Question

    <?php
session_start();


error_reporting(0);
include("config.php");
?>
<?php echo '<pre>' . print_r($_SESSION, TRUE) . '</pre>'; ?>

    <html>
    <head>
      <title>MR stempagina</title>
    </head>
    <body>
      <?php if( $_SESSION['user_info']['gestemd']>0){
        header("Location: logout.php");
      }
      ?>

      <?php echo $_SESSION['user_info']['name']  ?>, u kunt hier stemmen. 
      <form action="stemmen.php" method="POST">
        <p>Welke ouder wilt u als vertegenwoordiger van de ouders van de HBK afdeling in de medezeggenschapsraad?</p>
        <input type="radio" name="kandidaat" value="piet"> piet<br>
        <input type="radio" name="kandidaat" value="hein"> hein<br>
        <p><input type="submit" name="stem" value="stem"></p>

<?php

$error = '';
  if(isset($_POST['kandidaat'])){
      echo $_POST['kandidaat'];
      $_SESSION['user_info'] = $user;
      //$query = " UPDATE ".$SETTINGS["USERS"]." SET gestemd = gestemd+1 WHERE id=".$_SESSION['id'];
      //$query = " UPDATE ".$SETTINGS["USERS"]." SET gestemd = gestemd+1";
      //$query = " UPDATE ".$SETTINGS["kandidaat"]." SET aantal = aantal+1";
      //$query = " UPDATE ".$SETTINGS["USERS"]." SET gestemd = gestemd+1 WHERE id='{$_SESSION['id']}'";
      $query = "UPDATE {$SETTINGS["USERS"]} SET gestemd = gestemd+1 WHERE id={$_SESSION['id']}";
      mysql_query ($query, $connection ) or die ('request "Could not execute SQL query" '.$query . ': ' . mysql_error());


    }

?>

          </form>
    </body>
    </html>

当我使用时：

$query = " UPDATE ".$SETTINGS["USERS"]." SET gestemd = gestemd+1 WHERE id='{$_SESSION['id']}'";

table_column gestemd不会增加。

当我使用相同的查询时，没有显示它在哪里工作，但所有用户当然会增加。

printR和echo用于调试。

thx为您提供帮助

Answer 1

我找到了问题的答案

$ query =“ UPDATE”。$ SETTINGS [“ USERS”]。“ SET gestemd = gestemd + 1 WHERE id = {$ _ SESSION ['user_info'] ['id']}”;

使用此解决方案即可。

sql查询，包括不更新表在哪里呢？

1 个答案: