building json by using mysql fetch result

Question

hello i want to use my mysql result to build the json objects. my first query is fetching parent menu and another query is fetching child menu so i am going to put my child menu result under the parent menu.. i want to know what is wrong with my code and how to correct

<?php
    $selectparentMenu=mysql_query("SELECT `id`,`item_name`,`menu_type`,`parent` FROM `epic_master_menu` where parent=0");
    //echo $myfetch=mysql_fetch_array($selectMenu);
        if(mysql_num_rows($selectparentMenu)>1) {
    while($fetchparentMenu=mysql_fetch_array($selectparentMenu)) {
         //echo 'parent id is' .$parentId=$fetchparentMenu['id']. '<br/>';
        // echo 'parent name is' .$parentId=$fetchparentMenu['item_name']. '<br/>';


         $selectchildMenu=mysql_query("SELECT `id` , `item_name` , `menu_type` , `parent` FROM `epic_master_menu`
    WHERE parent >0 AND `menu_type` = 'item' AND `parent` ='".$fetchparentMenu['id']."'");
    if(mysql_num_rows($selectchildMenu)>1) {
         while($fetchchildMenu=mysql_fetch_array($selectchildMenu)) {

             $textjson = '{
    "dataSource": [{
            "id": "", "text": "Select All", "expanded": "true", "spriteCssClass": "rootfolder", "items": [
                {
                    "id": "'.$fetchparentMenu["id"].'", "text": "'.$fetchparentMenu["item_name"].'", "expanded": true,"spriteCssClass": "folder", "items": [
                        { "id": "'.$fetchchildMenu["id"].'", "text": "'.$fetchparentMenu["item_name"].'", "spriteCssClass": "html" },
                        { "id": "'.$fetchchildMenu["id"].'", "text": "'.$fetchparentMenu["item_name"].'", "spriteCssClass": "html" },
                        { "id": "'.$fetchchildMenu["id"].'", "text": "'.$fetchparentMenu["item_name"].'", "spriteCssClass": "image" }
                    ]
                }
            ]
        }]
    }';

             }  }

         /* $fetchMenu['item_name'];
           $fetchMenu['menu_type'];
           $fetchMenu['parent'];*/
        //print_r($fetchMenu);
        } }

// my json data is

 $textjson = '{
"dataSource": [{
        "id": 1, "text": "My Documents", "expanded": "true", "spriteCssClass": "rootfolder", "items": [
            {
                "id": 2, "text": "Project", "expanded": true,"spriteCssClass": "folder", "items": [
                    { "id": 3, "text": "about.html", "spriteCssClass": "html" },
                    { "id": 4, "text": "index.html", "spriteCssClass": "html" },
                    { "id": 5, "text": "logo.png", "spriteCssClass": "image" }
                ]
            },
            {
                "id": 6, "text": "New Web Site", "expanded": true, "spriteCssClass": "folder", "items": [
                    { "id": 7, "text": "mockup.jpg", "spriteCssClass": "image" },
                    { "id": 8, "text": "Research.pdf", "spriteCssClass": "pdf" }
                ]
            },
            {
                "id": 9, "text": "Reports", "expanded": true, "spriteCssClass": "folder", "items": [
                    { "id": 10, "text": "February.pdf", "spriteCssClass": "pdf" },
                    { "id": 11, "text": "March.pdf", "spriteCssClass": "pdf" },
                    { "id": 12, "text": "April.pdf", "spriteCssClass": "pdf" }
                ]
            }
        ]
    }]
}';

would you please clarify it i mean where and how should i implement — Sachin
– Sachin, Commented Feb 25, 2014 at 8:53

Community · Accepted Answer · 2023-11-17 19:55:19Z

4

Instead of creating your own version.. you could simply make use of json_encode()

while($fetchchildMenu=mysql_fetch_array($selectchildMenu))
 {
      $somearr[]=$fetchchildMenu;
 }

$jsonData = json_encode($somearr);
echo $jsonData; //<---- Prints your JSON data

edited Nov 17, 2023 at 19:55

CommunityBot

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answered Feb 25, 2014 at 8:53

Shankar Narayana Damodaran

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6 Comments

Sachin Over a year ago

i have my own json format and i just want to insert my query record on it

Robert Over a year ago

so format the array to your format and then use json_encode on it

Shankar Narayana Damodaran Over a year ago

@Sachin, Creating your own version is a pretty headache. Do like Robert mentioned.

Sachin Over a year ago

this is my json formta

Sachin Over a year ago

i know how to encode the array into json.. but i am actuly not getting how to build an array that will display in my json format

|

Community · Accepted Answer · 2023-11-17 19:32:19Z

0

Its much easier to use the included json functions from PHP. Here you can use json_encode to create a json string from an array.

$childArray = array();

while($fetchchildMenu=mysql_fetch_array($selectchildMenu)) {
   $childArray[] = array(
       'id' => $fetchchildMenu['id'],
       'text' => $fetchchildMenu['text']
   );
}

$jsonDataChilds = json_encode($childArray);
echo $jsonDataChilds;

edited Nov 17, 2023 at 19:32

CommunityBot

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answered Feb 25, 2014 at 8:56

René Höhle

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3 Comments

Sachin Over a year ago

i have edited my question i have put my json format there so if i want to use array then how i need to build the array by using the same format of json.

René Höhle Over a year ago

To avoid mistakes in your schema its easier to build an array and convert it to the final json string using the json functions. Don't reinvent the wheel its not necessary.

Sachin Over a year ago

i just want to know how to build an array which will show me my required json format..

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