Php 将sql表结果限制为仅选定内容_Php_Mysql_Database_Html Table

Php 将sql表结果限制为仅选定内容

php mysql database

Php 将sql表结果限制为仅选定内容,php,mysql,database,html-table,Php,Mysql,Database,Html Table,我觉得我很快就能弄明白这一点，但就是无法缩小差距。我有一个从SQL数据库中提取的列表，我希望能够从列表中选择一项，并且只显示表所连接的信息。除了列表中的所有项目都在拉，而不仅仅是选中的项目外，所有项目都在拉 <form action="#" method="post"> <table class="table"> <thead>Martial Weapon Name</thead> <t

我觉得我很快就能弄明白这一点，但就是无法缩小差距。我有一个从SQL数据库中提取的列表，我希望能够从列表中选择一项，并且只显示表所连接的信息。除了列表中的所有项目都在拉，而不仅仅是选中的项目外，所有项目都在拉

      <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 

      $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";

  $result = $mysqli->query($choose) or die($mysqli->error.__LINE__);

        if(isset($_POST['submit'])){
          $selected_weapon = $_POST['Choosen'];
        while($list = $result->fetch_assoc()){
                  //Display weapon
                  $show ='<tr>';
                  $show .='<td>'.$list['weapon_name'].'</td>';
                  $show .='<td>'.$list['weapon_type'].'</td>';
                  $show .='<td>'.$list['weapon_damage'].'</td>';
                  $show .='</tr>';

                  //Echo output
                  echo $show;
                  } 
                }
              ?>

      </form>


武名
武器
武器名称
武器类型
损坏

上面是表单和表的代码。同样，一切都在运转。数据库正在连接和调出一切很棒的东西。我唯一要做的就是确保输出的数据只是从下拉列表中选择的项目

下面是列表的屏幕截图。我目前在数据库中只有3项（故意的）。所以我选择了战斧而不是战斧线，3个都出现了

编辑1

根据要求，这里是完整的代码页。整个页面都在工作，数据库正在连接等等。只是输出不是放置1个选定的武器。根据之前建议的一些更改，我完全停止将武器输入表中，因此为了清晰起见，我发布了最初列出的页面代码

<?php
  include('includes/database.php'); ?>

  <?php
  //Create the select query
  $query = "SELECT   
        weapon_types_martial.id,
        weapon_types_martial.weapon_name,
        weapon_types_martial.weapon_type,
        weapon_types_martial.weapon_damage 
        FROM weapon_types_martial
        ORDER BY weapon_name";

  //Get results of query
  $result = $mysqli->query($query) or die($mysqli->error.__LINE__);

?>


<!DOCTYPE html>
<html lang="en">
  <head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<!-- The above 3 meta tags *must* come first in the head; any other head content must come *after* these tags -->
<meta name="description" content="">
<meta name="author" content="">
<link rel="icon" href="../../favicon.ico">

<title>App Test | Weapons</title>

<!-- Bootstrap core CSS -->
 <link href="css/bootstrap.min.css" rel="stylesheet">
<meta name="viewport" content="width=device-width, initial-scale=1">
      <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
      <script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js"></script>
       <style type="text/css"></style>


  </head>

  <body>

<div class="site-wrapper">

  <div class="site-wrapper-inner">

    <div class="cover-container">

      <div class="masthead clearfix">
        <div class="inner">
          <h3 class="masthead-brand">Cover</h3>
          <nav>
            <ul class="nav masthead-nav">
              <li><a href="index.php">Front Page</a></li>
              <li class="active"><a href="weapons.php">Weapons</a></li>
              <li><a href="armor.php">Armor</a></li>
              <li><a href="consumables.php">Consumables</a></li>
            </ul>
          </nav>
        </div>
      </div>
      <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 

  $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";

  $result = $mysqli->query($choose) or die($mysqli->error.__LINE__);

    if(isset($_POST['submit'])){
      $selected_weapon = $_POST['Choosen'];
    while($list = $result->fetch_assoc()){
              //Display weapon
              $show ='<tr>';
              $show .='<td>'.$list['weapon_name'].'</td>';
              $show .='<td>'.$list['weapon_type'].'</td>';
              $show .='<td>'.$list['weapon_damage'].'</td>';
              $show .='</tr>';

              //Echo output
              echo $show;
              } 
            }
          ?>

  </form>
  </div>
   </div>
    </div>
  </div>
</div>

<!-- Bootstrap core JavaScript
================================================== -->
<!-- Placed at the end of the document so the pages load faster -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script>window.jQuery || document.write('<script src="../../assets/js/vendor/jquery.min.js"><\/script>')</script>
<script src="../../dist/js/bootstrap.min.js"></script>
<!-- IE10 viewport hack for Surface/desktop Windows 8 bug -->
<script src="../../assets/js/ie10-viewport-bug-workaround.js"></script>


应用程序测试|武器
掩护






武名
武器
武器名称
武器类型
损坏
window.jQuery | | document.write（“”）

以上代码的结果显示在上面的屏幕截图中。下拉列表显示了正确的武器列表，但是我希望能够选择其中一种武器，点击生成，并且在下表中只显示一种选定的武器。现在它正在显示列表中的所有武器，而不仅仅是选定的武器。

它应该可以工作：

      <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 

      $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";

  $result = $mysqli->query($choose) or die($mysqli->error.__LINE__);

        if(isset($_POST['submit'])){
          $selected_weapon = $_POST['Choosen'];
        while($list = $result->fetch_assoc()){
                  //Display weapon
                  $show ='<tr>';
                  $show .='<td>'.$list['weapon_name'].'</td>';
                  $show .='<td>'.$list['weapon_type'].'</td>';
                  $show .='<td>'.$list['weapon_damage'].'</td>';
                  $show .='</tr>';

                  //Echo output
                  echo $show;
                  } 
                }
              ?>

      </form>

<?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;
    }
}
?>

在制定新问题时结束此问题。当前代码

<?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;
}
}
?>

Hmm，这太接近了！但现在我得到一个未知列错误。下面是我看到的图像。（我完全复制了你发布的内容）我确实想编辑一下，说明列表不再“技术上”显示在下面，但它显示了一行关于未知列的内容。你有列战斧吗？没有，上面的代码显示没有该列。但是“未知列”错误会根据我从下拉菜单中选择的选项而变化。所以在照片中它显示了战斧，但我可以尝试选择俱乐部，取而代之的是列俱乐部未知。我觉得这与WHERE子句有关（我的意思是，错误确实说明了这一点）@mfka我用页面代码编辑了上面的评论。我确实将代码还原为我的原始代码，因为它仍在输出到表中。我再次尝试了使用foreach添加的代码，但仍然得到未知列。它看起来像是专门说明了

$result

部分的行。