PHP error can not retrieve LIKE statements from android application

I am trying to search up the database with an android application, but when I search for a string element with the SQL syntaxes 'LIKE',

such as:

SELECT * FROM `test123` WHERE (`name` LIKE '%A%')

it would give me an error:

A JSONArray text must start with '[' at character 1 of ....

but if I search for:

SELECT * FROM `test123`

it'll work fine.

<?php

mysql_connect("website.com","root","password");
mysql_select_db("test123");

$q=开发者_Python百科mysql_query($_REQUEST['sql']);
while($e=mysql_fetch_assoc($q))
    $output[]=$e;

print(json_encode($output));

mysql_close();
?>

$output = array();
while ($e = mysql_fetch_assoc($q)) {
    $output[] = $e;
}

The problem may simply be that if there are no results, $output is never defined. When you then try to json_encode($output), PHP throws a warning about $output being undefined, which leads to final output like:

PHP Notice:  Undefined variable: output in ...
null

Which is invalid JSON. Even if your warnings are suppressed, the output will just be null, which is not a JSON array. If your client expects an array, then it will be disappointed by null.

Always initialize your variables.

It should be just

"SELECT * FROM test123 WHERE name LIKE '%A%'"

i think this can be work fine SELECT * FROM test123(name) WHERE (name LIKE '%A%')

When I user php json_encode it always returns a JSONObject, not a JSONArray. Try printing your raw result to Logcat and see what form it's in.