PHP while循環錯誤

我有一個PHPMySQL在我的腳本如下所示while循環獲取。我得到的結果是類似下面（JSON）：PHP while循環錯誤

[ 
    { 
     "productname":"Veg.Pizaa", 
     "price":"350", 
     "discount":"", 
     "discountprice":"350", 
     "imageURL":"http:\/\/farm8.staticflickr.com\/7154\/6694188161_9ee692d854_s.jpg", 
     "category":"Pizaa", 
     "configurablepone":[{"catname":"Extra","Pro0":"Extra 25g Cheese","Pro1":"Extra 75g Cheese"},"catname":"Nuts","Pro0":"Almonds","Pro1":"Peanuts","Pro2":"Pistachios"}], 
     "configurable":"yes" 
    }, 
    { 
     "productname":"Core i7 Pc", 
     "price":"48000", 
     "discount":"2", 
     "discountprice":"47040", 
     "imageURL":"http:\/\/www.4to40.com\/images\/science\/Basic_Computer_Parts\/Computer.jpg", 
     "category":"Pc", 
     "configurablepone":[{"catname":"Extra","Pro0":"Extra 25g Cheese","Pro1":"Extra 75g Cheese"},{"catname":"Nuts","Pro0":"Almonds","Pro1":"Peanuts","Pro2":"Pistachios"},{"catname":"Harddisk","Pro0":"Segate 500Gb","Pro1":"Samsung 250Gb"},{"catname":"Ram","Pro0":"8Gb Ram","Pro1":"4Gb Ram","Pro2":"2Gb Ram"}], 
     "configurable":"yes" 
    } 
]

但我需要的結果是象下面這樣：

[ 
    { 
     "productname":"Veg.Pizaa", 
     "price":"350", 
     "discount":"", 
     "discountprice":"350", 
     "imageURL":"http:\/\/farm8.staticflickr.com\/7154\/6694188161_9ee692d854_s.jpg", 
     "category":"Pizaa", 
     "configurablepone":[{"catname":"Extra","Pro0":"Extra 25g Cheese","Pro1":"Extra 75g Cheese"},"catname":"Nuts","Pro0":"Almonds","Pro1":"Peanuts","Pro2":"Pistachios"}], 
     "configurable":"yes" 
    }, 
    { 
     "productname":"Core i7 Pc", 
     "price":"48000", 
     "discount":"2", 
     "discountprice":"47040", 
     "imageURL":"http:\/\/www.4to40.com\/images\/science\/Basic_Computer_Parts\/Computer.jpg", 
     "category":"Pc", 
     "configurablepone":[{"catname":"Harddisk","Pro0":"Segate 500Gb","Pro1":"Samsung 250Gb"},{"catname":"Ram","Pro0":"8Gb Ram","Pro1":"4Gb Ram","Pro2":"2Gb Ram"}], 
     "configurable":"yes" 
    } 
]

正如你可以看到configurablepone得到重複JSON爲每個循環，使我得到在第二個產品也是第一個產品的價值，但我需要單獨象下面這樣：

第一製品像下面

"configurablepone":[{"catname":"Extra","Pro0":"Extra 25g Cheese","Pro1":"Extra 75g Cheese"},{"catname":"Nuts","Pro0":"Almonds","Pro1":"Peanuts","Pro2":"Pistachios"}]

其次產品像下面

"configurablepone":[{"catname":"Harddisk","Pro0":"Segate 500Gb","Pro1":"Samsung 250Gb"},{"catname":"Ram","Pro0":"8Gb Ram","Pro1":"4Gb Ram","Pro2":"2Gb Ram"}]

我試圖改變的循環，但我還沒有找到任何解決方案。請幫我解決這個問題。

來源

2012-09-14 Varun Sridharan

的foreach（$ childarr爲$ childarr）：< - 這樣不是會使問題？他們似乎和我有同樣的名字。 – Shadow

@shadow我已經嘗試更改每個名稱，但它導致相同的問題 –

歡迎來到堆棧溢出！請不要使用'mysql_ *'函數來編寫新的代碼。他們不再維護，社區已經開始[棄用程序]（http://goo.gl/KJveJ）。請參閱* [紅盒子]（http://goo.gl/GPmFd）*？相反，您應該瞭解[準備好的語句]（http://goo.gl/vn8zQ）並使用[PDO]（http://php.net/pdo）或[MySQLi]（http://php.net/ mysqli的）。如果你不能決定哪些，[這篇文章]（http://goo.gl/3gqF9）會幫助你。如果你選擇PDO，[這裏是很好的教程]（http://goo.gl/vFWnC）。 –

只需使用unset($subpro1);後array_push($totArr, $prodObj2);。下面是一個例子來源試試這個這可能工作

$result2211 = mysql_query("select * from products where is_config = 'yes' "); 
while($row2211 = mysql_fetch_assoc($result2211)) 
{ 
    $sn = $row2211['sn']; 
    $allparrentselectq = mysql_query("SELECT * FROM parrentpro where parrentsn = $sn"); 
    while($allparrentselect = mysql_fetch_assoc($allparrentselectq)) 
    { 
     $childarr = unserialize($allparrentselect['childsn']); 
     $subpro = '{"catname":"'.$allparrentselect['childname'].'",'; 
     $i = 0; 
     foreach($childarr as $childarr): 
      $subpro .= '"Pro'.$i++.'":"'.$childarr.'",'; 
     endforeach; 
     $subpro1[] = substr($subpro, 0, -1)."}"; 
     $subproa = "[".implode(",",$subpro1)."]"; 
    } 
    $prodObj2 = new ProductDetails(); 
    $prodObj2->productname = $row2211['productname']; 
    $prodObj2->price = $row2211['productprice']; 
    $prodObj2->discount = $row2211['discount']; 
    $prodObj2->discountprice = $row2211['discountprice']; 
    $prodObj2->imageURL = $row2211['productimageurl']; 
    $prodObj2->category = $row2211['productcat']; 
    $prodObj2->configurablepone = $subproa; 
    $prodObj2->configurable = 'yes'; 
    array_push($totArr, $prodObj2); 
    unset($subpro1); 
}

來源

2012-09-14 07:22:40

-1

while($allparrentselect = mysql_fetch_assoc($allparrentselectq)) 
{ 
    $childarr = unserialize($allparrentselect['childsn']); 
    $subpro = '{"catname":"'.$allparrentselect['childname'].'",'; 
    $i = 0; 
    foreach($childarr as $childarr): 
     $subpro .= '"Pro'.$i++.'":"'.$childarr.'",'; 
    endforeach; 
    $subpro1[] = substr($subpro, 0, -1)."}"; 
    $subproa = "[".implode(",",$subpro1)."]"; 
    $subpro1 = array(); 
}

嘗試這種在第二while循環

來源

2012-09-14 05:59:35 EbinPaulose

不能同樣的問題 –

我認爲，在線路$subpro1[] = substr($subpro, 0, -1)."}";您的問題。

因此，該行的第一個呼叫將「Veg.Pizaa」的數據保存到$subpro1[0]。該行的第二個調用將數據從「Core i7 Pc」保存到$subpro1[1]。

然後，行$subproa = "[".implode(",",$subpro1)."]";合併所有數組元素。

來源

2012-09-14 06:04:08 doktorgradus

你能解釋一下嗎？更多Sir –

'$ subpro'是當前項目的子產品列表。在'while'循環中，我們得到比薩的子過程，並將它保存在數組'$ subpro1'的第一個元素中（$ subpro1 [0]包含值：''Pro0「：」額外25g奶酪「，」Pro1「：」Extra 75g Cheese然後我們得到處理器的副產品，並將它保存在數組'subpro1'的第二個元素中（$ subpro1 [1]包含'「Pro0」：「Segate 500Gb」，「Pro1」：「Samsung 250Gb」'。函數'implode'將兩個元素都合併到第二個產品的一個字符串中。因此，在使用新數據填充之前，數組'$ subpro1'必須爲空。通過'$ subpro1 = array（）'或'unset（$ subpro1）'。 – doktorgradus

如何可以在動物中可以向我展示一個例子先生 –

PHP while循環錯誤

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