''where'子句中的未知列'

Question

我一直在淘洗堆棧溢出網站，並嘗試了大部分已被問及/回答的類似問題，但不幸的是迄今爲止，還沒有任何解決方案對我有效。我想要一個正在填充sql數據庫（正在工作）的列表，然後一旦選擇該項目，點擊「生成」按鈕並從表格中獲取數據並僅將所選數據發佈到表格中。以前我遇到過一個問題，那就是填充武器列表中所有數據的表格。在對它進行處理之後，我現在在第107條（我認爲是我的行號）中出現了「未知列（我選擇哪個武器名稱）」這樣的錯誤。從數據庫填充列表，以試圖從列表中選擇武器形成。

<form action="#" method="post"> 
    <table class="table"> 
     <thead> 
      Martial Weapon Name 
     </thead> 
     <tr> 
      <th> 
       <select name="Choosen"> 
        <?php 
        echo'<option>Select Weapon</option>'; 
        //Check if at least one row is found 
        if($result->num_rows >0){ 
         //Loop through results 
         while($row = $result->fetch_assoc()){ 
          //Display weapon info 
          $output = $row['weapon_name']; 
          echo '<option>'.$output.'</option>'; 
         } 
        } 
        ?> 
       </select> 
      </th> 
     </tr> 
    </table> 
    <input class="btn btn-default" type="submit" name="submit" value="Generate"> 
    <h3> 
     Weapon 
    </h3> 
    <table class="table table-striped"> 
     <tr> 
      <th> 
       Weapon Name 
      </th> 
      <th> 
       Weapon Type 
      </th> 
      <th> 
       Damage 
      </th> 
     </tr> 
     <?php 
     if (isset($_POST['submit'])) { 
      $selected_weapon = $_POST['Choosen']; 
      $choose = "SELECT 
      weapon_types_martial.id, 
      weapon_types_martial.weapon_name, 
      weapon_types_martial.weapon_type, 
      weapon_types_martial.weapon_damage 
      FROM weapon_types_martial WHERE weapon_types_martial.weapon_name = " . $selected_weapon; 
      $result = $mysqli->query($choose) or die($mysqli->error . __LINE__); 
      foreach ($result->fetch_assoc() as $item) { 
       //Display weapon 
       $show = '<tr>'; 
       $show .= '<td>' . $item['weapon_name'] . '</td>'; 
       $show .= '<td>' . $item['weapon_type'] . '</td>'; 
       $show .= '<td>' . $item['weapon_damage'] . '</td>'; 
       $show .= '</tr>'; 
       //Echo output 
       echo $show; 
      } 
     } 
     ?> 
    </table> 
</form> 

而且最後這裏是我得到

Answer 1

1）$ _ POST你不會得到數據[「choosen」如你沒有在下拉列表中傳遞值（選擇）

2）在你的數據庫表可以現場weapon_name爲varchar，所以你必須把它傳遞到單引號。

如下更改代碼：

<form action="#" method="post"> 
    <table class="table"> 
     <thead> 
      Martial Weapon Name 
     </thead> 
     <tr> 
      <th> 
       <select name="Choosen"> 
        <?php 
        echo '<option>Select Weapon</option>'; 
        //Check if at least one row is found 
        if($result->num_rows >0){ 
         //Loop through results 
         while($row = $result->fetch_assoc()){ 
          //Display weapon info 
          $output = $row['weapon_name']; 
          echo '<option value="'.$output.'">'.$output.'</option>'; //<--------------change here 
         } 
        } 
        ?> 
       </select> 
      </th> 
     </tr> 
    </table> 
    <input class="btn btn-default" type="submit" name="submit" value="Generate"> 
    <h3> 
     Weapon 
    </h3> 
    <table class="table table-striped"> 
     <tr> 
      <th> 
       Weapon Name 
      </th> 
      <th> 
       Weapon Type 
      </th> 
      <th> 
       Damage 
      </th> 
     </tr> 
     <?php 
     if (isset($_POST['submit'])) { 
      $selected_weapon = $_POST['Choosen']; 
      $choose = "SELECT 
      id, 
      weapon_name, 
      weapon_type, 
      weapon_damage 
      FROM weapon_types_martial WHERE weapon_name = '$selected_weapon'"; //<--------------change here 
      $result = $mysqli->query($choose) or die($mysqli->error . __LINE__); 
      if ($result->num_rows > 0) { 
      while($item = $result->fetch_assoc()) { 
       //Display weapon 
       $show = '<tr>'; 
       $show .= '<td>' . $item['weapon_name'] . '</td>'; 
       $show .= '<td>' . $item['weapon_type'] . '</td>'; 
       $show .= '<td>' . $item['weapon_damage'] . '</td>'; 
       $show .= '</tr>'; 
       //Echo output 
       echo $show; 
      } 
      } 
      else 
      { 
      echo "No data found"; 
      } 
     } 
     ?> 
    </table> 
</form> 

Answer 2

的你需要給值截圖裏面單引號，

$choose = "SELECT 
    weapon_types_martial.id, 
    weapon_types_martial.weapon_name, 
    weapon_types_martial.weapon_type, 
    weapon_types_martial.weapon_damage 
    FROM weapon_types_martial 
    WHERE weapon_types_martial.weapon_name = '" . $selected_weapon ."'"; 

Answer 3

您需要將引號放入SQL查詢。
嘗試：

$choose = "SELECT 
    weapon_types_martial.id, 
    weapon_types_martial.weapon_name, 
    weapon_types_martial.weapon_type, 
    weapon_types_martial.weapon_damage 
    FROM weapon_types_martial WHERE weapon_types_martial.weapon_name = '" . $selected_weapon . "'"; 

建議的字，這個查詢是非常不安全的。我建議使用框架或庫進行數據庫查詢。

Answer 4

$choose = "SELECT id, weapon_name, weapon_type, weapon_damage FROM weapon_types_martial WHERE weapon_name = '".$selected_weapon."'"; 

或

$choose = "SELECT weapon_types_martial.id, weapon_types_martial.weapon_name, weapon_types_martial.weapon_type, weapon_types_martial.weapon_damage FROM weapon_types_martial as weapon_types_martial WHERE weapon_types_martial.weapon_name = '" . $selected_weapon ."'"; 

Answer 5

在您選擇元素的options沒有value=..屬性，這意味着沒有什麼是出現在您的查詢。

所以$_POST['Choosen']將= ''。

更何況，你想傳遞一個字符串，那麼你的查詢需要被包裹在'：

$choose = "SELECT id, weapon_name, weapon_type, weapon_damage FROM weapon_types_martial WHERE weapon_name = '".$selected_weapon."'";