''where'子句中的未知列'

Question

我一直在淘洗堆栈溢出网站，并尝试了大部分已被问及/回答的类似问题，但不幸的是迄今为止，还没有任何解决方案对我有效。我想要一个正在填充sql数据库（正在工作）的列表，然后一旦选择该项目，点击“生成”按钮并从表格中获取数据并仅将所选数据发布到表格中。以前我遇到过一个问题，那就是填充武器列表中所有数据的表格。在对它进行处理之后，我现在在第107条（我认为是我的行号）中出现了“未知列（我选择哪个武器名称）”这样的错误。从数据库填充列表，以试图从列表中选择武器形成。

<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> 

而且最后这里是我得到

Answer 1

1）$ _ POST你不会得到数据[“choosen”如你没有在下拉列表中传递值（选择）

2）在你的数据库表可以现场weapon_name为varchar，所以你必须把它传递到单引号。

如下更改代码：

<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

的你需要给值截图里面单引号，

$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

您需要将引号放入SQL查询。
尝试：

$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 . "'"; 

建议的字，这个查询是非常不安全的。我建议使用框架或库进行数据库查询。

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

在您选择元素的options没有value=..属性，这意味着没有什么是出现在您的查询。

所以$_POST['Choosen']将= ''。

更何况，你想传递一个字符串，那么你的查询需要被包裹在'：

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