CODEWITHSUNDEEP
phpjqueryjsonajax
yank
yank

Reputation: 125

How to display json_encode value in html?

I am very new to ajax and json stuff. Here is jQuery:

$('#myHref').change(function(){     
    $.get('get_projectName.php',{id:value},function(data)
    {  
        data = JSON.parse(data); 
        $( '#detail' ).val(data.a);
        $( '#sector' ).val(data.b);
        $( '#unit' ).val(data.c);   
    });
});

In get_projectName.php

 $a=5;$b=10;$c=15;
 json_encode(array(
                   'a' => $project_code,
                   'b' => $b,
                   'c' => $c
                  ));

I want to display value of

$a, $b and $c

In div

enter image description here

detail, sector and unit

But I cannot display them.

Upvotes: 0

Views: 1749

Answers (2)

James
James

Reputation: 106

If I understand, you've managed to get the PHP to echo the results in JSON format and you're pulling this through to the page with the DIV via Ajax.

You'll want to:

1) Download the entire JSON response

2) Parse the response to split the three answers

3) Display the answers individually in the divs

//Step 1: Download the entire JSON response
$.get( "get_projectName.php", function( data ) {

  //Step 2: Parse the response
  result = JSON.parse(data);

  //Step 3: Load the responses into each div
  $( '#detail' ).html( result['a']);
  $( '#sector' ).html( result['b']);
  $( '#unit' ).html( result['c']);
});

If the "Detail", "Sector" and "Unit" elements are inputs and not divs, use .val instead of .html

If you need any more info, don't hesitate to let me know! :)

Upvotes: 1

Thamilhan
Thamilhan

Reputation: 13293

You are missing printing like echo. Your PHP should be:

$a=5;$b=10;$c=15;
echo json_encode(array(
    'a' => $a,
    'b' => $b,
    'c' => $c
    ));

JQuery:

$.get('1.php',{id:value},function(data)
    {  data = JSON.parse(data); 
      $( '#detail' ).html(data.a);
      $( '#sector' ).html(data.b);
      $( '#unit' ).html(data.c);  
    });

Upvotes: 4

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