上载数据并从两个表中选择它们，并在同一HTML表中显示它们

Question

I would like to ask you how to select data from two different tables and display them in the single HTML table. 我想问你如何从两个不同的表中选择数据并将它们显示在单个HTML表中。 I've tried many ways, but I can't find the right way. 我已经尝试了许多方法，但是找不到正确的方法。

First table 第一桌

Second table 第二张桌子

HTML: HTML：

<html>
  <body>
    <div id="reg-area">
     <form action="upload.php" method="post" align="center" id="register" enctype="multipart/form-data">
        <input type="text" name="username" minlength="5" maxlength="25" required/><br /><br />
        <input type="number" name="age" max="99" required/><br /><br />
        <input type="radio" name="gender" value="male" id="r1" checked="checked">
        <label for="r1">Muž</label>
        <input type="radio" name="gender" value="female" id="r2">
        <label for="r2">Žena</label><br /><br />
        <input type="file" name="file_img" /><br /><br />
        <input type="submit" name="btn_upload" value="Upload"/>
     </form>
    </div>
        <?php
            require_once('Db.php');         
            Db::connect('localhost', '***', '***', '***');

            if ($_POST)
            {   
        $existuje = Db::querySingle('
                  SELECT COUNT(*)
                  FROM table1
                  WHERE username=?
                  LIMIT 1
          ', $_POST['username']);
          if ($existuje)
                  die ('<p align="center" class="element-animation"><font color="red">Uživatel s touto přezdívkou již existuje.</font></p>');
        else

        /*INSERT USERNAME, age, gender into Table1 */

            $datum = date("Y-m-d H:i:s");
                Db::query('
                    INSERT INTO table1 (username,   age, gender, date)  
                    VALUES (?, ?, ?, ?)
                ', $_POST['username'], $_POST['age'], $_POST['gender'], $datum); 
        header("location: index.php"); 
        }                              

        /*DISPLAY CONTENT OF BOTH TABLES*/

            $snapy = Db::queryAll('
                SELECT *
          FROM table1
          JOIN table2 ON `ID` = `img_ID`
          ORDER BY ID;
            ');
            echo('<div align="center" id="content"><h1>Seznam se s:</h1><table width="100%" cellspacing="10" collspacing="10"></div>');
        echo('<th>Username</th><th>Age</th><th>Gender</th><th>Uploaded image</th>');
            foreach ($snapy as $s)
            {
                echo('<tr align="center"><td>' . htmlspecialchars($s['username']));
                echo('</td><td>' . htmlspecialchars($s['age']));
                echo('</td><td>' . '<img src="img/' . $s['gender'] .'.png' .'" width="25px"/>');
                echo('</td><td>' . '<img src="images/' . $filename . $filetype .'" width="25px"/>');
                echo('</td></tr>');
            }
                echo('</table>');     
        ?>
    </div>
    </body>
</html>

PHP Upload code: PHP上传代码：

<?php
if(isset($_POST['btn_upload']))
{
    $filetmp = $_FILES["file_img"]["tmp_name"];
    $filename = $_FILES["file_img"]["name"];
    $filetype = $_FILES["file_img"]["type"];
    $filepath = "images/".$filename;

    move_uploaded_file($filetmp,$filepath);

        $sql = "INSERT INTO snapcodes (img_name,img_path,img_type) VALUES (?, ?, ?), '$filename','$filepath','$filetype')";
  header("location: index.php");
}
?>

After clicking the Upload button is image uploaded to the "images" folder, but no data are uploaded to the database. 单击“上载”按钮后，图像被上载到“图像”文件夹，但是没有数据上载到数据库。 Tables are empty. 表是空的。

I would like to get table like this: 我想要这样的表：

********************************************
   username | age | gender | uploaded image
 1.user     | 15  | male   | image         
 2.user2    | 14  | female | image     
********************************************

Can someone help me please? 有人能帮助我吗？

Answer 1

I am not going to answer "the upload thing", as it is not the topic of the question. 我不会回答“上传内容”，因为这不是问题的主题。 I am not commenting on code structuring, SQL injections, use of Czech to name variables, and other issues, either. 我也没有评论代码结构，SQL注入，使用Czech来命名变量以及其他问题。 See @tadman's comments. 参见@tadman的评论。

As of JOIN, you join on IDs which are both autoincrement. 从JOIN开始，您加入都是自动递增的ID。 It does not make much sense. 这没有多大意义。 You should define a column like user_id in images table and then use something like this: 你应该这样定义列user_id在images表中，然后使用这样的：

SELECT *
FROM table1
JOIN table2 ON table1.ID = table2.user_id
ORDER BY table1.ID

Answer 2

I have already found a solution. 我已经找到了解决方案。

The error was in the tag. 错误出在标签中。 I used the action="upload.php" attribute, but upload.php cared only about uploading an image, not a recording of the entire form. 我使用action =“ upload.php”属性，但是upload.php只关心上传图像，而不关心整个表单的记录。

So I removed the action="upload.php" attribute and I put PHP ipmroved code from upload.php into the file that contains the function INSERT for entire form. 因此，我删除了action =“ upload.php”属性，并将来自上载PHP ipmroved的PHP代码放入了包含整个表格INSERT函数的文件中。

Thank you all for your advice and warnings :) 谢谢大家的建议和警告：）

上载数据并从两个表中选择它们，并在同一HTML表中显示它们

问题描述

2 个解决方案

解决方案1
0 2016-08-26 05:56:09

解决方案2
0 已采纳 2016-08-26 10:29:30

上载数据并从两个表中选择它们，并在同一HTML表中显示它们

问题描述

2 个解决方案

解决方案1 0 2016-08-26 05:56:09

解决方案2 0 已采纳 2016-08-26 10:29:30

解决方案1
0 2016-08-26 05:56:09

解决方案2
0 已采纳 2016-08-26 10:29:30