0

具有来自 mysql 数据库的多列的选择框

以下代码是来自 zgr024 的中奖代码

       <?php 
    include '../config.php';
    $sql = "SELECT * FROM megabase";  
    $resultaat = mysql_query($sql) or die (mysql_error());  
    $domains = array();
    while ($row = mysql_fetch_array($resultaat))       
    { 
       if (!empty($row['domeinnaam1'])) $domains[] = $row['domeinnaam1'];
       if (!empty($row['domeinnaam2'])) $domains[] = $row['domeinnaam2'];
    }
?>

   <select size="1" name="domeinnaam">
<?php
    foreach ($domains as $domain)
    {
        echo "<option>$domain</option>";
    } 
?>
    </select>

但是这段代码在我有以下代码的下一页上导致了一个预期的错误。

    <?php 

     //MySQL Database Connect
     include 'config.php';

    $domeinnaam=$_POST['domeinnaam']; 

    $data = 'SELECT * FROM megabase WHERE domeinnaam = "'.$domeinnaam.'"'; 
      $query = mysql_query($data) or die("Couldn't execute query. ". mysql_error()); 
      $data2 = mysql_fetch_array($query); 


    ?>

如何更改最后一个代码以使用第一个代码：

我试过了：

        <?php 

         //MySQL Database Connect
         include 'config.php';

        $domeinnaam=$_POST['domeinnaam'];
       $domeinnaamquerry='domeinnaam1'.'domeinnaam2'.'domeinnaam3'.'domeinnaam4'.'domeinnaam5'.'do meinnaam6'.'domeinnaam7'.'domeinnaam8'.'domeinnaam9'.'domeinnaam10';

        $data = 'SELECT * FROM megabase WHERE $domeinnaamquerry = "'.$domeinnaam.'"'; 
          $query = mysql_query($data) or die("Couldn't execute query. ".             mysql_error()); 
          $data2 = mysql_fetch_array($query); 


        ?>

但我收到以下错误：无法执行查询。“where 子句”中的未知列“$domeinnaamquery”

Okey 让它现在工作它很草率，但工作感谢大家的投入，这很有帮助。并感谢您的警告。

这是代码

     <?php 

      //MySQL Database Connect
      include 'config.php';

     $domeinnaam=$_POST['domeinnaam']; 

     $data = 'SELECT * FROM megabase WHERE domeinnaam1="'.$domeinnaam.'" OR domeinnaam2="'.$domeinnaam.'" OR domeinnaam3="'.$domeinnaam.'" OR domeinnaam4="'.$domeinnaam.'" OR domeinnaam5="'.$domeinnaam.'" OR domeinnaam6="'.$domeinnaam.'" OR domeinnaam7="'.$domeinnaam.'" OR domeinnaam8="'.$domeinnaam.'" OR domeinnaam9="'.$domeinnaam.'" OR domeinnaam10="'.$domeinnaam.'"    '; 
       $query = mysql_query($data) or die("Couldn't execute query. ". mysql_error()); 
       $data2 = mysql_fetch_array($query); 


     ?>
4

2 回答 2

1

尝试

<?php 

    //MySQL Database Connect
    include 'config.php';

    $domeinnaam=$_POST['domeinnaam']; 

    $data = "SELECT * FROM megabase WHERE domeinnaam1 = '$domeinnaam' OR domeinnaam2 = '$domeinnaam' "; 
    $query = mysql_query($data) or die("Couldn't execute query. ". mysql_error()); 
    $data2 = mysql_fetch_array($query); 

?>
于 2013-03-04T16:23:36.657 回答
0

如果你希望一个变量在 PHP 中被解释为它的值，你不能使用单引号。

$foo = 'World';

echo 'Hello $foo'; // Hello $foo
echo "Hello $foo"; // Hello World

因此，你想要

$data = "SELECT * FROM megabase WHERE $domeinnaamquerry = '".$domeinnaam."'";

尽管阅读此和此以了解您为什么真的不想要那个。

于 2013-03-04T16:10:05.600 回答