php while loop error

Question

I have a PHP MySQL fetch while loop as shown below in my script:

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

In that I have a problem. I get the result as like below (JSON):

[
    {
        "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"
    }
]

But I need the result as like below:

[
    {
        "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"
    }
]

As you can see configurablepone JSON is getting repeated for every loop so I am getting the value of the first product in second product also but I need to seperate like below:

First Product As Like Below

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

Second Product As Like Below

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

I have tried changing the loop but I haven't found any solutions. Kindly help me to solve this.

foreach($childarr as $childarr): <- wouldn't that cause problems? They seem to have the same name to me. — shadow, Sep 14 '12 at 6:00
@shadow i have tryed changing the name in for each but it causes the same problem — S.Varun, Sep 14 '12 at 6:05
Welcome to Stack Overflow! Please, don't use mysql_* functions to write new code. They are no longer maintained and the community has begun deprecation process. See the red box? Instead you should learn about prepared statements and use either PDO or MySQLi. If you can't decide which, this article will help you. If you pick PDO, here is good tutorial. — Madara Uchiha, Sep 14 '12 at 6:05
@Truth What Are You Saying Sir As I Dont Know About Mysqli I Wrote With Mysql... — S.Varun, Sep 14 '12 at 6:07
@vaahost: Like I said, mysql functions in PHP are old, and outdated. Please read the links in my comment. — Madara Uchiha, Sep 14 '12 at 6:09

doktorgradus · Answer 1 · 2012-09-14 06:04:08Z

up vote 2 down vote

I think, your problem in line $subpro1[] = substr($subpro, 0, -1)."}";.

So, first call of this line save data from "Veg.Pizaa" into $subpro1[0]. Second call of this line save data from "Core i7 Pc" into $subpro1[1].

Then, line $subproa = "[".implode(",",$subpro1)."]"; merged all array elements.

answered Sep 14 '12 at 6:04

doktorgradus
5361411

Can You Explain Me More Sir – S.Varun Sep 14 '12 at 6:06

1

$subpro is a list of subproducts for current item. In while loop we get subpro's for pizza and save it in first element of array $subpro1 ( $subpro1[0] contains value: "Pro0":"Extra 25g Cheese","Pro1":"Extra 75g Cheese". After that, we get subproducts for processor and save it in second element of array $subpro1 ($subpro1[1] contains "Pro0":"Segate 500Gb","Pro1":"Samsung 250Gb". Function implode merged both elements to one string for second product. So, array $subpro1 must be nulled before fill it with new data. By $subpro1 = array() or unset($subpro1). – doktorgradus Sep 14 '12 at 7:02

how it can be in dynimic can u show me an example sir – S.Varun Sep 14 '12 at 7:05

add a comment |

S.Varun · Accepted Answer · 2012-09-14 07:22:40Z

Just Use unset($subpro1); after the array_push($totArr, $prodObj2);. Below Is An example Source Try This One This May Work

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

harshit · Answer 3 · 2012-09-14 06:03:25Z

up vote -1 down vote

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();
}

try this in second while loop

edited Sep 14 '12 at 6:03

harshit
1,256933

answered Sep 14 '12 at 5:59

EbinPaulose
4131419

Not Working The Same Problem – S.Varun Sep 14 '12 at 6:04

add a comment |

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