PHP while循环错误

我有一个PHPMySQL在我的脚本如下所示while循环获取。我得到的结果是类似下面（JSON）：PHP while循环错误

[ 
    { 
     "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" 
    } 
]

但我需要的结果是象下面这样：

[ 
    { 
     "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得到重复JSON为每个循环，使我得到在第二个产品也是第一个产品的价值，但我需要单独象下面这样：

第一制品像下面

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

其次产品像下面

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

我试图改变的循环，但我还没有找到任何解决方案。请帮我解决这个问题。

来源

2012-09-14 Varun Sridharan

的foreach（$ childarr为$ childarr）：< - 这样不是会使问题？他们似乎和我有同样的名字。 – Shadow

@shadow我已经尝试更改每个名称，但它导致相同的问题 –

欢迎来到堆栈溢出！请不要使用'mysql_ *'函数来编写新的代码。他们不再维护，社区已经开始[弃用程序]（http://goo.gl/KJveJ）。请参阅* [红盒子]（http://goo.gl/GPmFd）*？相反，您应该了解[准备好的语句]（http://goo.gl/vn8zQ）并使用[PDO]（http://php.net/pdo）或[MySQLi]（http://php.net/ mysqli的）。如果你不能决定哪些，[这篇文章]（http://goo.gl/3gqF9）会帮助你。如果你选择PDO，[这里是很好的教程]（http://goo.gl/vFWnC）。 –

只需使用unset($subpro1);后array_push($totArr, $prodObj2);。下面是一个例子来源试试这个这可能工作

$result2211 = mysql_query("select * from products where is_config = 'yes' "); 
while($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); 
}

来源

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(); 
}

尝试这种在第二while循环

来源

2012-09-14 05:59:35 EbinPaulose

不能同样的问题 –

我认为，在线路$subpro1[] = substr($subpro, 0, -1)."}";您的问题。

因此，该行的第一个呼叫将“Veg.Pizaa”的数据保存到$subpro1[0]。该行的第二个调用将数据从“Core i7 Pc”保存到$subpro1[1]。

然后，行$subproa = "[".implode(",",$subpro1)."]";合并所有数组元素。

来源

2012-09-14 06:04:08 doktorgradus

你能解释一下吗？更多Sir –

'$ subpro'是当前项目的子产品列表。在'while'循环中，我们得到比萨的子过程，并将它保存在数组'$ subpro1'的第一个元素中（$ subpro1 [0]包含值：''Pro0“：”额外25g奶酪“，”Pro1“：”Extra 75g Cheese然后我们得到处理器的副产品，并将它保存在数组'subpro1'的第二个元素中（$ subpro1 [1]包含'“Pro0”：“Segate 500Gb”，“Pro1”：“Samsung 250Gb”'。函数'implode'将两个元素都合并到第二个产品的一个字符串中。因此，在使用新数据填充之前，数组'$ subpro1'必须为空。通过'$ subpro1 = array（）'或'unset（$ subpro1）'。 – doktorgradus

如何可以在动物中可以向我展示一个例子先生 –

PHP while循环错误

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