獲取PHP中生成的每行JSON數據

這是一個艱難的，至少對我而言。所以基本上我想要做的是讓這個JSON數據的每一行的每一個值，通過Javascript：獲取PHP中生成的每行JSON數據

{"id":2,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":3,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":4,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":5,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":6,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":7,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":8,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":9,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":10,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":31,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":32,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":33,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":34,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":35,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":36,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":37,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":38,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":39,"url":"image.png","x":19,"y":10,"user_id":20} 
{"id":40,"url":"image.png","x":19,"y":10,"user_id":20}

我產生這種方式（在我的PHP文件）：

$db->where('user_id', '20'); 
$results = $db->get('profile_stickers'); 

foreach ($results as $parameters => $values) { 
     echo json_encode($values); 
}

我已經嘗試了很多與jQuery的$.each相關的代碼，但沒有運氣，代碼從未工作。我希望獲得每一個價值，因爲那樣我就會在頁面上顯示它們，具有正確的位置和圖像。有誰知道正確的函數來獲取每一行？就像PHP的$row['foo']一樣。

-----編輯------

我做了這樣的：

$.get('/Application/Ajax/__ajaxProfile.php?a=GetWidgets', function(data) { 

var json = data; 

for (i=0;i<json.length;i++){ 
    var obj = json[i]; 
    for (var key in obj) { 
     document.getElementById("log").innerHTML += key+": "+obj[key]+"<br />"; 
    } 
    document.getElementById("log").innerHTML += "<br />"; 
} 

});

但問題是，它返回的數據是這樣的：

0：「 0：[ 0：{ 0：I 0：d 等。

---- ----- EDIT

現在修好了，它是「$ .get」的東西。

來源

2013-07-31 Denis

這不是有效的JSON。 – MightyPork

當從PHP生成JSON時，您只應該一次調用'json_encode' * *！生成你想要的結構，然後在最後調用'json_encode'。 –

這可能是你想要的。

$db->where('user_id', '20'); 
$results = $db->get('profile_stickers'); 

$arr = array(); 

foreach ($results as $parameters => $values) { 

    $arr[] = $values; 

} 

echo json_encode($arr);

來源

2013-07-31 18:37:47 MightyPork

這可能不是你正在尋找的東西，但是這是一種通過JavaScript獲取所有內容的方式。

你基本上需要存儲作爲JSON對象的數組：

var json = [ 
{"id":2,"url":"image.png","x":19,"y":10,"user_id":20}, 
{"id":3,"url":"image.png","x":19,"y":10,"user_id":20}, 
...etc 
]

然後，您可以通過數組和對象這樣的循環：

for (i=0;i<json.length;i++){ 
    var obj = json[i]; 
    for (var key in obj) { 
    document.write(key+": "+obj[key]); 
    } 
}

http://jsfiddle.net/Y4uAx/

來源

2013-07-31 18:43:09

獲取PHP中生成的每行JSON數據

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