Mysql query variable syntax - php

Question

I am trying to pass a variable to a very basic mysql query. but php doesnt return a true value. nothing.
i have checked everything
the problem is here.
the syntax of $a varible typing into mysql query
$result = mysql_query("SELECT id,floatingnumber FROM posts WHERE id='$a' LIMIT 1");
when i change $a to 22 it returns a value otherwise nothing.
exact query is here...
$a=$this->post_id;
$result = mysql_query('SELECT floatingnumber FROM posts WHERE id="'.$a.'" LIMIT 1')or die(mysql_error());
$row = mysql_fetch_row($result);
$sdfa=$a.'-'.$row[0];
$sdfa returns "86 - " without quotes 86 - space
so the problem is on the mysql fetch row please help

Answer 1

Have you tried echoing the query to see what the real value of $a is?
echo "SELECT id,floatingnumber FROM posts WHERE id='$a' LIMIT 1";
Have you tried checking for errors?
$result = mysql_query("SELECT id,floatingnumber FROM posts WHERE id='$a' LIMIT 1") or die(mysql_error());
Also, you shouldn't even be using mysql_* as it's deprecated.
This is how you'd do it in PDO:
$stmnt = $db->prepare("SELECT id,floatingnumber FROM posts WHERE id=:id LIMIT 1");
$stmnt->bindValue( ':id' , $a , PDO::PARAM_INT );
$stmnt->execute();
$result = $stmnt->fetchAll(PDO::FETCH_ASSOC);

Answer 2

typically when I'm writing in double quotes, simply putting in the variable works:
"... $1 ..."
but also, I originally learned it with brackets
"... {$1} ..."
you can try that. also, a handy way to write queries is store the query string in its own variable so you can easily print out the query and see what you wrote before submitting.
$query = "SELECT id,floatingnumber FROM posts WHERE id=$a LIMIT 1";
$result = mysql_query( $query );
This helps identify things like this.

Answer 3

try this
$result = mysql_query("SELECT id,floatingnumber FROM posts WHERE id='".$a."' LIMIT 1");
if your $a is a number then do like that
$result = mysql_query("SELECT id,floatingnumber FROM posts WHERE id= $a LIMIT 1");
EDIT :
your code is right
$row = mysql_fetch_row($result);
$sdfa=$a.'-'.$row[0];
the problem is in your sql or table because there is no floatingnumber where id is 86 .