dongqiao3833 2016-06-27 08:14
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从下拉列表中删除值

I'm trying to Code a function that allow me to delete a selected value from a Dropdown list.

<?php
require_once("db.inc.php");
?>
</head>
<body>
<form action="" method="POST">
 <?php





$stmt = $mysqli->prepare("SELECT anr, name FROM artikel");
$stmt->execute();
$stmt->bind_result($anr, $name);


echo "<select name='selected_name'><br />";

while ($stmt->fetch()) {

echo '<option value='.$anr.'>'.$anr.' | '.$name.'</option>';


if(isset($_POST['loeschen'])){

    $stmt = $mysqli->prepare("DELETE FROM artikel WHERE anr=?");
        $stmt->bind_param('i', $anr);
        $stmt->execute();
        $stmt->close();
        $mysqli->close();

    }

}   


  ?>
  <input type="submit" value="Datensatz löschen" name="loeschen">
</form>
</body>
</html>

My Problem is that the values are going to be deleted even i don't press the submit button. Thank you in advance for your suggestions.

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2条回答 默认 最新

  • dscojuxf69080 2016-06-27 08:21
    关注

    Use Post value instead of $anr.

    <?php
    require_once("db.inc.php");
    ?>
    </head>
    <body>
    <?php
    if(isset($_POST['selected_name'])){
    
        $stmt = $mysqli->prepare("DELETE FROM artikel WHERE anr=?");
            $stmt->bind_param('i', $_POST['selected_name']);
            $stmt->execute();
            $stmt->close();
            $mysqli->close();
    
        }
    
    }
    ?>
    <form action="" method="POST">
     <?php
    
    $stmt = $mysqli->prepare("SELECT anr, name FROM artikel");
    $stmt->execute();
    $stmt->bind_result($anr, $name);
    echo "<select name='selected_name'><br />";
    while ($stmt->fetch()) {
    echo '<option value='.$anr.'>'.$anr.' | '.$name.'</option>';
    }
    echo "</select>";
      ?>
      <input type="submit" value="Datensatz löschen" name="loeschen">
    </form>
    </body>
    </html>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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