PAssing PHP variables inside a mysql_query function

Question

I want to pass a php variable in mysql_query such as:

$tty = '217';   

$num_of_comments = mysql_query("SELECT count(*) FROM comments WHERE img_id =  '.$tty.'");
$num_of_comments1 = mysql_fetch_array($num_of_comments);
$num_of_comments2 = $num_of_comments1[0];
echo $num_of_comments2 ;

However, I am not able to get the value needed in num_of_comments2. It returns a 0 on echo.

Use PDO or mysqli and prepared statements: SELECT count(*) FROM comments WHERE img_id = ? — jeroen
– jeroen, Commented Apr 4, 2014 at 15:53
You are entirely misusing the mysql functions. What tutorial are you using to teach you? — Andy Lester
– Andy Lester, Commented Apr 4, 2014 at 15:53
@Jeroien: Could you provide some more insights on using PDO/mysqli function for this problem. How can I write the code for the same? — Hardik Dave
– Hardik Dave, Commented Apr 4, 2014 at 15:54
You should start with a tutorial or the php manual, there are plenty of examples there. — jeroen
– jeroen, Commented Apr 4, 2014 at 15:57

Niet the Dark Absol · Accepted Answer · 2014-04-04 15:54:13Z

2

As the colour coding will show you, your query is wrong. You could also debug it by just echoing your query:

SELECT count(*) FROM comments WHERE img_id =  '.217.'

Clearly incorrect!

$tty = '217';   

$sql = mysql_query("SELECT count(*) FROM comments WHERE img_id = ".intval($tty));
$row = mysql_fetch_row($sql);
$number = $row[0];
echo $number;

Alternative one-liner for getting the value:

list($number) = mysql_fetch_row(mysql_query("select count(*) from `comments` where `img_id`=".intval($tty)));

answered Apr 4, 2014 at 15:54

Niet the Dark Absol

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2 Comments

Hardik Dave Over a year ago

How can I echo the list($number) i.e the count that I get from the sql query?

Hardik Dave Over a year ago

This still returns a 0

Companjo · Accepted Answer · 2014-04-04 15:53:51Z

0

This should work:

$tty = '217';   

$num_of_comments = mysql_query("SELECT count(*) FROM comments WHERE img_id =  '".$tty."'");
$num_of_comments1 = mysql_fetch_array($num_of_comments);
$num_of_comments2 = $num_of_comments1[0];
echo $num_of_comments2 ;

Use '".$tty."' instead of '.$tty.'

answered Apr 4, 2014 at 15:53

Companjo

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Comments

Marc B · Accepted Answer · 2014-04-04 15:54:45Z

0

Basic PHP syntax:

$num_of_comments = mysql_query("[[...snip...]]=  '.$tty.'");

You never "closed" your string, so you're trying to execute a PHP concatenation INSIDE your string, which won't work. Your query string is literally going to be

WHERE imd_id = '.217.'
                ^---^--- note the concatentation operators

For a "-quoted string, you do NOT need to concatenate:

$num_of_comments = mysql_query([[..snip..] =  '$tty'");
                                              ^^^^^^^---note: no dots

is all you need.

answered Apr 4, 2014 at 15:54

Marc B

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2 Comments

Niet the Dark Absol Over a year ago

is all you need... if you want a visit from Bobby Tables.

jeroen Over a year ago

Tempted to post one of your own enjoy having your server pwn3d comments ;-)

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