如果在MySQL中ID相同，则合并两行

Question

如果产品ID相同，如何将产品的数量合并为一行？这是我客户的购物车。当我向购物车添加产品时，我希望它组合成一行而不是单独的行。

编辑：我有两个表，其中包含一个单独的数量字段（tblproducts，tblcart）。

EDIT2：PHP给我一个错误

  

mysqli_fetch_array（）期望参数1为mysqli_result boolean   在我的第61行给出

              <?php 
            $query = "SELECT * FROM products";
            $exec = mysqli_query($connection, $query);
            $a = 1;
            $b = 1;
            while ($row = mysqli_fetch_array($exec)) {

但我的代码工作正常。有谁能找到问题？

我客户购物车的代码

        <table class="table table-hover">
      <thead>
        <tr>
          <th>Product ID</th>
          <th>Product Name</th>
          <th>Description</th>
          <th>Quantity</th>
          <th>Price per Unit</th>
          <th>Total Amount</th>
          <th>Remove</th>
        </tr>
      </thead>
      <tbody>


      <?php 

        $selectCart = "SELECT * FROM cart INNER JOIN products ON products.product_id = cart.product_id";
        $execSelectCart = mysqli_query($connection, $selectCart);

        while ($row = mysqli_fetch_array($execSelectCart)) {

          $cartProId = $row['product_id'];
          $cartProName = $row['product_name'];
          $cartProDesc = $row['description'];
          $cartSellPrice = $row['sell_price'];
          $cartQty = $row['quantityCart'];

          $compute = $cartSellPrice * $cartQty;
          $totalAmount = number_format((float)$compute, 2, '.', '');
       ?>

        <tr>
          <td><?php echo $cartProId; ?></td>
          <td><?php echo $cartProName; ?></td>
          <td><?php echo $cartProDesc; ?></td>
          <td><?php echo $cartQty; ?></td>
          <td><?php echo $cartSellPrice; ?></td>
          <td><?php echo $totalAmount ?></td>
          <td class="text-center">
            <div class="btn-group">
              <a href="edit_brand.php?brand_id=<?php echo $brand_id; ?>"  class="btn btn-xs btn-info" data-toggle="tooltip" title="Edit">
                <span class="glyphicon glyphicon-edit"></span>
              </a>
              <a href="manage_brands.php?delete=<?php echo $brand_id; ?>"  class="btn btn-xs btn-danger" data-toggle="tooltip" title="Remove">
                <span class="glyphicon glyphicon-trash"></span>
              </a>
            </div>
          </td>
        </tr>

      <?php } ?>
      </tbody>
    </table>

添加产品的代码

<?php
   if (isset($_POST['addCart']) && $_POST['addCart']=="Add Items to Cart") {
     foreach($_POST['qtyBuy'] as $index=>$value){
       if($value > 0){
        $cartID = $_POST['product_id'][$index];
           $addQuery = "INSERT INTO cart (product_id, quantityCart) VALUES (".$_POST['product_id'][$index].", ".$value.");";
           $addQuery .= "UPDATE products SET quantity = quantity - $value WHERE product_id = $cartID;";
           $execQuery = mysqli_multi_query($connection, $addQuery);
        }  
     }
     header('Refresh: 0; url=add_sales.php');
  }
?>

<div class="row">
  <div class="col-md-12"> <!-- Product List Info Start -->
    <div class="panel panel-default">
      <div class="panel-heading">
        <strong>
          <span class="glyphicon glyphicon-th"></span>
          <span>Select Products</span>
       </strong>
      </div>

      <div class="panel-body">
        <div class="form-group pull-left">
           <input type="text" class="search form-control" placeholder="Search products">
         </div>
         <span class="counter pull-left"></span>
          <div class="span3">

              <form action="add_sales.php" method="POST">
            <table class="table table-striped table-bordered table-hover results table-fixed">
              <thead>
                <tr>
                    <th class="text-center">#</th>
                    <th>Product Name</th>
                    <th>Description</th>
                    <th>Price</th>
                    <th>In Stock</th>
                    <th style="width: 20%">Quantity</th>
                </tr>
                <tr class="warning no-result">
                  <td colspan="8"><i class="fa fa-warning"></i> No Product Found</td>
                </tr>
              </thead>


              <tbody>
              <?php 
                $query = "SELECT * FROM products";
                $exec = mysqli_query($connection, $query);
                $a = 1;
                $b = 1;
                while ($row = mysqli_fetch_array($exec)) {

                  $product_id = $row['product_id'];
                  $product_name = $row['product_name'];
                  $product_price = $row['sell_price'];
                  $description = $row['description'];
                  $product_quantity = $row['quantity'];

               ?>
              <tr>
              <td class="text-center"><?php echo $product_id; ?>
                <input type="hidden" name="product_id[]" value="<?php echo $product_id; ?>">
              </td>
                  <td><?php echo $product_name; ?></td>
                  <td><?php echo $description; ?></td>
                  <td><?php echo $product_price; ?></td>
                  <td><input type="number" name="hehe" value="<?php echo $product_quantity; ?>" id="<?php echo "qtyResult" . $a++; ?>" disabled></td>
                  <td><input type="number" name="qtyBuy[]" id="<?php echo "qtyBuy" . $b++; ?>" onkeyup="updateStock(this, event)"></td>
              </tr>
              <?php } ?>
              </tbody>
          </table>


          </div>
          <div class="form-group">
              <input type="submit" name="addCart" value="Add Items to Cart" class="btn btn-info pull-right">

          </div>
          </form>

表格图片

Answer 1

如果我理解，您的基本问题是您允许数据​​库在cart表中的同一购物车下多次存储相同的产品ID。我的解决方案依赖于同一购物车中所有商品具有相同价值的字段。我使用cart_id但它可能是其他东西。

由于您希望产品只在同一个购物车中出现一次，因此您应该做两件事：

1。添加UNIQUE INDEX(cart_id, product_id) cart表。完成此操作后，任何将同一产品添加到同一购物车的尝试都将失败并显示错误。只运行一次此查询：

ALTER TABLE `cart` ADD UNIQUE INDEX (cart_id, product_id);

2. 更改用于将商品添加到购物车的查询。正如我上面所指出的，如果产品已经在该购物车中，则常规INSERT将失败。而是使用INSERT...ON DUPLICATE KEY UPDATE...查询。如果找不到，则此查询将插入产品ID。但是，如果购物车已有该产品，则查询将更新现有记录。

变化：

INSERT INTO cart (product_id, quantityCart) 
    VALUES (".$_POST['product_id'][$index].", ".$value.")

要：

INSERT INTO cart (cart_id, product_id, quantityCart) 
    VALUES ($cartID, {$_POST['product_id'][$index]}, $value)
ON DUPLICATE KEY UPDATE quantityCart = quantityCart + $value

我在我的解决方案中使用cart_id作为确定属于同一购物车的所有商品的方式。如果您有另一个字段执行此操作（例如：user_id），请使用该字段而不是cart_id。基本解决方案是一样的。

Answer 2

假设您有此表

id employee_id contract_id month year d1 d2 d3 1 25 1 11 2011 1 01 01 2 16 5 11 2011 1 11 0 3 29 3 11 2011 1 001 100 1 25 4 11 2011 0 11 011

您希望在Id的基础上合并它，在您的情况下是product_id

    id  employee_id contract_id      month    year    d1     d2      d3
    1          25        1,4            11       2011    1,0    01,11   01,011
    2          16        5              11       2011    1      11      0
    3          29        3              11       2011    1      001     100

您可以通过此查询

执行此操作

SELECT
    id,
    employee_id,
    GROUP_CONCAT(contract_id SEPARATOR ',') AS contract_ids,
    `month`,
    `year`,
    GROUP_CONCAT(d1 SEPARATOR ',') AS d1s,
    GROUP_CONCAT(d2 SEPARATOR ',') AS d2s,
    GROUP_CONCAT(d3 SEPARATOR ',') AS d3s
FROM
    `table`
WHERE
    `month` = 11 AND `year` = 2011
GROUP BY
    employee_id