How to query MySQL from a web browser URL ?

[karthikeyanck]

How to query MySQL from a web browser URL.

I 've a Apache server running on my Ubuntu machine which has PHP and MySQL installed.

I 've an assignment to demonstrate how SQL Injection works, I need to pass SQL Queries through my PHP Page so that i can INSERT, UPDATE, DROP data from the MySQL Database.

I need to complete the project at the earliest

Any help Appreciated !!!!!!!!!!!!!

[pbmods]

Heya, karthikeyanck.

You posted this in the Articles section. I'll go ahead and move it to the Forum where an Expert will be more likely to find it.

[ronverdonk]

When it is just simply passing an SQL query to the db and echoing the output resource ID and the array of rows returned (e.g. after a select), this snippet will do. You'll have to adapt it to your own needs.
[php]
<?php
if (isset($_POST['sql']) ) {
$sql=$_POST['sql'];
// Make a MySQL Connection
$conn = mysql_connect(" localhost", "ronverdonk ", "ronnie09")
or die("Could not connect to the db server: ".mysql_error() );
mysql_select_db ("vwso",$con n)
or die("Could not select the db: " . mysql_error());
$result=mysql_q uery($sql);
echo $result.'<br />';
if ($_POST['out'] == 'y') {
while ($row = mysql_fetch_ass oc($result)) {
echo '<pre>'; print_r($row);
}
}
}
?>
<form name="MyForm" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
Type your mySQL query:<br />
<input type="text" name="sql" size="70" value="<?php echo (isset($_POST['sql'])) ? $_POST['sql'] : ""; ?>"><br />
Do you want query output displayed?<br />
<input type="radio" name="out" value="y"<?php if ($_POST['out'] == 'y') echo " checked"; ?> />yes&nbsp;
<input type="radio" name="out" value="n"<?php if ($_POST['out'] == 'n') echo " checked"; ?> />No<br />
<input type="submit" value="submit query" />
</form>
</body>
</html>
[/php]
Ronald

[karthikeyanck]

When it is just simply passing an SQL query to the db and echoing the output resource ID and the array of rows returned (e.g. after a select), this snippet will do. You'll have to adapt it to your own needs.
[php]
<?php
if (isset($_POST['sql']) ) {
$sql=$_POST['sql'];
// Make a MySQL Connection
$conn = mysql_connect(" localhost", "ronverdonk ", "ronnie09")
or die("Could not connect to the db server: ".mysql_error() );
mysql_select_db ("vwso",$con n)
or die("Could not select the db: " . mysql_error());
$result=mysql_q uery($sql);
echo $result.'<br />';
if ($_POST['out'] == 'y') {
while ($row = mysql_fetch_ass oc($result)) {
echo '<pre>'; print_r($row);
}
}
}
?>
<form name="MyForm" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
Type your mySQL query:<br />
<input type="text" name="sql" size="70" value="<?php echo (isset($_POST['sql'])) ? $_POST['sql'] : ""; ?>"><br />
Do you want query output displayed?<br />
<input type="radio" name="out" value="y"<?php if ($_POST['out'] == 'y') echo " checked"; ?> />yes&nbsp;
<input type="radio" name="out" value="n"<?php if ($_POST['out'] == 'n') echo " checked"; ?> />No<br />
<input type="submit" value="submit query" />
</form>
</body>
</html>
[/php]
Ronald

The script works fine, but when I query the server like the one below

SELECT * FROM employee WHERE username = 'admin'

I get Warning: mysql_fetch_ass oc(): supplied argument is not a valid MySQL result ***********

[ronverdonk]

This was just a simple sample and you'll have to play with it. At my site the error is probably the insertion of backslashes. At your site I assume you have no result set. So I changed the snippet code to get rid of backslashes and test the result. Next is the first part.
[php]
<?php
if (isset($_POST['sql']) ) {
$sql=$_POST['sql'];
$sql=str_replac e('\\','',$sql) ;
$db=$_POST['db'];
// Make a MySQL Connection
$conn = mysql_connect(" localhost", "ronverdonk ", "ronnie09")
or die("Could not connect to server: ".mysql_error() );
mysql_select_db ($db,$conn)
or die("Could not select db $db: " . mysql_error());
$result=mysql_q uery($sql) or
die('Error: '.mysql_error() );
echo "<b>stateme nt:</b> $sql<br />";
if ($_POST['out'] == 'y') {
if (mysql_num_rows ($result) > 0) {
while ($row = mysql_fetch_ass oc($result)) {
echo '<pre>'; print_r($row);
}
}
else
echo 'No results';
}
}
?>
[/php]
Ronald