Search user using part of the name - php

Question

I have this PHP code:
function getusers($user) {
$result = query("SELECT IdUser, username FROM login WHERE username='%s'",$user);
if (count($result['result'])>0) {
//authorized
print json_encode($result);
} else {
errorJson('Actualization failed');
}
}
But this only returns the user that matches the name exactly.
I'd like to return all users containing that name string, for example:
dani -> daniel, dani_56, dani563, elnenedani, ...
It is usually done by putting in PHP: %dani% but as I have put the %s to grab the variable $user, I do not know how to put it.
Any idea?

Answer 1

It is not a great Question. If you have searched well in Stackoverflow you would have go it the answer.. As you asked the Question the answer is.. Instead of Equal use LIKE:
function getusers($user) {
$result = query("SELECT IdUser, username FROM login WHERE username LIKE %'%s'%",$user);
if (count($result['result'])>0) {
//authorized
print json_encode($result);
} else {
errorJson('Actualization failed');
}
}
It seems the PHP code and DB is working well. Checkout the below links for the error:
iOS 5 JSON Parsing Results in Cocoa Error 3840
Cocoa error 3840 using JSON (iOS)
The Operation couldn't be completed. (Cocoa error: 3840.)
Cocoa Error 3840 - NSJSONSerialization

Answer 2

You should use the LIKE syntax. Make sure to include % to indicate wildcards:
query('SELECT IdUser, username FROM login
WHERE username LIKE "%' . $user . '%"')

Answer 3

My query() function is
function query() {
global $link;
$debug = false;
//get the sql query
$args = func_get_args();
$sql = array_shift($args);
//secure the input
for ($i=0;$i<count($args);$i++) {
$args[$i] = urldecode($args[$i]);
$args[$i] = mysqli_real_escape_string($link, $args[$i]);
}
//build the final query
$sql = vsprintf($sql, $args);
if ($debug) print $sql;
//execute and fetch the results
$result = mysqli_query($link, $sql);
if (mysqli_errno($link)==0 && $result) {
$rows = array();
if ($result!==true)
while ($d = mysqli_fetch_assoc($result)) {
array_push($rows,$d);
}
//return json
return array('result'=>$rows);
} else {
//error
return array('error'=>'Database error');
}
}

Answer 4

I do not get fixed. This code is for an ios app that uses AFNetworking, might please that helps you know what happens because I do not get it