Home/ Questions/Q 6055321
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-23T08:15:07+00:00

<html> <form action=update.php method=POST name=ID> <input type=text name=ID> <input type=Submit value=Submit> </form> </html> Up

  • 0
<html>
<form action="update.php" method="POST" name="ID">
<input type="text" name="ID"">
<input type="Submit" value="Submit">
</form>
</html>

Up there is the submit form to get an ID number.
I try to get that ID entered by user ( NOTE: It’s a number) and show mysql table row coresponding to that ID.
Example : User enter 2 and row number 2 from database is shown.
My problem is that all rows are shown and not only wanted one.
– Extra Question : How can I show user an error if he entered a NULL value ?

<?php
    $id=$_POST['ID'];
    .
    .
    .
    mysql_connect($host,$username,$password);

    if (!mysql_select_db($database))
        die("Can't select database");
    $query="SELECT * FROM table WHERE ID= '$id'";
    $result = mysql_query("SELECT * FROM vbots");
    $num=mysql_num_rows($result) or die("Error: ". mysql_error(). " with query ". $query);

    mysql_close();
    .
    .
    .
?>
Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    You’re not running your query.

    You have this:

    $query="SELECT FROM table WHERE ID= '$id'";
$result = mysql_query("SELECT * FROM vbots");

    You want this:

    $query="SELECT FROM table WHERE ID= '$id'";
$result = mysql_query( $query);

    **Insert nag about SQL injection**

    • 0