What's wrong with this sql query and mysql_fetch_array?

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

Did you define a custom function query()? If not, you have not called mysql_query(). Turn up error reporting and display errors on screen, always when developing code. error_reporting(E_ALL); ini_set('display_errors', 1); — Michael Berkowski
– Michael Berkowski, Commented May 1, 2014 at 14:19
The problem is not with the query; it's with your JSON being parsed. You probably have malformed JSON. — Alex W
– Alex W, Commented May 1, 2014 at 14:21
Also unless query() is a custom function that creates the sql command it looks like you are using prepared statements which is mysqli or pdo and you can't use mysql_fetch_array with it. — Pitchinnate
– Pitchinnate, Commented May 1, 2014 at 14:22

Pitchinnate · Accepted Answer · 2014-05-01 14:32:10Z

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.

edited May 1, 2014 at 14:32

answered May 1, 2014 at 14:26

Pitchinnate

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Comments

Zakari · Accepted Answer · 2014-05-01 14:28:13Z

0

maybe you have to json_encode the result?

//return json
return json_encode(array('result'=>$rows));

answered May 1, 2014 at 14:28

Zakari

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