how to checked the checkbox after fetching the data from database using jquery and php

Question

when user click on a button it displays the dynamic data fetched from the database along with the appropriate check box ..so user has select or check the checkbox of some data and submit the result then those data are stored in database. again that user clicked the same button then again those data are displayed but this time the data which are selected buy the user previously should be checked. I am trying to do that but enable to write code for it in jquery.

below is my code...

javascript

 $(document).ready( function() {    
        $('#location_choose').click(function(e){
                branch();
                document.getElementById('brn_light').style.display='block';

                document.getElementById('fade').style.display='block';

        });

        function branch()
        {               
            //alert(user_id);
            $.ajax({
                        url: "<?php echo base_url(); ?>/login/branch",
                        data: {},
                        type: "post",
                        cache: false,
                        success: function (data)
                        {
                            //alert(data);
                            var obj = $.parseJSON(data);
                            var result = "";
                            <?php foreach($resultsb as $rows){?>


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

                                result+="<table id='branchtable'><tr height='25px' ><td>&nbsp&nbsp</td><td ><input class='test' type='checkbox' name='branch_name' value="+obj[i].id+"<?php echo ($rows['branch_id']=="+obj[i].id+"? 'checked' : '');?>></td><td width='15px'></td><td width='630px'><b>"+obj[i].branch_name+
                                    "</b></td></tr><tr><table id='branch_address' style='background-color:#EBF5FB'><tr><td>&nbsp&nbsp</td></tr><tr><td width='60px'></td><td width='440px'>"+obj[i].address+"</td><td width='40px'></td></tr><td>&nbsp</td></table></tr><tr></tr></table>";

                            }
                            <?php }?>
                            //alert(result);

                            document.getElementById("branch_table").innerHTML=result;
                        },
                        error: function (xhr, ajaxOptions, thrownError) 
                        {
                            //alert(thrownError);
                        }
                });
        }

    });

i tried using

    <?php echo ($rows['branch_id']=="+obj[i].id+"? 'checked' : '');?>

in above code but still it doesnt checked the previous selected one.

You can't compare a php variable with a javascript variable like that. PHP get's parsed on the server. De result will then be sent to the browser that runs the js. — M. Eriksson, Aug 23 '17 at 11:37
then how can i compare it? because data are printing dynamically..even instead of javascript value i directly wrote some number id still its not working — Archana Gupta, Aug 23 '17 at 11:38
You need to rethink this completely. Why don't you generate everything with PHP on the page your calling with ajax and just return the result? — M. Eriksson, Aug 23 '17 at 11:41
Possible duplicate of [What is the difference between client-side and server-side programming?](https://stackoverflow.com/questions/13840429/what-is-the-difference-between-client-side-and-server-side-programming) — M. Eriksson, Aug 23 '17 at 11:43
your php is not close where your if condition is start in table — Mayank Vadiya, Aug 23 '17 at 11:44
It's easy to check a checkbox on data generated dynamically via ajax code.. but you have mixed php unnecessarily just to comapre obj[i].id in success function and never used anywhere while generating table element using js. — Chaitanya Ghule, Aug 23 '17 at 11:48
By using ajax it displayed all the object but i have to checked only the slected one then how can i do it using ajax....by calling another function in usse ajax in that? @ChaitanyaGhule — Archana Gupta, Aug 23 '17 at 11:53
what does your checkbox actually refers to w.r.t. branch_name — Chaitanya Ghule, Aug 23 '17 at 11:55
Severity: Parsing Error Message: syntax error, unexpected '?>' this error has encountered @MayankVadiya — Archana Gupta, Aug 23 '17 at 11:57

Samir Lakhani · Answer 1 · 2019-07-03T12:02:14.483

Give The Same Class of Checkbox and Write Jquery Code

<html>
<head>

<script src="https://ajax.aspnetcdn.com/ajax/jQuery/jquery-3.2.1.min.js"></script>
<script>
$( document ).ready(function() {

                $( '.select_all_text').click(function() {

                         var is_check = $(".select_type").prop('checked');
                                        if(is_check == false)
                                        {
                                            $(".select_type").prop('checked', true);
                                            $(''.select_all_text").text('De-select All');
                                        }
                                        else
                                        {
                                            $(".select_type").prop('checked', false);
                                            $('.select_all_text").text('Select All');
                                        }


                    });


});

</script>
</head>
<body>
<p style="position: sticky; z-index: 100; width: 50%;"><a  id="select_all" onclick="check_uncheck" style="margin-right: 2%;" href="javascript:void(0)">Select all</a> 
 <?php
        foreach($data_list as $data)
              {
              ?>
              <input type="checkbox" name="select_type[]" value="<?php echo $item_id; ?>" class="select_type" id="select_type"/>

            <?php
              }
 ?>
</body>
</html>

how to checked the checkbox after fetching the data from database using jquery and php

1 Answers1