bu json dan songtitle (9 eleman) almak nasıl - ajax/jquery/javascript

Question

json

örnek:

{"currentlisteners":0,"peaklisteners":0,"maxlisteners":5, 
"uniquelisteners":0,"averagetime":0,"servergenre":"Bollywood", 
"servergenre2":"","servergenre3":"","servergenre4":"","servergenre5":"", 
"serverurl":"http:\/\/www.website.in","servertitle":"RadioName", 
"songtitle":"Udit Narayan - Bheed Mein - www.Songspk.name","dj":"encoder","songurl":"","streamhits":21514,"streamstatus":1,"backupstatus":0,"streamlisted":1,"streamsource":"203.xxx.xxx.xxx","streampath":"\/stream","streamuptime":0,"bitrate":"64","content":"audio\/mpeg","version":"2.4.7.256 (posix(linux x64))"} 

new Request.JSONP({ 
 
    url: 'http://ip:port/stats?sid=1&json=1', 
 
    onSuccess: function(response) { 
 
     show_response(response, $('post')); 
 
    } 
 
}).send(); 
 
show_response = function(obj, result) { 
 
    $H(obj).each(function(v, k) { 
 

 
     new Element('div', { 
 
      html: v 
 
     }).inject(result); 
 
      
 
    }); 
 
    
 

 
};

<div class='wrapper'> 
 
    <div id='post'></div> 
 
</div>

Answer 1

Sadece döndü nesnesini kullanın:

show_response = function(obj, result) { 
     new Element('div', { 
      html: obj.songtitle 
     }).inject(result); 
}; 

Bkz: `v.songtitle` gibi

http://jsfiddle.net/s7trsu9r/2/

Answer 2

Sen erişmek mümkün olmalıdır doğrudan nesne olduğu result["songtitle"];

show_response = function(obj, result) { 
    if (result["songtitle"]) 
     console.log(result["songtitle"]); 
    else 
     console.log("no title availble");    
    }); 

Answer 3

onay tuşu "songtitle"

new Request.JSONP({ 
url: 'http://ip:port/stats?sid=1&json=1', 
onSuccess: function(response) { 
    show_response(response, $('post')); 
} 
}).send(); 
show_response = function(obj, result) { 
$H(obj).each(function(v, k) { 
    if(k == 'songtitle)'{ 
     new Element('div', { 
      html: v 
     }).inject(result); 
    } 

}); 

} eşitse;