在jquery中獲取parsererror

我試圖執行以下操作：在jquery中獲取parsererror

從一個html頁面，按一個按鈕將調用一個php腳本，查詢數據庫和回聲json。 Php頁面可以在http://vscreazioni.altervista.org/prova.php找到，工作正常。什麼不工作是jQuery的一面，因爲我得到parsererror作爲迴應。這裏是我的代碼：

<!DOCTYPE html> 
<html> 
<head> 
<meta charset="UTF-8" /> 
<meta name="viewport" content="initial-scale=1, maximum-scale=1" /> 
<style type="text/css"> 

</style> 
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8/jquery.min.js"></script> 
<script type="text/javascript"> 
$(document).ready(function() { 
$('#button_1').click(function(e){ 
    e.preventDefault(); 
    e.stopPropagation(); 
    favfunct(); 
}); 
}); 

function favfunct() { 
$.ajax({ 
      type: 'GET', 
      url: 'prova.php', 
      dataType: 'json', 
      success: function (json) { 
      alert("SUCCESS!!!"); 

      }, 
      error: function (xhr, status) { 
     alert(status); 
    }, 
     }); 
} 
</script> 
</head> 
<body> 
<input id="button_1" type="button" value="push" /> 
</body> 
</html>

我完全新的這個東西...任何幫助，將不勝感激

編輯：從prova.php PHP代碼

<?php 

$conn = mysql_connect("localhost", 「username」, 「passwd」); 

if (!$conn) 
{ 
mysql_close($conn); 
die("Problemi nello stabilire la connessione"); 
} 

if (!mysql_select_db("my_vscreazioni")) 
{ 
mysql_close($conn); 
die("Errore di accesso al data base utenti"); 
} 

$queryIcostanza = "SELECT SUM(iCostanza) FROM apps"; 
$resultIcostanza = mysql_query($queryIcostanza) or die(mysql_error()); 
$rowIcostanza = mysql_fetch_array($resultIcostanza); 

$queryIversi = "SELECT SUM(iVersi) FROM apps"; 
$resultIversi = mysql_query($queryIversi) or die(mysql_error()); 
$rowIversi = mysql_fetch_array($resultIversi); 

$queryI10numeri = "SELECT SUM(i10Numeri) FROM apps"; 
$resultI10numeri = mysql_query($queryI10numeri) or die(mysql_error()); 
$rowI10numeri = mysql_fetch_array($resultI10numeri); 

$queryIcostanza4x = "SELECT SUM(iCostanza4x) FROM apps"; 
$resultIcostanza4x = mysql_query($queryIcostanza4x) or die(mysql_error()); 
$rowIcostanza4x = mysql_fetch_array($resultIcostanza4x); 

$queryOndanews = "SELECT SUM(OndaNews) FROM apps"; 
$resultOndanews = mysql_query($queryOndanews) or die(mysql_error()); 
$rowOndanews = mysql_fetch_array($resultOndanews); 

$queryFarmachimica = "SELECT SUM(FarmaChimica) FROM apps"; 
$resultFarmachimica = mysql_query($queryFarmachimica) or die(mysql_error()); 
$rowFarmachimica = mysql_fetch_array($resultFarmachimica); 

$queryIcarrano = "SELECT SUM(iCarrano) FROM apps"; 
$resultIcarrano = mysql_query($queryIcarrano) or die(mysql_error()); 
$rowIcarrano = mysql_fetch_array($resultIcarrano); 

$totale = 0; 
$totaleIcostanza = $rowIcostanza['SUM(iCostanza)']; 
$totaleIversi = $rowIversi['SUM(iVersi)']; 
$totaleI10numeri = $rowI10numeri['SUM(i10Numeri)']; 
$totaleIcostanza4x = $rowIcostanza4x['SUM(iCostanza4x)']; 
$totaleOndanews = $rowOndanews['SUM(OndaNews)']; 
$totaleFarmachimica = $rowFarmachimica['SUM(FarmaChimica)']; 
$totaleIcarrano = $rowIcarrano['SUM(iCarrano)']; 

$totale = $totaleIcostanza + $totaleIversi + $totaleI10numeri + $totaleIcostanza4x +  $totaleOndanews + $totaleFarmachimica + $totaleIcarrano; 

$comando = "select * from apps"; 

$result = mysql_query($comando) or die(mysql_error()); 

$ultima_data=""; 

while ($dati = mysql_fetch_assoc($result)) 
{ 
    $ultima_data = $dati['data']; 
} 

$response = array(); 

$posts = array('icostanza'=> $totaleIcostanza, 'iversi'=> $totaleIversi, 'i10numeri'=> $totaleI10numeri, 'icostanza4x'=> $totaleIcostanza4x, 'ondanews'=>$totaleOndanews, 'farmachimica'=> $totaleFarmachimica, 'icarrano'=> $totaleIcarrano, 'totale'=>$totale, 'ultimo'=>$ultima_data); 

$response['posts'] = $posts; 

$json = json_encode($response); 
echo $json; 

mysql_close($conn); 
?>

編輯2 ：我有一個拼寫錯誤的問題。現在我成功了！如在

success: function (json) { 
      alert("SUCCESS!!!");   
}

如何提醒json內容？我試着用

alert(json);

，但我得到的翻譯：

來源

2014-02-15 Enzoses

請張貼您的PHP代碼。 – Felix

發表了，請看看 – Enzoses

嘗試用'console.log（json）'查看javascript控制檯中的對象 – anurupr

在成功塊警報不喜歡下獲得的職位。

success: function (json) { 
      alert(json.posts.icostanza); 

      },

這將提醒「icostanza」值。

來源

2014-02-15 11:05:17

這樣你就可以獲得所有的帖子數據。例如alert（json.posts.iversi）; –

優秀。如果我想顯示所有值，該怎麼辦？ – Enzoses

您必須按照我所說的收集所有的值json.posts.iversi，json.posts.icostanza4x，json.posts.ondanews等 –

在jquery中獲取parsererror

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