Question

I have a script I wrote to return records for cases out of the database. I am returning one record for my mysql query when there are actually two records. This is what I am returning:

{ "cases": [ {"name":"Test Case for App","number":"3846"}] } 

I should see:

{ "cases": [ {"name":"Test Case for App","number": "2903"}, {"name":"Test Case 2","number": "2856"} ] }

Here is my source:

$sql = "select * from cases as c join contacts_cases as conc on c.id = conc.case_id where conc.contact_id = '1b360507'";
$query = mysql_query($sql);  

// If we find a match, create an array of data, json_encode it and echo it out  
if (mysql_num_rows($query) > 0)  
{  
    $row = mysql_fetch_assoc($query);  
    $response = array(  
        'name' => $row['name'],  
        'number' => $row['case_number']
    );  

    echo '{ "cases": [ ',  json_encode($response), "] }";

Answer 1

If you are expecting more than one result you should try

if (mysql_num_rows($query) > 0)  
{  
    $responses = array();
    while($row = mysql_fetch_assoc($query)) {
        $responses[] = array(  
            'name' => $row['name'],  
            'number' => $row['case_number']
        );  
    }
    echo '{"cases": ' . json_encode($responses) . '}';
}