以JSON格式返回Mysql SQL结果

Question

I am trying to implement Typeahead.js to my site.

The typeahead.js will take from a remote page that would return JSON,
something like: http://example.org/search?q=%QUERY

For my site, this is what I've wrote for the PHP:

$q=mysql_real_escape_string($_GET['q']);

$getship= @mysql_query('SELECT * FROM `tbl` WHERE data1 LIKE \'%'.$q.'%\' OR schar LIKE \'%'.$q.'%\';');

while($tbl=mysql_fetch_array($getship)){
    $id=$tbl['id'];
    $data1=$tbl['data1'];
    $fplod=explode(" ",$data1);
    $data2=$tbl['data2'];
    $splod=explode(" ",$data2);
    $data3=$tbl['data3'];
    $data4=$tbl['data4'];
    echo '{
            "value":'.$id.',
            "tokens":["'.$fplod[0].'","'.$fplod[1].'","'.$splod[0].'","'.$splod[1].'"],
            "data1" :"'.$data1.'",
            "data2":"'.$data2.'",
            "data3":"'.$data3.'",
            "data4":"'.$data4.'"
        }';
}

But when ever I ask that typeahead thing to return, it seems to return in text/html and not application/json .

How can I make this to work?

Thanks in advance

Answer 1

You can set the Content-Type header yourself. Before any output is sent, call header :

header('Content-Type: application/json');

Answer 2

That is not valid JSON. You do not have quotes around the names. PHP has built in json encoding/decoding functions already so you don't have to build the string yourself.

echo json_encode(array("value" => $id /* etc

Answer 3

It is not a good practice to convert your data into json manually, rather use json_encode

$data = array();
while($tbl=mysql_fetch_array($getship)){
   $data[] = $tbl;
}
$return = array("data"=>$data);
echo json_encode($return);

Answer 4

try something like:

while($row = mysql_fetch_array($getship, MYSQL_ASSOC))
{
    $row_set[] = $row;
}

echo json_encode($row_set);

you can also use mysql_fetch_assoc.