gpt4 book ai didi

php - 合并两个表并将它们的数据显示在一个数组中

转载 作者:可可西里 更新时间:2023-11-01 06:30:52 24 4
gpt4 key购买 nike

我有2个表:product和cart,我想把这2个表合并起来,按照特定的条件在数组中显示数据,如下:

All the products under a specific category should get displayed and if a particular user have purchased any product among the given products then its details should also get displayed in front of that product

到目前为止我完成的代码是

$catid = $_REQUEST['catid'];
$userid = $_REQUEST['userid'];

$sql = "select * from productsize where catid = '".$catid."' GROUP BY productid";
$result = mysqli_query($con, $sql);
if (mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_assoc($result))
{
$rows['catid'] = $row['catid'];
$rows['catname'] = $row['catname'];
$rows['productid'] = $row['productid'];
$rows['prodname'] = $row['prodname'];
$rows['prodimg'] = $row['prodimg'];
$row2[]=$rows;
}
}

echo "<pre>";
print_r($row2);
echo "</pre>";

它给出了这样一个数组

 Array
(
[0] => Array
(
[catid] => 2
[catname] => C1
[productid] => 13
[prodname] => P1
[prodimg] =>
)

[1] => Array
(
[catid] => 2
[catname] => C1
[productid] => 14
[prodname] => P1
[prodimg] =>
)

[2] => Array
(
[catid] => 2
[catname] => C1
[productid] => 15
[prodname] => P3
[prodimg] =>
)

)

但是我想要代替上述数组的最终数组是

Array
(
[0] => Array
(
[catid] => 2
[catname] => C1
[productid] => 13
[prodname] => P1
[prodimg] =>
[size] => Array
(
[0] => small
[1] => medium
[2] => large
[3] => perpiece
)
[cost] => Array
(
[0] => 10
[1] => 20
[2] => 30
[3] => 12
)
[purchasedsize] => Array
(
[0] => small
[1] => 0
[2] => large
[3] => 0
)
[purchasedquantity] => Array
(
[0] => 2
[1] => 0
[2] => 1
[3] => 0
)
[userid] => 1
)

[1] => Array
(
[catid] => 2
[catname] => C1
[productid] => 14
[prodname] => P1
[prodimg] =>
[size] => Array
(
[0] => small
[1] => medium
[2] => large
[3] => 0
)
[cost] => Array
(
[0] => 15
[1] => 20
[2] => 25
[3] => 0
)
[purchasedsize] => Array
(
[0] => 0
[1] => medium
[2] => 0
[3] => 0
)
[purchasedquantity] => Array
(
[0] => 0
[1] => 1
[2] => 0
[3] => 0
)
[userid] => 1
)

[2] => Array
(
[catid] => 2
[catname] => C1
[productid] => 15
[prodname] => P3
[prodimg] =>
[size] => Array
(
[0] => 0
[1] => medium
[2] => 0
[3] => perpiece
)
[cost] => Array
(
[0] => 0
[1] => 20
[2] => 0
[3] => 18
)
[purchasedsize] => Array
(
[0] => 0
[1] => 0
[2] => 0
[3] => 0
)
[purchasedquantity] => Array
(
[0] => 0
[1] => 0
[2] => 0
[3] => 0
)
[userid] => 0
)

)

产品 TableView (如您所见,产品表带有一个产品 ID,每个产品 ID 下最多有 4 个尺寸(不会超过 4 个))

id  catid  catname  productid  prodsize   cost  prodname  prodimg
1 2 C1 13 small 10 P1
2 2 C1 13 medium 20 P1
3 2 C1 13 large 30 P1
4 2 C1 13 perpiece 12 P1
5 2 C1 14 small 15 P2
6 2 C1 14 medium 20 P2
7 2 C1 14 large 25 P2
8 2 C1 15 perpiece 18 P3
9 2 C1 15 medium 20 P3

购物车 TableView

id   catid  catname  userid  productid  prodname  prodsize   quantity  prodcost
1 2 C1 1 13 P1 large 1 30
2 2 C1 1 13 P1 small 2 10
3 2 C1 1 14 P2 medium 1 20

谁能帮我得到所需的数组作为结果?

最佳答案

试试这个

 $catid = $_REQUEST['catid'];
$userid = $_REQUEST['userid'];

$sql= "SELECT p.catid, p.catname, p.productid, p.prodimg,
GROUP_CONCAT(p.prodsize ORDER BY p.id ASC) as size,
GROUP_CONCAT(p.cost ORDER BY p.id ASC) as cost, p.prodname,
GROUP_CONCAT(c.prodsize,'-',c.quantity) as cart_details, GROUP_CONCAT(DISTINCT(c.userid)) as user_id
FROM products p
LEFT JOIN cart c ON(c.productid = p.productid AND c.userid = '$userid' AND p.prodsize = c.prodsize)
WHERE p.catid ='$catid'
GROUP BY p.productid
ORDER BY user_id DESC, p.productid ASC";


$result = mysql_query($sql);


if (mysql_num_rows($result) > 0)
{
$i = 0;
while($row = mysql_fetch_assoc($result))
{
$rows[$i]['catid'] = $row['catid'];
$rows[$i]['catname'] = $row['catname'];
$rows[$i]['productid'] = $row['productid'];
$rows[$i]['prodname'] = $row['prodname'];
$rows[$i]['prodimg'] = $row['prodimg'];
$final_size = array_fill(0, 4, '0');
$final_cost = array_fill(0, 4, '0');

$size = explode(',', $row['size']);
$cost = explode(',', $row['cost']);
foreach($size as $k=>$sizecol) {
switch($sizecol) {
case 'small':
$array_key = '0';
break;
case 'medium':
$array_key = '1';
break;
case 'large':
$array_key = '2';
break;
case 'perpiece':
$array_key = '3';
break;
}
$final_size[$array_key] = $sizecol;
$final_cost[$array_key] = $cost[$k];

}


$cart_details = explode(',', $row['cart_details']);
$purchasedsize = array_fill(0, 4, '0'); //Since you displayed this array has 4 values only
$purchasedquantity = array_fill(0, 4, '0');
foreach($cart_details as $cart) {
if($cart != '') {
$details = explode('-', $cart);
$key = array_search($details[0], $size);

$purchasedsize[$key] = $details[0];
$purchasedquantity[$key] = $details[1];
}

}

$rows[$i]['size'] = $final_size;
$rows[$i]['cost'] = $final_cost;
$rows[$i]['purchasedsize'] = $purchasedsize;
$rows[$i]['purchasedquantity'] = $purchasedquantity;
$rows[$i]['userid'] = $row['user_id'];

$i++;
}
}

echo "<pre>";
print_r($rows);
echo "</pre>";

输出数组

Array
(
[0] => Array
(
[catid] => 2
[catname] => c1
[productid] => 13
[prodname] => P1
[prodimg] =>
[size] => Array
(
[0] => small
[1] => medium
[2] => large
[3] => perpiece
)

[cost] => Array
(
[0] => 10
[1] => 20
[2] => 30
[3] => 12
)

[purchasedsize] => Array
(
[0] => small
[1] => 0
[2] => large
[3] => 0
)

[purchasedquantity] => Array
(
[0] => 2
[1] => 0
[2] => 1
[3] => 0
)

[userid] => 1
)

[1] => Array
(
[catid] => 2
[catname] => c1
[productid] => 14
[prodname] => P2
[prodimg] =>
[size] => Array
(
[0] => small
[1] => medium
[2] => large
[3] => 0
)

[cost] => Array
(
[0] => 15
[1] => 20
[2] => 25
[3] => 0
)

[purchasedsize] => Array
(
[0] => 0
[1] => medium
[2] => 0
[3] => 0
)

[purchasedquantity] => Array
(
[0] => 0
[1] => 1
[2] => 0
[3] => 0
)

[userid] => 1
)

[2] => Array
(
[catid] => 2
[catname] => C1
[productid] => 15
[prodname] => P3
[prodimg] =>
[size] => Array
(
[0] => 0
[1] => medium
[2] => 0
[3] => perpiece
)

[cost] => Array
(
[0] => 0
[1] => 20
[2] => 0
[3] => 18
)

[purchasedsize] => Array
(
[0] => 0
[1] => 0
[2] => 0
[3] => 0
)

[purchasedquantity] => Array
(
[0] => 0
[1] => 0
[2] => 0
[3] => 0
)

[userid] =>
)

)

关于php - 合并两个表并将它们的数据显示在一个数组中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32758649/

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com