Question

2个问题......

情景：

我想通过表单查询我的数据库表，然后显示发生的结果（如果有结果），我目前的情况是它确实有效，但它完全清除了表单并将我留给了一个空页面只有结果才有意义。我想在我输入表单数据的同一页面上看到我的结果。我希望以基于表格的格式看到我的数据，有点像......

|Name   | Age|
|-------|----|
|Anthony| 20 |

我有一个单独的设计，以使其看起来很晚。这是我到目前为止的设置......

displayform.html：

<html>
<form method="post" name="display" action="display.php" />
Enter the name you like to display the data from MySQL:<br>
<input type="text" name="name" />
<input type="submit" name="Submit" value="display" />
</form>
</html>

和display.php

<?php
mysql_connect("localhost", "toor", "FakePassword") or die("Connection Failed");
mysql_select_db("FakeDatabase")or die("Connection Failed");
$name = $_POST['name'];
$query = "select name, age from test WHERE name = '$name'";
$result = mysql_query($query);
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
    echo "<tr>";
echo "<td>".$line['name']."</td>";
echo "<td>".$line['age']."</td>";
echo "<br>\n";
echo "</tr>";
}
?>

Answer 1

注意：

您可以从中制作单个文件以实现您想要的输出
使用mysql_real_escape_string()清理传递的值以阻止SQL injections
您应该使用mysqli_*代替deprecated mysql_* API

将它们组合在一个像这样的文件中display.php）：

<html>
  <form method="post" name="display" action="display.php" />
  Enter the name you like to display the data from MySQL:<br>
    <input type="text" name="name" />
    <input type="submit" name="Submit" value="display" />
  </form>

  <?php
    mysql_connect("localhost", "toor", "FakePassword") or die("Connection Failed");
    mysql_select_db("FakeDatabase")or die("Connection Failed");

    if(!empty($_POST["name"])){ /* WE ADD THIS PART SO WHEN NO PASSED DATA IS FOUND, IT WILL NOT GENERATE "UNIDENTIFIED VARIABLE" ERROR */

      $name = mysql_real_escape_string($_POST['name']); /* SANITIZE THE VALUE OF THIS VARIABLE */
      $query = "select name, age from test WHERE name = '$name'";
      $result = mysql_query($query);
      echo "<table>";
      while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
        echo "<tr>";
        echo "<td>".$line['name']."</td>";
        echo "<td>".$line['age']."</td>";
        echo "<br>\n";
        echo "</tr>";
      } /* END OF WHILE LOOP */
      echo "</table>";
    } /* END OF NOT EMPTY NAME */
  ?>

</html>

如果您对mysqli_*感兴趣，建议您使用prepared statement。只需将您的php部分替换为：

  <?php

    /* ESTABLISH YOUR CONNECTION FIRST */
    $con = new mysqli("localhost", "toor", "FakePassword", "FakeDatabase");
    if (mysqli_connect_errno()) {
      printf("Connect failed: %s\n", mysqli_connect_error());
      exit();
    }

    if(!empty($_POST["name"])){

      if($stmt = $con->prepare("SELECT name, age FROM test WHERE name = ?")){
        $stmt->bind_param("s",$_POST["name"]); /* BIND THE PASSED-ON VALUE TO THE QUERY */
        $stmt->execute(); /* EXECUTE THE QUERY */
        $stmt->bind_result($name,$age); /* BIND THE RESULT TO THESE VARIABLES */
        echo "<table>";
        while($stmt->fetch()){ /* FETCH ALL RESULTS */
          echo "<tr>";
          echo "<td>".$name."</td>";
          echo "<td>".$age."</td>";
          echo "<br>\n";
          echo "</tr>";
        } /* END OF WHILE LOOP */
        echo "</table>";
        $stmt->close();
      } /* END OF PREPARED STATEMENT */

    } /* END OF NOT EMPTY NAME */
  ?>

Answer 2

如果您想要列名称，则必须使用此查询：SHOW COLUMNS FROM test

这是一个有效的例子。

另外，请注意SQL injection。如果你没有收到它，我的网络服务器会向我的屏幕发出警告：

不推荐使用mysql扩展程序，将来会将其删除：改为使用mysqli或PDO

<?php
mysql_connect("localhost", "toor", "FakePassword") or die("Connection Failed");
mysql_select_db("FakeDatabase")or die("Connection Failed");


$query = "SHOW COLUMNS FROM test";
$result = mysql_query($query);
echo "<tr>";
while ($header = mysql_fetch_array($result, MYSQL_ASSOC)) {
    echo "<td>". $header['Field']."</td>";
}
echo "<br>\n";
echo "</tr>";

$name = $_POST['name'];
$query = "select name, age from test WHERE name = '$name'";
$result = mysql_query($query);
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
    echo "<tr>";
    echo "<td></td>";
    echo "<td>".$line['name']."</td>";
    echo "<td>".$line['age']."</td>";
    echo "<br>\n";
    echo "</tr>";
}
?>

修改

我回答了问题标题中描述的主要问题。下次，你应该发一个问题......每个问题。祝你好运。

如何在php / html输出中显示我的mysql表列标题？

2 个答案:

注意：