Passing A parameter From an HTML to an PHP page

[Kawal]

Hello all,

I am a student, new to PHP and have a small simple question. How do I properly pass a parameter from an HTML page to an PHP page.

HTML:
...
<form action="search.php" method="POST">
<input type="text" name="search">
<input type="submit" value="Search It">
</form>
...
(where search.php is the name of my PHP file)

PHP:

$result1 = mysql_query("SELECT * FROM keyboards where name like '$search'", $link);

where the variable $search should be the parameter passed from the HTML. I keep getting the error:

Warning: Undefined variable: search in C:\Program Files\Apache Group\Apache\htdocs/search.php on line 17

Thanks for any help in advance,

[Rohan_Hill]

Instead of using $search, you should use $POST['search']. $POST is an array of variables passed to the script from an HTML page 🙂

[Rodders]

Try this:

# Add the wildcard '%' to the search variable
$searchvar = $_POST['search'] . "%";

# Build the SQL statement
$sql = "SELECT * FROM keyboards WHERE name like \"$searchvar\" ";

# Run the query
$result1 = mysql_query($sql, $link);

Hope it works !

[Kawal]

Thanks Guys for your replies,

I've tried your suggestions but now I get another error.

Warning: Undefined variable: _POST in C:\Program Files\Apache Group\Apache\htdocs/search.php

Do you think there maybe something wrong with my setup?

Im using PHP4.0,
apache webserver,
and MySQL

Thanks again for your help

[jonsof]

You should use $HTTP_POST_VARS['search'] because the $_POST introduced in 4.1(or)2.x and over, if i am right

[Kawal]

Thanks Guys,

($HTTP_POST_VARS['search']) got that part working. I have a couple of questions if anybody has some time.

Question 1.

I have an HTML page with a link that looks like the following:

<a href="display.php?=cpus" ...

What im trying to do is pass the parameter cpus to my PHP page, when watered down looks like the following:

$tables = $HTTP_POST_VARS['table'] . "%";
$sql = "SELECT * \"$tables\" ";
$result1 = mysql_query($sql, $link); //error 2 lies here
$num_rows = mysql_num_rows($result1);
$num_fields = mysql_num_fields(mysql_db_query($dbname, $sql, $link));

I get two errors:

Error #1
Warning: Undefined index: cpus in C:\Program Files\Apache Group\Apache\htdocs/display.php on line 6

Error #2
Warning: Supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Group\Apache\htdocs/display.php on line 19

I believe for error 1 I shouldnt be using $HTTP_POST_VARS to accept the parameter from a link. Hopefully someone can show me what I should be using

Question #2

Im trying to perform a search in MySQL. I tried

$sql = "SELECT * FROM keyboards WHERE name like \"$searchvar\" ";

and it worked fine. How would I modify this statement to include more then one table?

Again, im just a student learing PHP, any help would be greatly appreciated.

[BuzzLY]

Ok... I'm a PHP newbie, so bear with me if I am wrong 🙂

First off... your URL looks wrong. Should it be:

<a href="display.php?table=cpus" ... ?

Second... if that's true, then your SQL is wrong, too. It should be:

SELECT * FROM '$tables' ...

I don't know if you just typed it in wrong or if your code is wrong. I hope I was able to help.

[Kawal]

I typed them in wrong in the forum, im still having the same problems. My coding is correct, but im still getting the errors

[Rodders]

You should replace $HTTP_POST_VARS['cpus'] with $HTTPGETVARS['cpus'].

$HTTP_POST_VARS is for use when the FORM METHOD=POST.
$HTTP_GET_VARS is for use when the FORM METHOD=GET OR to access variables passed in the URL after the '?'.

To access more than one table with the same code could be awkward. You would probably need to make all your tables have the same field definitions. In doing this, I would suggest that you improve your database design and use one table for the whole thing with a unique key field to distinguish between the rows.
e.g.

TABLE 'KEYBOARDS' - FIELDS 'MAKE', 'MODEL', 'PRICE'
TABLE 'MONITORS' - FIELDS 'MAKE', 'MODEL', 'PRICE'

could be replaced with
TABLE 'HARDWARE' - FIELDS 'TYPE', 'MAKE', 'MODEL', 'PRICE'

Then your query would just need to say 'where type= $chosentype'

OK?

[Kawal]

Thanks Rodders, the $HTTP_GET_VARS worked to get the variable from the URL. Also, for all my tables, I have it setup the way you have quoted below, where all the fields have the same name.

However im getting this one error with this line, and I dont seem to understand why.

Error: Warning: Supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Group\Apache\htdocs/cpus.php on line 19
...
$tables = $HTTP_GET_VARS['table'] . "%";
...
$sql = "SELECT * FROM \"$tables\" ";
$result1 = mysql_query($sql, $link);
$num_rows = mysql_num_rows($result1); //line 19
....
it was working fine before when I manually typed the table name in the select statement.

Thanks..

[Rodders]

I think you need to change this line:

$sql = "SELECT * FROM \"$tables\" ";

to....
$sql = "SELECT * FROM \"$table\" ";

[Kawal]

I tried that, I believe the syntax is correct, but there is a problem with "GET"ing a parameter and the functions:

mysql_query()
mysql_num_rows()

[paulg]

Originally posted by Kawal
...
$tables = $HTTP_GET_VARS['table'] . "%";
...
$sql = "SELECT * FROM \"$tables\" ";
$result1 = mysql_query($sql, $link);
$num_rows = mysql_num_rows($result1); //line 19
....

I think you're getting confused here. Adding the "%" to the $HTTP_GET_VARS[] is good for SEARCHING in a "...where var = '$tables' ... line.

Try this:-
...
$table = $HTTP_GET_VARS['table'];
$sql = "SELECT * FROM \"$table\"";
$result1 = mysql_query($sql,$link);
...

Good luck 😃

[Kawal]

Thanks for the reply,

However it did not work and I am getting the same errors. Im attaching my code if that helps. The error is on line 19 and 21.

$num_rows = mysql_num_rows($result1); //line19
$num_fields = mysql_num_fields(mysql_db_query($dbname, $sql, $link))😉 //line 21

Error: Warning: Supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Group\Apache\htdocs/cpus.php on line 19

This script works if I type the name of the table into the select statement in the script. For Example: "Select * from monitors"

Thanks