Using PHP variable in mysql

Question

I want to pass a php variable to 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. Also I tried hardcoding 217 in the place of $tty in the query. There it works.

I am also aware that this problem can be solved by mysqli/PDO. Can someone be kind enough to help me with the code in that case?

Duplicate of http://stackoverflow.com/questions/7537377/how-to-include-a-php-variable-inside-a-mysql-insert-statement/7537500#7537500 — Your Common Sense, Apr 05 '14 at 05:42
Not to mention that you posted this question already but for some reason deleted it — Your Common Sense, Apr 05 '14 at 05:43

score 0 · Answer 1 · edited Apr 05 '14 at 06:40

0

you have adding extra dots in query try below

$num_of_comments = mysql_query("SELECT count(*) FROM comments WHERE img_id = '$tty'");

This (mysql_*) extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the Prepared Statements of MySQLi or PDO_MySQL extensions should be used to ward off SQL Injection attacks !

edited Apr 05 '14 at 06:40

Shankar Narayana Damodaran

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answered Apr 05 '14 at 05:39

Rakesh Sharma

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i haven't dowvoted but probably because `mysql_query` – itachi Apr 05 '14 at 06:18

score 0 · Answer 2 · edited Apr 05 '14 at 06:41

0

You have error in your sql. Try with this.

$num_of_comments = mysql_query("SELECT count(*) FROM comments WHERE img_id ='$tty'");

This (mysql_*) extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the Prepared Statements of MySQLi or PDO_MySQL extension should be used to ward off SQL Injection attacks !

edited Apr 05 '14 at 06:41

Shankar Narayana Damodaran

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answered Apr 05 '14 at 05:39

Jenz

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Why downvote for this solution? .The query will definitely won't work due to the extra dot in it. – Jenz Apr 05 '14 at 05:45

score 0 · Answer 3 · edited Apr 05 '14 at 06:41

Follow this

Replace

 $num_of_comments = mysql_query("SELECT count(*) FROM comments WHERE img_id =  '.$tty.'");

with this

 $num_of_comments = mysql_query("SELECT count(*) FROM comments WHERE img_id ='$tty'");

This (mysql_*) extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the Prepared Statements of MySQLi or PDO_MySQL extension should be used to ward off SQL Injection attacks !

score 0 · Answer 4 · answered Apr 05 '14 at 05:45

0

try this

$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 ;

answered Apr 05 '14 at 05:45

user1844933

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score 0 · Answer 5 · answered Apr 05 '14 at 06:42

you have to make difference between fetch_array or fetch_num

try this:

 $tty = '217';   

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

Using PHP variable in mysql

5 Answers5