How to get data from mysql db which saved as an array

Question

I want to get the data from my mysql db , but one column is saved data as an array ... How can i get the values separately . Please advice...

This is how Data saved in db for the dropdown values

{"value":["Authority ","Boards"],"order":["1","2"]}

mysql query

SELECT a.select FROM sltdb_cddir_fields a WHERE a.categories_id=81

What i want is to get Authrity and Boards as two data values instead of this array ...

Please advice

 $searchg=$_GET["term"];
// $query=mysql_query("SELECT * FROM sltdb_cddir_content where title like '%".$searchg."%'AND categories_id=81 order by title ASC ");
$query=mysql_query ("SELECT a.select FROM sltdb_cddir_fields a WHERE  a.select like '%".$searchg."%'  a.categories_value_id=19 ");
$json=array();
    while($display=mysql_fetch_array($query)){
         $json[]=array(
                    'value'=> $display["title"],
                    'label'=>$display["title"]
                        );
    }

 echo json_encode($json);
?>

@Karthick Kumar Ganesh : i wrote a query to get data but its output the same as which saved in array . you ask me to unserialize the query results is it ? — fdz
– fdz, Commented Feb 25, 2014 at 12:55
@Karthick Kumar Ganesh : im using this for my jquery autocomplete and this json array is passed as source url — fdz
– fdz, Commented Feb 25, 2014 at 13:06

ceciliajoanna · Accepted Answer · 2014-02-25 13:03:15Z

1

The data format stored is in JSON format. Hence, you could obtain the data and parse them using the function json_decode(); For instance:

    $data = '{"value":["Authority ","Boards"],"order":["1","2"]}';
    $objJSON = json_decode($data);
    $values = $objJSON->value;  //contains "Authority ","Boards"
    $orders = $objJSON->order; //contains "1","2"

edited Feb 25, 2014 at 13:03

answered Feb 25, 2014 at 13:01

ceciliajoanna

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Comments

Community · Accepted Answer · 2023-11-17 19:53:52Z

0

you can use implode to change array to string like this

while($display=mysql_fetch_array($query)){
         $json[]=array(
                    'value'=> implode(",", $display["title"]),
                    'label'=>implode(",", $display["title"])
                        );
    }

or even you can use serialize() instead of implode

edited Nov 17, 2023 at 19:53

CommunityBot

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answered Feb 25, 2014 at 12:59

krishna

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1 Comment

fdz Over a year ago

:can you please check this kidly stackoverflow.com/questions/22163466/…

krish · Accepted Answer · 2014-02-25 13:01:47Z

0

try this

Use the FIND_IN_SET function:

SELECT t.*
  FROM YOUR_TABLE t
 WHERE FIND_IN_SET(3, t.ids) > 0

answered Feb 25, 2014 at 13:01

krish

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Comments

Peter · Accepted Answer · 2014-02-25 13:13:42Z

0

Mysqli offers several result types, mysqli_fetch_all() for instance returns an associative array. If you want the data formatted in a non standard way, chances are you'll need to loop through it and build your data structures yourself.

$json = array();

while($row = mysqli_fetch_assoc($result)) {
    $json[$row['order']] = $row['value'];
}

return json_encode($json);

In JavaScript:

$.each(json, function(key, value) {
    html += '<option value="' + key + '">' + value + '</option>';
});

edited Feb 25, 2014 at 13:13

answered Feb 25, 2014 at 12:56

Peter

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4 Comments

fdz Over a year ago

i want this to on my json encode code, i have update my question as well , how to use this for that code . please advice

Peter Over a year ago

You should make an associative array, json_encode will convert that to a javascript object, then you can use $.each() from jquery to iterate through your object.

fdz Over a year ago

im using this for jquery autocomplete jQuery(function(){ jQuery("#searchp").autocomplete({ , how can i put $.each, pls advice

fdz Over a year ago

you please check this kindly stackoverflow.com/questions/22163466/…

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