 2, 
     'end': new Date(2013, 4, 27, 18, 38), 
     'start': new Date(2013, 4, 27, 18, 38), 
     'title': 'test' 
    }, { 
     'id': 3, 
     'end': new Date(2013, 4, 23, 11, 0), 
     'start': new Date(2013, 4, 23, 14, 15), 
     'title': 'testing23' 
    } 
] } 

任何人都可以请 帮我。我是新的PHP。任何帮助将不胜感激。

Answer 1

Php有一个称为json_encode的函数，它将处理json输出的生成。

下面是一个例子：

$o = array(); 
$o['events'] = array(); 
while ($row = mysql_fetch_assoc($q)) 
{ 
    $event = array(); 
    $event['id'] = $row['id']; 
    $event['start'] = "new Date(".$row['startyear'].",".$row['startMonth'].",".$row['startDay']."," .$row['startHour'].",".$row['startMin'].")"; 
    $event['end'] = "new Date(".$row['endyear'].",".$row['endMonth'].",".$row['endDay'].",".$row['endHour'].",".$row['endMin'].")"; 
    $event['title'] = $row['title']; 

    $o['events'][] = $event; 
} 

return json_encode($o); 

而且javscript代码可能是这样的：

function returnJson() 
{ 
    var events = $.getJSON("../PHP/PhpAction.php?f=fetchCalendarEvent"); 

    for (var i = 0; i < events.length; i++) { 
     var event = events[i]; 
     event.start = eval(event.start); 
     event.end = eval(event.end); 

     events[i] = event; 
    } 

    return events; 
} 

Answer 2

发送AJAX数据到浏览器时，应添加适当的标题和MIME类型：

header('Cache-Control: no-cache, must-revalidate'); 
header('Expires: Mon, 26 Jul 1997 05:00:00 GMT'); 
header('Content-type: application/json'); 

下面是关于如何使用发送给浏览器的数据的一些示例代码：

function requestData() { 
     $.ajax({ 
      url: url, 
      dataType: 'json', 
      success: function(json) { 
       events = json.events; 

       useData(events); 
      }, 
      error: function (xhr, status, error) { 
       alert('Status: ' + status +' Error: ' + error); 
      } 
     }); 
    } 

错误信息在你的案例中说了什么？

Answer 3

你可以在JSON格式不日期对象。一个选项是使用时间戳并在解析json时将其转换。

如果你检查http://jsonlint.com/你会看到你的json给出错误。

您的语法是Javascript对象表示法，不是JSON表示法。 Javascript可以使用json.stringify函数生成json。

我只是双重检查，如果你的字符串化JavaScript对象下面的转换将发生：

var date = new Date(2013, 4, 26, 12); 
var string = JSON.stringify(date); 
console.log(string); 
//output = 2013-05-26T10:00:00.000Z 
console.log(JSON.parse(string)); 
//will again output 2013-05-26T10:00:00.000Z 

所以，你可以约会对象转换为字符串JSON.stringify。但是你不能将它们格式化回Date对象。

Answer 4

$data = preg_replace('@new Date\(([^\)]*)\)@', '"$1"', $data); 
$data = json_decode($data, true); 

结果：

新日期（2008,3,30,0,31,26）=> “2008,3,30,0,31,26”