How can i get data from mysql table - php

Question

I have a couple of easy problems.
First I am trying to get names from database where surname='lion'. I wrote php a file but it didn't work:
$con = mysql_connect("localhost","yata_ali","password");
if (!$con){
die('error: ' . mysql_error());
}
mysql_select_db("yatanada_iBess", $con);
$degisken = mysql_query("select name from people where surname LIKE '%lion%'");
if(mysql_query){
return "$degisken";
}
mysql_close($con);
?>
I wrote this code and tried to use $degisken in my xcode project. But it didn't work.
shortly i am trying to use the names whichs surname =lion in my ios project and i know i should use url.but i couldn find the code part that return name what shall i write at the end of php code ? return or something else to use in xcode.
how can i send response in php? i wonder that. what shall i write "return $name" or something else. i know call url. but i dont know whats the full php code that i shall use

Answer 1

You can't use PHP in an iOS project. You'll need to write some objective-c to call a URL on a server which returns this data in some sort of format (xml? json?) and then have the iOS app parse the response.

Answer 2

I don't think you understand how to use the mysql_* functions in PHP. Take a look at the examples on this page for guidance: http://www.php.net/manual/en/function.mysql-query.php

Answer 3

$degisken= mysql_query("select name from people where surname='lion'");
if ($degisken){
while($row = mysql_fetch_assoc($degisken))
{
echo $row["name"] . "<br/>";
}
}

Answer 4

There are a lot of errors, in your code, but the most serious are that
(a) you are running an invalid test:
if (mysql_query){ //YOU CANNOT DO THIS
(b) You cannot return "$degisken"; because $degisken is a MySQL
resource, not a string.
(c) You should not close your mysql
connection after returning something. You don't necessarily need to
close it at all, but if you're going to, close it after the query
because anything after the return won't be evaluated (assuming the
return is triggered).
(d) If you're looking for cases where the surname='lion' then don't use wildcards in the MySQL query. where surname LIKE '%lion%' will match 'scalion','lioness','slioner', etc.
Your code should look something like this:
$con = mysql_connect("localhost","yatanada_ali","sifre");
if (!$con) {
die('error: ' . mysql_error());
}
mysql_select_db("yatanada_iBess", $con);
$degisken = mysql_query("select name from people where surname LIKE '%lion%'") or die('Error: '. mysql_error());
if (mysql_num_rows($degisken)){
//your query could return lots of results, so you may want to loop through results:
while($row = mysql_fetch_array($query)){
$name = $row['name'];
//do something with the name... I'm going to echo it.
echo $name . "<br />";
}
}