new to php, have a noob php question!

starkes · 09-12-2008, 10:34 AM

Hey there.

I'm trying to make a webpage that, through the use of php, will let me enter some information into forms and pass it on to a bash script.

I have a really simple page working where I can enter information into a form and submit it, and I am able to run a bash script through php. I am also able to print out what was entered in the form, but when I try to run the bash script and give it what was entered in the form, I only get the variable name, not its contents.

the whole thing goes like this

text from form goes to bash script ---> bash script relays this to a file to be checked ---> file contains only the word name.

this is my index.html with the forms
---------------------------
<html>
<body>
<form action="test.php" method="post">
Name: <input type="text" name="name" />
Age: <input type="text" name="age" />
<input type="submit" />
</body>
</html>

this is my test.php that relays the form's stuff to the bash script
-----------------------------
<html>
<body>
<?php
// outputs the username that owns the running php/httpd process
// (on a system with the "whoami" executable in the path)
echo exec('test.sh' . " " . '$_POST["name"]');
echo 'You entered ' . $_POST["name"] . "!";
?>
</form>
</body>
</html>

the bash script IS executing and passing the first argument to it into a file I can check. But instead of the file containing what you entered in the form, it only contains "name". What throws me off is that I can print the contents of name in the webpage just fine.

in short:

echo exec('test.sh' . " " . '$_POST["name"]'); ends up sending just "name" to test.sh

echo 'You entered ' . $_POST["name"] . "!"; prints the contents of name.

how is this possible? how do i get the contents of name passed into the bash script?

thanks.

keefaz · 09-12-2008, 10:51 AM

I think the quotes are not correct, try:

PHP Code:



echo exec('test.sh' . ' ' . $_POST['name']);

starkes · 09-12-2008, 11:03 AM

damn it! thank you, that was definitely it.

graemef · 09-12-2008, 05:39 PM

In php the double quote is a suggestion to the interpreter to expand any variable, whilst the single quote will never expend any variable. Thus $i = 7; echo '$i'."=$i"; will result in $i=7 being displayed. Things get a little more complicated when you have arrays, since (if I recall correctly) the interpreter will expand the first thing that it can hence, "$POST['name']" would be expanded to Array['name'], since $POST is an array. To force the array element to be expanded in double quotes you can include them in braces thus "{$POST['name']}". This means that you exec call could be exec("test.sh {$_POST['name']}");