Reputation: 25
Okay, so I'm trying to create an app for iOS.
My problem is that, what I think, when using JSON and combine it with sql, somehow it does not work.
So this is my code
<?php
$user='Admin';
echo $user;
$host='HOST';
$user='USER';
$password='PASSWORD';
$connection = mysql_connect($host,$user,$password);
if(!$connection){
die('Connection Failed');
}
else{
$dbconnect = @mysql_select_db('DATABASE', $connection);
if(!$dbconnect){
die('Could not connect to Database');
}
else{
$query = "SELECT * FROM AppLogin WHERE Username = '".$user."' ";
$resultset = mysql_query($query, $connection);
$records= array();
while($r = mysql_fetch_assoc($resultset)){
$records[] = $r;
}
echo json_encode($records);
}
}
?>
The output I get is
Admin[]
If I do it like this
<?php
$host='HOST';
$user='USER';
$password='PASSWORD';
$connection = mysql_connect($host,$user,$password);
if(!$connection){
die('Connection Failed');
}
else{
$dbconnect = @mysql_select_db('DATABASE', $connection);
if(!$dbconnect){
die('Could not connect to Database');
}
else{
$query = "SELECT * FROM AppLogin WHERE Username = 'Admin' ";
$resultset = mysql_query($query, $connection);
$records= array();
while($r = mysql_fetch_assoc($resultset)){
$records[] = $r;
}
echo json_encode($records);
}
}
?>
I get the output I want:
[{"Username":"Admin","Password":"Password"}]
Why is the sql not working when I'm using WHERE = '".$value."'
Albin
Upvotes: 0
Views: 70
Reputation: 74216
As I mentioned in comments:
"$user='Admin'; and $user='USER';. The second one is overwriting the other."
Therefore, you need to rename one of those.
Plus, you should really switch to PDO with prepared statements or mysqli_*
with prepared statements if your server supports them, as the mysql_*
functions are deprecated.
Upvotes: 2
Reputation: 5334
Try this $query = "SELECT * FROM AppLogin WHERE Username = '$user' ";
Upvotes: 0