Access Javascript Variable in php base_url

Question

I am using AJAX and I'm getting the data from the database as Javascript variables. Now, in a table, I want to print an image using php base_url. The problem is the image is not showing. Here is my code: everything is printing except the picture

<script>
    $(function(){
       showAllEmployee();
        function showAllEmployee(){
        $.ajax({
                type: 'ajax',
                url: '<?php echo site_url('Welcome/showAllEmployee') ?>',
                async: false,
                dataType: 'json',
                success: function(data){
                                var html = '';
                    var i;

                    for(i=0; i<data.length; i++){


                        html +='<tr>'+
                                    '<td> <img class="size1"  src="<?php echo base_url("/uploads/university/data[i].image")?>" /></td>'+

                                    '<td>'+data[i].name+'</td>'+
                                    '<td>'+data[i].division+'</td>'+
                                    '<td>'+data[i].location+'</td>'+
                                    '<td>'+data[i].min_hsc+'</td>'+
                                    '<td>'+data[i].ad+'</td>'+
                                    '<td>'+data[i].email+'</td>'+
                                '</tr>';
                    }
                    $('#showdata').html(html);
                },
                error: function(){
                    alert('Could not get Data from Database');
                }
            });
        }
        }
        )
        </script>

Can you show your image url when it is printed on a page? You can verify this using firebug or developer tools in any popular web browser. — Artur Poniedziałek
– Artur Poniedziałek, Commented Aug 24, 2017 at 20:01
'<td> <img class="size1" src="<?php echo base_url("/uploads/university/")?>data[i].image" /></td>'+ wrote like that, but not working — Ahsan Habib Himel
– Ahsan Habib Himel, Commented Aug 24, 2017 at 20:14

yezzz · Accepted Answer · 2017-08-24 20:37:56Z

1

I'd say something like this:

'<td> <img class="size1" src="<?php echo base_url("/uploads/university/")?>' + data[i].image + '"/></td>'

answered Aug 24, 2017 at 20:37

yezzz

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Comments

Sandeep K. · Accepted Answer · 2018-09-25 05:43:52Z

You have not use javascript code directly with php. Please see below code. I hope it will works for you.

for(i=0; i<data.length; i++){
       document.cookie = "image =" + data[i].image;
     <?php
       $image= @$_COOKIE['image'];
     ?> 
                    html +='<tr>'+
                                '<td> <img class="size1"  src="<?php echo base_url("/uploads/university/".$image)?>" /></td>'+

                                '<td>'+data[i].name+'</td>'+
                                '<td>'+data[i].division+'</td>'+
                                '<td>'+data[i].location+'</td>'+
                                '<td>'+data[i].min_hsc+'</td>'+
                                '<td>'+data[i].ad+'</td>'+
                                '<td>'+data[i].email+'</td>'+
                            '</tr>';
                }

@shawn Check your console and find what error occured? and let me know!

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