Question

用错误404将我射到/index.php。我不希望它导航到另一页，只需将结果显示在表中即可。

我不太确定自己在做什么错，我刚启动php，并一直使用w3导航语法。我设法制作了一个可以显示整个表格的插件，但是现在我试图使用一种形式来搜索表格并显示结果。

...
if( isset($_GET['submit']) ){
    $ssterm = $_POST["name"];
    $query = "SELECT  ID, Retailer, City, State, Contact, Address1, Address2, Country, Zip, Email, Website FROM Stores WHERE Retailer LIKE '%" . $rssterm .  "%' OR City LIKE '%" . $ssterm ."%'";
    $result = $conn->query($query);

    if($result->num_rows > 0){
    echo "<table>
               <tr>
                  <th>ID</th>
                  <th>Retailer</th>
                  <th>City</th>
                  <th>State</th>
                  <th>Contact</th>
                  <th>Address1</th>
                  <th>Address2</th>
                  <th>Country</th>
                  <th>Zip</th>
                  <th>Email</th>
                  <th>Website</th>
              </tr>";


      while($row = $result->fetch_assoc()){   //Creates a loop to loop through results
      echo "<tr><td>" . $row['ID'] . "</td><td>" . $row['Retailer'] . "</td><td>" . $row['City'] . "</td><td>" . $row['State'] . "</td><td>" .
      $row['Contact'] . "</td><td>" . $row['Address1'] . "</td><td>" . $row['Address2'] . "</td><td>" . $row['Country'] . "</td><td>" . $row['Zip'] .
      "</td><td>" . $row['Email'] . "</td><td>" . $row['Website'] . "</td></tr>";  //$row['index'] the index here is a field name
      }

      echo "</table>"; //Close the table in HTML
    }
else {
  echo "No Results Found...";
}
}
?>

<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" id="searchform">
  <input  type="text" name="name">
  <input  type="submit" name="submit" value="search">
</form>

<?php
$conn->close(); //Make sure to close out the database connection
} // end of a shortcode brace
?>

我认为这肯定可以工作，但是它只是将我射向/index.php并出现错误404。我不希望它导航到另一页，只在表中显示结果。我在wordpress网站上使用它，并作为一个使用简码的插件将其实现到页面中。

表单条目表搜索结果导致打开新页面和错误

0 个答案: