Making variable from HTML-file work in PHP-file

Question

I have two rather simple files. The first is an HTML-file like this:

<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<html>
<body>

<form action="test.php" method="post">
Search for: <input type="text" name="search" />
<input type="submit" name="submit" value="Search" />
</form>

</body>
</html>

The second file is test.php and it looks like this:

<?
       $filepath = "/usr/sbin";
       exec("ONE $search -command $filepath ");
       fopen($filepath, rw);
?>

My problem is that I want to use the argument "search" given in the HTML-form as value of variable in the PHP-script. ONE is a search script I made that takes one argument and I want it to be "$search".

Could this be done and if so how?

Many thanks in advance.

It should be in $_POST["search"], I think. But be very wary about taking user-entered data and passing it into an exec call. — andrewsi
– andrewsi, Commented Sep 19, 2012 at 17:48
Do not post the same question more than once - it will be considered as SPAM. — Nir Alfasi
– Nir Alfasi, Commented Sep 21, 2012 at 23:17

Horen · Accepted Answer · 2012-09-19 17:49:19Z

1

You can access that variable by using $_POST['search']

Before you use it in a shell call you should make sure that it won't damage your system.

answered Sep 19, 2012 at 17:49

Horen

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1 Comment

Paul Bergström Over a year ago

Thanks but it doesn't work. Do I need to put it in a new variables like this: search1 = $_POST['search'] or search1 = $($_POST['search']) ?

Nelson Benítez León · Accepted Answer · 2012-09-19 19:03:00Z

1

Just do this:

exec("ONE " . $_POST['search']. " -command $filepath ");

UPDATE:
You have another error, your following line:

fopen($filepath, rw);

should be like this:

fopen($filepath, "rw");

cause the second parameter to fopen() should be passed as a string.

UPDATE 2:
So as it seems, the OP want the output of the command to be printed on the page, for that you have to use passthru() instead of exec() , like so:

passthru("ONE " . $_POST['search']. " -command $filepath ");

edited Sep 19, 2012 at 19:03

answered Sep 19, 2012 at 17:49

Nelson Benítez León

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11 Comments

Ashley Davies Over a year ago

"it doesn't seem to work" isn't very helpful. Does it error? Do nothing?

Paul Bergström Over a year ago

The PHP-file should find a couple of files containing the word given in "search", but it gives a blank page. I guess the variable isn't taken.

Paul Bergström Over a year ago

PHP Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in test.php on line 3

Nelson Benítez León Over a year ago

It seems an error in your fopen call, see the update to my answer.

Paul Bergström Over a year ago

Thank you! I've done the corrections and if I run php test.php in a terminal there's no more errors except that an argument is missing and so it should be. When I fill in a searchword in the HTML-file then clicking "Search" it pops right over to the test.php - but seemingly without getting any variable since the output is blank just like when I run php test.php.

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