Cannot print database value using PHP

Question

I've got a datbase with a simple table called "channels" which has 2 fields: id and link. I use this code in my php page to get the link of the channel with id 1:

<!DOCTYPE html>
<html>
<body>

<?php
$db = mysql_connect("localhost", "username", "");
mysql_select_db("my_database", $db);
$result = mysql_query("SELECT link FROM channels WHERE id='1'");
$data = mysql_fetch_array($result);

print $data;
?>
</body>
</html>

but what I get as output is a page with the text "Array" and anything else. id field is an INT value and link is TEXT. How can get my data correctly and print it?

it's an expected result. You need to loop over your results. — Funk Forty Niner, Nov 24 '15 at 16:37
Did you misspell channels there? Also, take a look at mysqli and prepared statements, as your approach is not secure at all. There are some examples here: http://php.net/manual/en/mysqli.prepare.php — Chris Evans, Nov 24 '15 at 16:39
You should use mysqli to connect to your database, "mysql" is deprecated. Then search in the documentation how to loop throught a resulset array. — leonardo_palma, Nov 24 '15 at 16:40
You need to [stop using `mysql_` functions](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) as they are deprecated and will soon be removed — Machavity, Nov 24 '15 at 16:46

score 1 · Accepted Answer · answered Nov 24 '15 at 16:41

Apart from the insecurity stuff (probably look at mysqli and PDO) this is how you access data

<?php 
$db = mysql_connect("localhost", "username", "");
mysql_select_db("my_database", $db);
$result = mysql_query("SELECT link FROM chhannels WHERE id='1'");
$data = mysql_fetch_array($result);

print $data["link"]; //access the first column of the first row (in this case link)
?>

This only returns the first row. To get all the rows, you have to call mysql_query repeatedly

while ($row = mysql_fetch_array($result)) {
    print $row["link"];
}

Martin Ibert · Answer 2 · 2015-11-24T17:43:53.677

What you get back from mysql_fetch_array($result) is an array that represents one row from the result set. It is an array because you could have many columns on your result row, not just one as in your example.

Try printing

$data['link']

instead.

Some posters have suggested looping over the result as a SELECT statement can return more than one row. Good idea in principle but misleading here. mysql_fetch_array() gets only one result row at a time (it is an array, indexed by column name). If you are expecting more than one result, call mysql_fetch_array() repeatedly until the result is empty (mysql_num_rows($result) is 0).

Just seeing another issue. You say that the id column is an integer value. In that case you need to compare it to an integer literal, not to a string/varchar/text literal:

SELECT link FROM channels WHERE id = 1

score 0 · Answer 3 · answered Nov 24 '15 at 16:44

0

You should create a while loop

while ($data = mysql_fetch_array($result)) {
    print $data['link'];
}

Looking at your query you probably only expect one result, you could just do print $data['link'];

answered Nov 24 '15 at 16:44

I. Bielderman

43
7

Cannot print database value using PHP

3 Answers3