PHP - 如何使用POST将单选按钮选择传递到另一个页面？ [重复]

Question

This question already has an answer here:

• “Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP 28 answers

This is my form in page 1, the radio buttons are generated using php it is a list of 5 names of the tables in the database:

<form action="showtable.php" method="post">
    <table>
        while ($fieldInfo = mysqli_fetch_field($results)) { 
            <input type="radio" name="tableNames" value="<?php echo $tempName; ?>" > <?php echo $tempName ?> <br/>
        <?php } ?>
        <tr> 
            <td><input type="submit" value="submit"></td>
        </tr>
    </table>
</form>

this is my action page in page 2, it needs to receive the table name selection and create a query, it does not:

<?php
$tName = $_POST["tableNames"];

require_once("conn.php");

$sql = "SELECT * FROM $tName";

$results = mysqli_query($conn, $sql)
or die ('Problem with query' . mysqli_error($conn));

if (mysqli_num_rows($results) >= 1) {
    echo "query success";
} else {
    echo "query fail";
}
?>

I get an error that states:

Undefined index: tableNames

in the second page

</div>

Answer 1

Write your code as

<?php
     if(isset($_POST["tableNames"])){
         $tName = $_POST["tableNames"];
         require_once("conn.php");
         $sql = "SELECT * FROM $tName";
         $results = mysqli_query($conn, $sql) or die ('Problem with query' . mysqli_error($conn));

         if (mysqli_num_rows($results) >= 1) {
             echo "query success";
         } else {
             echo "query fail";
         }
     }

?>