How to design an Array Structure and Encode to Json using PHP?

Question

I have 2 tables in MySql

Section

ID       Name
=====================
1        Section1
2        Section2

Category

ID        SectionID     Name
=========================================
1           1           Category1
2           1           Category2
3           2           Category3

This is what I have now:

$sql_section = "select * from section";<br>
$sql_category = "select * from category";<br>
$result_section = mysql_query($sql_section) or die("Could not execute query.");
$result_category = mysql_query($sql_category) or die("Could not execute query.");

echo json_encode(???????);

And I would like to Encode JSON in PHP to get the result that looks like this:

{sections:[
{sectionName: "Section1", categoryList: [{categoryName: "category1"},
              {categoryName: "category2"}]},
{sectionName: "Section1", categoryList: [{categoryName: "category3"}]}<br>
]}

Any clue of how can I design an array that looks like this?

Please don't use the deprecated mysql_XXX functions, use mysqli or PDO. — Barmar, Aug 16 '12 at 07:30
Read http://stackoverflow.com/questions/548986/mysql-vs-mysqli-in-php — Barmar, Aug 16 '12 at 15:00

TigerTiger · Accepted Answer · 2012-08-16T08:27:28.693

$arr = array('sections' => array());
$arr['sections'][] = array('sectionName' => array('categoryList' => array( array('categoryName' => 'Category 1'), array('categoryName' => 'Category 2'))));
$arr['sections'][] = array('sectionName' => array('categoryList' => array( array('categoryName' => 'Category 3'), array('categoryName' => 'Category 4'))));
echo json_encode($arr);

output://

{"sections":[
   {"sectionName":
      {"categoryList":
         [{"categoryName":"Category 1"},
          {"categoryName":"Category 2"}]}
      },
    {"sectionName":
      {"categoryList":
         [{"categoryName":"Category 3"},{"categoryName":"Category 4"}]}}]}

You'll just need to replace the string values with variables and put it in the loop to create the required data set.

score 0 · Answer 2 · answered Aug 16 '12 at 07:39

Something like this should work.

$sections = mysql_query("select * from section") or die("Could not execute query.");
$result = array();
if(mysql_num_rows($sections)>0) {
    while($section = mysql_fetch_assoc($sections))   {
        $result['sections'][$section['ID']] = $section['Name'];
        $categories = mysql_query("select * from category where SectionID='".mysql_real_escape_string($section['ID'])."'");
        if(mysql_num_rows($categories)>0) {
            while($category = mysql_fetch_assoc($categories))  {
                    $result['sections'][$section['ID']]['categoryList'][$category['ID']] = $category['Name']; 
            }
        }
    }
}

echo json_encode($result);

It will output like below, instead of sectionName as index I used section ID which is better. Same for the categories.

{sections:[
{sectionID: "SectionName", categoryList: [{categoryID: "categoryName"},
              {categoryName: "category2"}]},
{sectionID: "SectionName", categoryList: [{categoryID: "categoryName"}]}<br>
]}

score 0 · Answer 3 · answered Aug 16 '12 at 07:42

$sections = array();
$categories = array();
while ($row = mysql_fetch_object($result_section))
  $sections[$row->ID] = array('sectionName' => $row->Name, 'categoryList' => array());
while ($row = mysql_fetch_object($result_category))
  $sections[$row->sectionID]['categoryList'][] = array('categoryName' => $row->Name);

How to design an Array Structure and Encode to Json using PHP?

3 Answers3

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