cannot select a row in mysql - php

Question

EDIT1 : used double quotes and single quotes but I am getting same error.
EDIT2 : same query is returning me result in mysql shell
I am selecting a row from a table.
if(!isset($_GET['title']) || !isset($_GET['user'])){
echo "hi"; //something come here
}
else{
$title = $_GET['title'];
$title = mysqli_real_escape_string($conn,$title);
$user = $_GET['user'];
$user = mysqli_real_escape_string($conn,$user);
echo $title ;
echo $user ;
// tried giving value directly to test but no luck
$query = "SELECT * FROM site WHERE client=\"Chaitanya\" && title=\"werdfghb\" ";
$result5 = mysqli_query($conn,$query) or die(mysqli_error());
$count = mysqli_num_rows($result5);
echo $count ;
while($result9 = mysqli_fetch_array($result5)){
$kk=$result9['url'];
echo $kk ;
}
$page = $kk;
include ( 'counter.php');
addinfo($page);
}
In my database there is a row with columns title and client and the values I entered are in that row but when I echo count(no of rows) it is showing zero.
Is there anything wrong with code ?

Answer 1

The error you are getting is due to the line
$page = $kk;
in this code $kk is not declared previously. The defined $kk is in the while loop scope.
declare the variable like this in the outer scope from the while loop
...
$kk = null;
while($result9 = mysqli_fetch_array($result5)) {
$kk = $result9['url'];
echo $kk ;
}
$page = $kk;
...
Error on Fetching Data
You have to crack you SQl into smaller pieces and test the code like this.
run the query SELECT * FROM site without any where and get the count
run the query SELECT * FROM site WHERE client='Chaitanya' and get the count
SELECT * FROM site WHERE title='werdfghb' and check the count
Then run the whole query
And see the results. This way u can find out in where the issue is in your SQL code. I prefer you use the mysql client to execute this queries

Answer 2

As I pointed out in my comment, $kk is undefined in the $page = $kk;, since it is declared in the while loop.
Do something like:
$kk = ''; //can also do $kk=NULL; if you want.
while($result9 = mysqli_fetch_array($result5)) {
$kk=$result9['url'];
echo $kk ;
}
$page = $kk;

Answer 3

try this one
$client = "Chaitanya";
$title = "werdfghb";
$query="SELECT * FROM site WHERE client='".$client."' and title='".$title."' ";
you can also use this
$query="SELECT * FROM site WHERE client={$client} and title={$title} ";