super easy questions

[walterbyrd]

I am trying to get user input from a php form. Problem is: the form
doesn't wait for the user to press the submit button, but falls right
through. Therefore the variable I am trying to get, is never set. I
know it has to be something simple, but I'm missing it. Here is the
code:

echo "<form action=\"$PHP_SELF\" method=\"post\">";
echo "Name of new database to create: ";
echo "<br><input type=\"text\" name=\"new_db\">";
echo "<input type=\"submit\" value=\"submit\"><br><br>";
if ($new_db)
{
$create_db = 'CREATE DATABASE '.$new_db;
if (mysql_query($create_db, $conn))
{
echo "Database $new_db created successfully\n";
}
else
{
echo 'Error creating database: ' . mysql_error() . "\n";
}
}
else
{
echo ("new database not created");
};

[noone]

wa********@iname.com wrote:

I am trying to get user input from a php form. Problem is: the form
doesn't wait for the user to press the submit button, but falls right
through. Therefore the variable I am trying to get, is never set. I
know it has to be something simple, but I'm missing it. Here is the
code: echo "<form action=\"$PHP_SELF\" method=\"post\">";
echo "Name of new database to create: ";
echo "<br><input type=\"text\" name=\"new_db\">";
echo "<input type=\"submit\" value=\"submit\"><br><br>";
if ($new_db)
{
$create_db = 'CREATE DATABASE '.$new_db;
if (mysql_query($create_db, $conn))
{
echo "Database $new_db created successfully\n";
}
else
{
echo 'Error creating database: ' . mysql_error() . "\n";
}
}
else
{
echo ("new database not created");
};

You need something that stops at the form - I have used either a _SESSION
variable or a hidden variable.
like this:

<?php

if (isset($_POST['hiddenvar']))
// hidden variable set - process input form
{
if ($new_db)
{
$create_db = 'CREATE DATABASE '.$new_db;
if (mysql_query($create_db, $conn))
{
echo "Database $new_db created successfully\n";
}
else
{
echo 'Error creating database: ' . mysql_error() . "\n";
}
}
else
{
echo ("new database not created");
};
}
// hidden variable not set - display form
else {
echo "<form action=\"$PHP_SELF\" method=\"post\">";
echo "Name of new database to create: ";
echo "<br><input type=\"text\" name=\"new_db\">";
echo "<input type=hidden value=1 name=hiddenvar>";
echo "<input type=\"submit\" value=\"submit\"><br><br>";

}
?>

[fiziwig]

You have it structured as if you expect it to wait for the form the
first time through. Remember, the first time the page is displayed
there is no input, so ALL the php ont the page will be executed, but
there will be no data for it to work on. Only AFTER the user clicks the
submit button and THE PAGE IS RELOADED will the data be available.

Most often, code to process the data is put before the table itself
since you will need to get that data THE SECOND TIME THROUGH to make
the form entries sticky.

For example:

<?
if (isset($_POST['submitted'])) {
$input_value=$_POST['stuff'];
}
?>
<FORM ACTION="thispage.php" METHOD='post'>
<INPUT TYPE='text' name="stuff" VALUE="<? echo $input_value;?>">
<INPUT TYPE="hidden" NAME="submitted" VALUE="TRUE">
<INPUT TYPE="submit" VALUE="DO IT">
</FORM>

Notice that by picking the value out of the $_POST array before you get
to the FORM tage then the value that was entered in the form THE FIRST
TIME THROUGH, gets stuck in as the default value in the form THE SECOND
TIME THROUGH.

--gary