onclick get data from database using ajax laravel

Question

What I am trying to do is to load data-id in drop-down menu when I click the list item, it need retrieve the data from db and append in table.

So far, I loaded the data id in drop-down menu and while on-click the list I send ajax call to retrieve data.

My code (controller page):

public function calcList($id)
{

    $designId=design::find($id);
    $design = design::select( `design_no`, `design_name`, `weight_length`, `quantity`, `design_image`, `des2`, `des3`, `des4`, `des5`, `des6`, `des7`)->where('id','=',$designId);
    // return Response::json(array('datas' => $design));
    // return response()->json(['data'=>$design]);
    return Response::json($design);
}

Ajax Request:

$(document).ready(function() {
    $('.designList').click(function() {
        $.ajaxSetup({
            headers: {
                'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
            }
        });

        var id =($(this).attr("data-id"));
        $.ajax({
             type:"POST",
             url:"/calc_list/"+id,
             success : function(results) {
                 console.log(results);
             }
        }); 
    });
});

On the console log, the results print like {..}
How to built the json to table?

@SagarGautam i dont want to reload the current page,without reloading i want to load data multiple time . if there is any way to archive the same pls suggest me — Harish Karthick
– Harish Karthick, Commented Sep 3, 2017 at 2:03

KalyanLahkar · Accepted Answer · 2017-09-03 05:21:39Z

2

You don't need to find designId. You just need to do

    public function calcList($id) {


    $design = design::select( `design_no`, `design_name`, `weight_length`, `quantity`, `design_image`, `des2`, `des3`, `des4`, `des5`, `des6`, `des7`)->where('id',$id)->get();  

    return Response::json($design);
}

You should not make another query to database by first finding a single design with id then using it again to get design_no, design_name, ... etc. It seems redundant.

answered Sep 3, 2017 at 5:21

KalyanLahkar

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Comments

shukshin.ivan · Accepted Answer · 2017-09-03 02:36:47Z

2

First, you'd better use GET method with cache: false ajax setting. In this case you don't need to pass csrf token.

Second, you forgot to use get() method to get data. Third, you can use where without =, it is by default.

use Response; // don't forget or use response()

public function calcList($id)
{
    $designId=design::find($id);
    $design = design::select( 'design_no', 'design_name', 'weight_length', 'quantity', 'design_image', 'des2', 'des3', 'des4', 'des5', 'des6', 'des7')->where('id', $designId)->get();
    return Response::json($design);
}

You can display your results as a table as follows.

<div id="destination"></div>
...
$.ajax({
         type:"POST",
         url:"/calc_list/"+id,
         success : function(results) {
             var $table = $('<table></table>');
             $('#destination').html('');

             for(var i=0;i<=results.length;i++) {
                 $table.append('<tr><td>No</td><td>'+results[i].design_no+'</td></tr>');
                 $table.append('<tr><td>Name</td><td>'+results[i].design_name+'</td></tr>');
             }
             $('#destination').append($table);
         }
    });

edited Sep 3, 2017 at 2:36

answered Sep 3, 2017 at 2:16

shukshin.ivan

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10 Comments

Harish Karthick Over a year ago

I am getting error Like this: QueryException SQLSTATE[42S22]: Column not found: 1054 Unknown column '' in 'field list' (SQL: select , , , , , , , , , , , , , , , , , , , , , , , , `` from designs where id = {"id":5,"design_no":"005","design_name":"Design-5","weight_length":"10gms","quantity":"10gms","design_image":null,"balls_7_pcs":"7","ravi_die":"2","balls_8_pcs":"3","rak_die":"2","balls_10_pcs":"4","deepak_die":"2","other_balls":"4","die_Gold":"2","kambi_2":"4","others":"2","kambi_3":"4"})

shukshin.ivan Over a year ago

Which line? Do you pass wrong $id to the first line of the method?

Harish Karthick Over a year ago

i fix the above by replacing (design ->'design' )

Harish Karthick Over a year ago

iam getting result like this [..] how to built it to html @shukshin.ivan

Harish Karthick Over a year ago

It Showing -> calculation:37 Uncaught TypeError: Cannot read property 'design_no' of undefined

|

Sreejith BS · Accepted Answer · 2017-09-03 13:22:56Z

0

You are checking the condition the wrong way. find($id) retrieve single RECORD, not the primary key. Modify your controller method.

public function calcList($id) {

    $design = design::select( `design_no`, `design_name`, `weight_length`, `quantity`, `design_image`, `des2`, `des3`, `des4`, `des5`, `des6`, `des7`)->where('id','=', $id)->get();   //*** change to id

    // return Response::json(array('datas' => $design));
    // return response()->json(['data'=>$design]);
    return Response::json($design);
}

More: https://laravel.com/docs/5.4/eloquent#retrieving-single-models

Also make sure you add this in your head

<meta name="_token" content="{{ csrf_token() }}" />

edited Sep 3, 2017 at 13:22

answered Sep 3, 2017 at 4:16

Sreejith BS

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