編輯/更新逗號分隔值到mysql行並添加新

Question

編輯/更新逗號將值分隔成行並添加新的

我正在嘗試使用不同行的Ajax編輯產品，然后從使用相同名稱創建動態字段的表單中更新和添加新行[]

但不知道如何拉動它們進行編輯，然后在同一位置更新它們，因為它們位於我需要更新的不同行和表中

我能想到的編輯它們的唯一方法是像以下示例：http://www.infotuts.com/ajax-table-add-edit-delete-rows-dynamically-jquery-php/，但這是每行使用1行線，我有不同行的產品

不知道是否應該像這樣編輯它們，但是我將如何拉舊的行並添加新的行以及鏈接到相同順序的http://www.aquim.com/web-article-229.html

我試圖編輯它們，然后更新它們，例如，但是我如何更新舊產品以及添加新產品並將它們鏈接到不同的表

編輯表格

產品1 [產品1行] [價格1] [數量1] [總計1]

產品2 [產品2行] [價格2] [數量2] [總計2]

然后用表格添加新產品

<pre>`
  <form name="test" method="post" action="">
    <div id="container">
    <p id="add_field"><a href="#"><span>&raquo; Add New Bid</span>    </a></p>
    </div>
  </form>
 $('p#add_field').click(function(){
    count += 1;
    $('#container').append(
        '<strong>Link #' + count + '</strong><br />' 
        + '<input id="field_' + count + '" name="fields[]' + '" type="text" placeholder="Product Name" /><br />' 
        + '<input id="field_' + count + '" name="price[]' + '" type="text"  placeholder="Price"/><br />' 

        );

   });`

選擇舊產品並更新，然后獲取ID並添加新產品並鏈接到表格

我考慮可能進行更新的唯一方法是擦除所有內容並重新插入它，但我認為如果多個人這樣做，將來會引起沖突

             quotes TABLE 1

---

| quoteID | bid_name | 狀態| 總計

| ------------------------------ |

| 1 | bid1 | 待處理| 100 |

| 2 | bid2 | stuff_y | 200 |

              quote_products_link TABLE 2

---

| quoteID | productID |
| ------------------------------ |

| 1 | 4 |

| 1 | 4 |

| 2 | 5 |

| 2 | 5 |

              quote_products_link TABLE 3

---

| productID | 產品名稱| 價格| 數量| bid_name |

| 4 | pro_1 | price_1 | quan_1 | bid_1 |

| 4 | pro_2 | 價格_2 | quan_2 | bid_1 |

| 5 | pro_3 | price_3 | quan_3 | bid_2 |

| 5 | pro_4 | price_4 | quan_4 | bid_2 |

Answer 1

您可以拉---檢查---更新---刪除

偽代碼，不起作用！ 這是顯示流程：

// Get the current database rows that are being edited
$results = mysql_query('SELECT * FROM entries WHERE table_id=10');
$itemsInDB = [];
while($itemsInDB[] = mysql_fetch_row($results)) {}

// Iterate over the fields sent to the server
foreach($_POST['fields'] as $index=>$value) {
    // New Values that were submitted
    $field = $value;
    $price = $_POST['prices'][$index];

    // Get Rid Of Updated Rows From Pulled Array
    $found = false;
    foreach($itemsInDB as $dbindex=>$dbvalue) {
       // Check if the row exists in the database
       if ($dbvalue['field'] == $field) {
           $found = true;

           // Update the values that have changed
           if ($price !== $dbvalue['price'])
               UpdateDatabase('SET price=$price WHERE field=$field');

           // Remove the row from the pulled database rows
           array_shift($itemsInDB($dbindex,1);
       }
    }

    // insert it because it wasnt in the database
    if (!$found) 
       InsertDatabase('Values($field,$price)');
}

// Now that you have updated all the rows that were submitted
// The rows that are in the database that were not updated
// Need to be deleted because the user has removed them from the table
foreach($itemsInDB as $row) {
    $field = $row['field'];
    DeleteDatabaseRow('WHERE field='.$field.' LIMIT 1');
}

可能有更好的方法來執行此操作...這是我前一段時間遇到此問題時所采用的方法。

Answer 2

拉產品

$sqlString = "SELECT  
 GROUP_CONCAT(productID) AS id,
GROUP_CONCAT(productName) AS productsname,
GROUP_CONCAT(price) AS prices,
bidid FROM products GROUP BY bidid";

while($row = $result->fetch_array(MYSQLI_ASSOC)){
   $currentbid[$row['bidid']][] = $row;  

}

foreach ($currentbid as $bid => $itemList){

   foreach($itemList as $itemInfo => $loas){

 // count ids, by create a rande array or counting the wors
foreach(explode(",",$loas['id']) as $itemInfoss){

then inside the foreach - create the table rows

   $str.='<tr id="'.$itemInfoss.'" 

更新它們取決於您如何更新它們，我正在使用序列化功能和一些自定義數據來更新它們

要獲取ID，我使用JavaScript將其推入數組中

  mysqlrow.push($js(this).attr('data-rowid'));

首先刪除它們

   if(!empty($deleteoldproductsfrom)){

                foreach($erasecount as $deleteold){

                 $deleteProducts = "DELETE FROM products WHERE 

productID ='$ deleteold'“;

然后更新

    $sql_quote_add = "SELECT * FROM quotes WHERE bid_name ='$quotename' "; 

    $client_data = $con->query($sql_quote_add); 

獲取ID以查看其是否與任何提交的ID匹配

   $getquoteid= mysqli_fetch_assoc($client_data);   

    $oldquoteid = $getquoteid['quoteID']; 

//因為這取決於您如何提交，所以我正在為（$ for = 0; $ for <count（$ data ['productName']）; $ for ++）{
然后在for lopp內部，我們檢查ID是否與任何當前產品匹配

我分解了一個提交的數組，這是我想到的最簡單的方法，然后在里面做一個計數，這樣它就可以循環

 if($tagArray[$for] == $pro_old_id){

    $updateold = "UPDATE products SET price='$priceNew', 

   } else
 $sql_bid = sprintf("INSERT INTO products

 }