What's wrong with this sql query and mysql_fetch_array? - php

Question

These lines are from a php function on a web server, the client is an ios app,
I get an error on the result2 query
$result = query("SELECT field1 FROM table1 WHERE field2='%s' limit 1", $value);
$row = mysql_fetch_array($result);
$result2 = query("SELECT field2 FROM table1 WHERE field3='%s' limit 1", $row['field1']);
on Xcode I get the error (in json):
{
error = "The operation couldn't be completed. (Cocoa error 3840.)";
}
here is the definition of the function query
//executes a given sql query with the params and returns an array as result
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');
}
}
what's wrong with that query?
thanks

Answer 1

You are using mysqli_ functions in your query() function yet you are trying to use mysql_fetch_array() to get the results. You need to use: mysqli_fetch_array()
http://www.php.net/manual/en/mysqli-result.fetch-array.php
Actually it looks like your query() function does it for you. You shouldn't need to use a function at all. Look at this section of your query() function:
$rows = array();
if ($result!==true)
while ($d = mysqli_fetch_assoc($result)) {
array_push($rows,$d);
}
//return json
return array('result'=>$rows);
However not sure why it says it is json cause it is just a normal array not a json array. So you should just use this in your code:
$row = $result['result'][0];
This will get the first row.

Answer 2

maybe you have to json_encode the result?
//return json
return json_encode(array('result'=>$rows));