how to chose MySQL table via drop down list - php

Question

for example i have
i have two table in my datbase 1st is list1 and 2nd is list2
<select>
<option name='select' value="0">Select</option>
<option value="list1">List1</option>
<option value="List2">List2</option>
</select>
suppose user in drop down option chose list 1 then data insert in list1 option
if user select list2 then data insert in to list2 how to do this please help me to fix this issue
thanks
mysql_query("INSERT list1 SET title='$titile', subject='$subject'")
and here is complete code
<?php
}
//connect to database
mysql_connect('localhost','root','');
mysql_select_db('pdsd');
// check if the form has been submitted. If it has, start to process the form and save it to the database
if (isset($_POST['submit']))
{
// get form data, making sure it is valid
$title = mysql_real_escape_string(htmlspecialchars($_POST['title']));
$subject = mysql_real_escape_string(htmlspecialchars($_POST['subject']));
// check to make sure both fields are entered
if ($title == '' || $subject == '')
{
// generate error message
$error = 'ERROR: Please fill in all required fields!';
// if either field is blank, display the form again
renderForm($title, $subject,$date, $error);
}
else
{
// save the data to the database
$tables = array('list1', 'list2');
if (in_array($_POST['select'], $tables)) {
mysql_query("INSERT {$_POST['select']} SET title='$title',subject='$subject'");
}
or die(mysql_error());
echo "<center>Succesfully add</center>";
echo "<script>setTimeout(\"location.href = 'login.php';\",1500);</script>";
// once saved, redirect back to the view page
}
}
else
// if the form hasn't been submitted, display the form
{
renderForm('','','','','','','','','','','','','','','','','','','','');
}
?>

Answer 1

As I stated in comments, use a conditional statement and 2 separate queries based on the conditional statement and what the choice equals to the value chosen.
For example and assuming you are using as pure PHP and using a form:
Sidenote: You will need use your own queries here, as seen in commented // query for LIST X.
Another sidenote: The name attribute belongs to <select> and not <option>.
Last sidenote: My omission of action="" is equal to "self". So, you can add action="handler.php" to it if you wish to use separate files.
<form method="post">
<select name="select">
<option value="0">Select</option>
<option value="list1">List1</option>
<option value="list2">List2</option>
</select>
<input type = "submit" name = "submit" value = "Submit">
</form>
<?php
if(isset($_POST['submit'])){
if(isset($_POST['select']) && $_POST['select'] == 'list1'){
// query for LIST 1
}
if(isset($_POST['select']) && $_POST['select'] == 'list2'){
// query for LIST 2
}
if(isset($_POST['select']) && $_POST['select'] == '0'){
// Do nothing
}
}
This is but an example and the use of a prepared statement should be taken into account.
https://en.wikipedia.org/wiki/Prepared_statement
Other references you should read related to MySQL:
http://dev.mysql.com/doc/en/insert.html
http://dev.mysql.com/doc/en/update.html
Edit:
You can also use a switch/case statement:
if(isset($_POST['submit'])){
switch($_POST['select']) {
case 'list1':
// query for LIST 1
break;
case 'list2':
// query for LIST 2
break;
case '0':
// Do nothing
break;
}
}

Answer 2

You should first verify that the input is valid, then you can just substitute it into the SQL.
$tables = array('list1', 'list2');
if (in_array($_POST['select'], $tables)) {
mysql_query("INSERT INTO {$_POST['select']} SET title='$titile', subject='$subject'") or die(mysql_error());
}
Make sure you're properly escaping the variables $titile and $subject if they derive from user input, to protect against SQL-inject (use mysql_real_escape_string()). It would be even better if you used MySQLI or PDO so you could use a prepared statement instead of substituting variables into the query.