首页 \ 问答 \ 从mySQL获取到HTML TABLE(Fetch from mySQL to HTML TABLE)

从mySQL获取到HTML TABLE(Fetch from mySQL to HTML TABLE)

 我已经使用过这段代码，但除了在网页上显示表格之外，它还没有取得任何其他内容。 
 

<!doctype html>
    <html lang="en">
    <head>
      <meta charset="UTF-8">
      <title>database connections</title>
    </head>
    <body>
      <?php
      $username = "root";
      $password = " ";
      $host = "localhost";

      $connector = mysql_connect($host,$username,$password)
          or die("Unable to connect");
        echo "Connections are made successfully::";
      $selected = mysql_select_db("ticad", $connector)
        or die("Unable to connect");

      //execute the SQL query and return records
      $result = mysql_query("SELECT * FROM users ");

      ?>
      <table border="2" style= "background-color: #84ed86; color: #761a9b; margin: 0 auto;" >
      <thead>
        <tr>
          <th>id</th>
          <th>Names</th>
          <th>Company</th>
          <th>Position</th>
          <th>Email</th>
          <th>Event</th>
          <th>Comments</th>
        </tr>
      </thead>
      <tbody>
        <?php
          while( $row = mysql_fetch_assoc( $result ) ){
            echo
            "<tr>
              <td>{$row\['id'\]}</td>
              <td>{$row\['Names'\]}</td>
              <td>{$row\['Company'\]}</td>
              <td>{$row\['Position'\]}</td>
              <td>{$row\['Email'\]}</td>
              <td>{$row\['Event'\]}</td> 
              <td>{$row\['Comments'\]}</td> 
            </tr>\n";
          }
        ?>
      </tbody>
    </table>
     <?php mysql_close($connector); ?>
    </body>
    </html>

 

 这是我的表和数据库 
 

 当我将代码作为Web运行时，这就是我得到的。 我希望当我们从表用户获取时，所有记录都显示在表上 

I have used this code but its not fetching anything other than displaying the table on the web. 
 

<!doctype html>
    <html lang="en">
    <head>
      <meta charset="UTF-8">
      <title>database connections</title>
    </head>
    <body>
      <?php
      $username = "root";
      $password = " ";
      $host = "localhost";

      $connector = mysql_connect($host,$username,$password)
          or die("Unable to connect");
        echo "Connections are made successfully::";
      $selected = mysql_select_db("ticad", $connector)
        or die("Unable to connect");

      //execute the SQL query and return records
      $result = mysql_query("SELECT * FROM users ");

      ?>
      <table border="2" style= "background-color: #84ed86; color: #761a9b; margin: 0 auto;" >
      <thead>
        <tr>
          <th>id</th>
          <th>Names</th>
          <th>Company</th>
          <th>Position</th>
          <th>Email</th>
          <th>Event</th>
          <th>Comments</th>
        </tr>
      </thead>
      <tbody>
        <?php
          while( $row = mysql_fetch_assoc( $result ) ){
            echo
            "<tr>
              <td>{$row\['id'\]}</td>
              <td>{$row\['Names'\]}</td>
              <td>{$row\['Company'\]}</td>
              <td>{$row\['Position'\]}</td>
              <td>{$row\['Email'\]}</td>
              <td>{$row\['Event'\]}</td> 
              <td>{$row\['Comments'\]}</td> 
            </tr>\n";
          }
        ?>
      </tbody>
    </table>
     <?php mysql_close($connector); ?>
    </body>
    </html>

 

This is my Table and dabase
 

This is what am getting when I run the code as a web. I want all the records to be displayed on the table as we fetch from the table users 

原文：https://stackoverflow.com/questions/38952437
更新时间：2023-08-18 18:08

最满意答案

 使用 
 

TRY_CONVERT(UNIQUEIDENTIFIER, parameter2) AS customerguid

 

 代替 
 

 CONVERT(UNIQUEIDENTIFIER, parameter2) AS customerguid

 

 视图内联到查询中， CONVERT可以在WHERE之前运行。 
 

 对于其他一些讨论，请参阅SQL Server不应引发不合逻辑的错误 

Use
 

TRY_CONVERT(UNIQUEIDENTIFIER, parameter2) AS customerguid

 

instead of
 

 CONVERT(UNIQUEIDENTIFIER, parameter2) AS customerguid

 

Views are inlined into the query and the CONVERT can run before the WHERE.
 

For some additional discussion see SQL Server should not raise illogical errors

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