如何將多個複選框選擇值放入SQL表中的單獨列中？

Question

概述 我是新來的PHP，所以忍受着我。我似乎無法在這裏找到答案，儘管我可能不知道它是否會打擊我。

我正在構建Knowladgebase系統。我有一個表單將新的KB文章插入到數據庫中。我希望用戶能夠使用多個複選框來選擇新KB所適用的類別。

我有這兩個相關的表 - 第一個是類別這是類別列表和秒category_kb_members這將保持其知識庫文章與每個類別相關的記錄。

這是到目前爲止我的PHP代碼：

if ($_SERVER['REQUEST_METHOD'] == 'POST') { 
    if (isset($_POST['title']) && isset($_POST['type']) && isset($_POST['category']) && isset($_POST['article'])){ 

      $title = mysqli_real_escape_string($dbc,$_POST['title']); 
      $type = mysqli_real_escape_string($dbc,$_POST['type']); 
      $category = mysqli_real_escape_string($dbc,$_POST['category']); 
      $article = mysqli_real_escape_string($dbc,$_POST['article']); 
      $now = gmdate(DATE_RFC822); 
      $mod = "Never"; 
      $author = $_SESSION['username']; 

      // Insert into article database 
      $q1 = $dbc->prepare("INSERT INTO articles (title, type, author, createdate, moddate, article) VALUES (?, ?, ?, ?, ?, ?)"); 
      $q1->bind_param("ssssss", $title, $type, $author, $now, $mod, $article); 

      $q1->execute(); 
      $q1->close(); 

      //Query the KB Number and store in variable 

      $kbnumber = mysqli_query($dbc, "SELECT idarticles FROM articles WHERE createdate = '$now'"); 

      //Get Categories as array and insert KB id's into relevent columns. 
      $query = mysqli_query($dbc,"SELECT * FROM categories"); 
      while($row = mysqli_fetch_assoc($query)) 
        { 

         mysqli_query($dbc,"INSERT INTO category_kb_members (".$row['name'].") VALUES ($kbnumber)"); 

      } 




    }else{$error = "Isset is false"; 
    } 

}else{$error = "Request methos is false"; 
} 

mysqli_close($dbc); 

我與PHP的一些HTML內嵌：

<div id="kbform"> 
        <?php echo $error ?> 
        <form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" method="post"> 
        Title/Description: <input type="text" name="title" maxlength="255" placeholder="Please enter a Title/Description" required/><br /> 
        KB Type:      
        <select name="type"> 
        <option value="Fault" selected="selected" maxlength="20">Fault</option> 
        <option value="Guide" maxlength="20">Guide</option> 
        </select> 
        <br /> 
        <br /> 
        Please select any categories that this KB matches:    
        <div class="form-group"> 
        <?php 
        require "admin/connection.php"; 
        $query = mysqli_query($dbc,"SELECT * FROM categories"); 

        while($row = mysqli_fetch_array($query)) 
        { 
        echo "<input type='checkbox' name='category' value=".$row['name'].">".$row['name']." | "; 
      } 
      mysqli_close($dbc); 
      ?> 
        </div> 
        <br /> 
        <p>Enter you KB Info. HTML is supported.<textarea name="article" rows="10" cols="40" required placeholder="Describe the fault, any errors, relevent environmental factors and the steps took to resolve. Try to add as much useful information as possible."></textarea></p> 
        <p>Entered By: <?php echo $_SESSION['username'] ?></p> 
        <input type="submit" name="submit" value="Submit" /> 
        </form> 
        </div> 

列在類別表 'idcategories' 和「名」。第二個表中的列是'idcategory_kn_members'，然後是'category1'和'category2'等等，它們將被添加。

我目前得到的錯誤是「PHP可恢復的致命錯誤：類mysqli_result的對象無法轉換爲字符串」，我堅持如何實現這一目標。我發現每行都有這樣的文章，但不是每欄。

任何幫助非常感謝。

Answer 1

忽略所有不好的編碼習慣，我會進入你的問題，錯誤說，這一切，你要使用的sqlian object爲一個字符串在這裏：

$kbnumber = mysqli_query($dbc, "SELECT idarticles FROM articles WHERE createdate = '$now'"); 

從先得到你的結果作爲字符串結果集：

$row = mysqli_fetch_assoc($kbnumber); 
$kbnumberString = $row['idarticles']; 

然後在你的下一個查詢中使用$kbnumberString代替$kbnumber