PHP passing variable through href

Question

I'm having a little problem with passing a parameter for a query to another page.

I want to make the name of the restaurant a link that would pass the name of the product to process a query on the next page.

echo "<p class='p2'><a class='link' href='restaurant.php?name=". $row['name'] ."'><strong>". $row['name'] ."</strong></a>

on the restaurant page

<?php
require ("db.php");
$name = $_GET['name'];

$query = "SELECT * FROM restaurant WHERE name =\"$name\"";
$result = mysql_query($query);
$row = mysql_fetch_array($result);   
?>

but nothing is displayed.

Any idea what I'm doing wrong?

Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information. — Madara's Ghost
– Madara's Ghost, Commented Jul 26, 2012 at 20:28
you need to learn about SQL injection, especially when you are a beginner: bobby-tables.com — Jakub
– Jakub, Commented Jul 26, 2012 at 20:28
What name are you trying to pass? You may have to use urlencode() to pass the name in the url. — Johnnyoh
– Johnnyoh, Commented Jul 26, 2012 at 20:29
Stupid question, but... did you var_dump or print_r your $row? I mean, on the restaurant page. — Palladium
– Palladium, Commented Jul 26, 2012 at 20:34

Madara's Ghost · Accepted Answer · 2012-07-26 20:30:19Z

4

First, a note: you should pass an ID rather than the name, since certain characters aren't great in URLs.

Second, try using urlencode() on the name.

echo "<p class='p2'><a class='link' href='restaurant.php?name=". urlencode($row['name']) ."'><strong>". $row['name'] ."</strong></a>

edited Jul 26, 2012 at 20:30

Madara's Ghost

176k52 gold badges274 silver badges314 bronze badges

answered Jul 26, 2012 at 20:29

Matt

7,0504 gold badges31 silver badges50 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Matt Over a year ago

@user1354275 you should also make a note of what everyone else is saying about your code being wide open to SQL injection and mysql_* functions being deprecated.

Nikola K. · Accepted Answer · 2012-07-26 20:39:37Z

1

Of course it won't display anything, because you don't have any print functions (echo, print, var_dump, ...) in your file.

Anyways, you probably thought that your query doesn't work. If so, try to echo your $row['name']. If everything's OK, check if your variable is set, but it probably isn't because you get null.

To fix that issue, use isset() or empty().

Example:

if(!empty($_GET['name'])) $name = $_GET['name'];
else die('Variable name is empty');

Try also to add ini_set('display_errors', true) to the top of your pages to see if there's any errors.

Note that your code is very insecure and vulnerable. Use mysql_real_escape_string() before executing queries.

Example:

$name = mysql_real_escape_string($_GET['name']);

edited Jul 26, 2012 at 20:39

answered Jul 26, 2012 at 20:31

Nikola K.

7,15313 gold badges33 silver badges39 bronze badges

Collectives™ on Stack Overflow

PHP passing variable through href

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related