Returns Mysql SQL results in JSON format

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 screenshot from Chrome browser .

How can I make this to work?

Thanks in advance

Also, I'm not familiar with JSON so I'm not really sure my JSON structure is correct for the TypeAhead to work... But I'll work on that later (as soon as this is solved, which I think the reason this snippet won't run) — potasmic
– potasmic, Commented Sep 22, 2013 at 15:30
Isn't the answer type set by the client application and then just used by the server for the response? — Kokozaurus
– Kokozaurus, Commented Sep 22, 2013 at 15:33

Ry- · Accepted Answer · 2013-09-22 15:29:09Z

2

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

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

answered Sep 22, 2013 at 15:29

Ry-♦

226k56 gold badges496 silver badges504 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

potasmic Over a year ago

Thanks! I used this code because I don't feel like using the json encoder.

Ry- Over a year ago

@KoiFish: Er, you should actually use both. What’s wrong with a JSON encoder? It will be a lot safer. And as it is, you have SQL injection problems, too…

Explosion Pills · Accepted Answer · 2013-09-22 15:31:21Z

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

answered Sep 22, 2013 at 15:31

Explosion Pills

192k56 gold badges341 silver badges417 bronze badges

Comments

zzlalani · Accepted Answer · 2013-09-22 15:41:16Z

0

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);

answered Sep 22, 2013 at 15:41

zzlalani

24.6k16 gold badges48 silver badges73 bronze badges

Comments

Web Hkp · Accepted Answer · 2013-09-22 15:51:45Z

0

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.

answered Sep 22, 2013 at 15:51

Web Hkp

6026 silver badges5 bronze badges

Collectives™ on Stack Overflow

Returns Mysql SQL results in JSON format

4 Answers 4

2 Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

2 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related