Data not insert and update through ajax and jQuery in admin page?

Question

I made custom plugin and done crud operation, display all data in admin page, used ajax and jquery. Data successfully deleted but not inserted or updated. Data successfully pass through ajax but not inserted. Also What I saw if input block is empty and I put some data and updated it. It got first row data.
Error- https://prnt.sc/wnzqjr
ajax for insert the data

jQuery('.ins_btn').click(function(){
        var id = jQuery(this).attr('data-id');
        var question = jQuery('#question').val();
        var answer = jQuery('#answer').val();
        // alert(id);
        $.ajax({
            url: '<?php echo admin_url('admin-ajax.php');?>', 
            type: 'POST',
            data:{ 
              action: 'insert_records', 
              insert_record : id,
              insert_question: question,
              insert_answer: answer
            },
            success: function( data ){
                alert("Records are successfully insert");
                location.reload();
            }
         });
    });

insert query

 function insert_records(){
  global $wpdb;
 $id = $_POST['insert_record'];
 $question = $_POST['insert_question'];
 $answer = $_POST['insert_answer'];


  $db_inserted = $wpdb->insert( $wpdb->prefix.'faqq', 
        array( 'ID' => $id, 
               'question' => $question, 
               'answer' => $answer) 
    );
}
add_action( "wp_ajax_insert_records", "insert_records" );
add_action( "wp_ajax_nopriv_insert_records", "insert_records" );

ajax for update the data

jQuery('.upd_btn').click(function(){
        var id = jQuery(this).attr('data-id');
        var question = jQuery('#question').val();
        var answer = jQuery('#answer').val();
        alert(question);
        $.ajax({
            url: '<?php echo admin_url('admin-ajax.php');?>', 
            type: 'POST',
            data:{ 
              action: 'update_records', 
              update_record : id,
              update_question : question,
              update_answer : answer

            },
            success: function( data ){
                alert("Records are successfully updated");
                location.reload();
            }
         });
    });

update query

function update_records(){
  global $wpdb;
  // $table_name = $wpdb->prefix.'faqq';
  $id = $_POST['update_record'];
  $question = $_POST['update_question'];
  $answer = $_POST['update_answer'];
  $db_updated = $wpdb->update( $wpdb->prefix.'faqq', 
        array('question'    => $question,
              'answer'   => $answer, array( 'ID' => $id ) )
          ); 
}

this is html code

        <form method="POST" action=""  > 
            <table>
                 <td><?php echo $print->id ?></td>
                 <td>
                    <input type="text" name="question" id="question"  value="<?php echo $print->question ?>" ></td>
                 <td>
                 <input type="text" name="answer" id="answer"  value="<?php echo $print->answer ?>" > </td>
                 <td>
                    <input type="button" value="Insert" id="insert" data-id = "<?php echo $print->id ?>" name="insert" class="ins_btn">
                </td>
                <td>
                    <input type="button" value="Update" id="update" data-id = "<?php echo $print->id ?>" name="update" class="upd_btn">
                </td>
                 <td>
                    <input type="button" value="Delete" id="delete" data-id = "<?php echo $print->id ?>" name="delete" class="del_btn">
                </td>
            </table>
        </form>

Here are some errors. 1)Getting error when update the data through ajax- https://prnt.sc/wnymkx, https://prnt.sc/wnyos5, https://prnt.sc/wnyuhk

You don’t specify which table to update and insert the data to, that the first thing that jumped out to me, just the db prefix. — Tony Djukic
– Tony Djukic, Commented Jan 15, 2021 at 6:47

Sally CJ · Accepted Answer · 2021-01-15 10:15:52Z

1

I see that there's an internal server error (see screenshot) when you tried to update a record in the database, and the error is likely because you incorrectly called $wpdb->update() which has the syntax of — and note that the first three parameters ($table, $data and $where) are required:

wpdb::update( string $table, array $data, array $where, array|string $format = null, array|string $where_format = null )

But in your code, you did not set the third parameter ($where):

// In update_records():
$db_updated = $wpdb->update( $wpdb->prefix.'faqq',
    array( 'question' => $question,
           'answer'   => $answer, array( 'ID' => $id ) )
);

And I guess the array( 'ID' => $id ) is the $where part, but you probably made a typo which resulted in that part being part of the second parameter instead.

So fix that and your code would work properly. Example:

$db_updated = $wpdb->update( $wpdb->prefix.'faqq',
    array( 'question' => $question,
           'answer'   => $answer,
    ),
    array( 'ID' => $id )
);

As for your insert_records() function, it looks fine to me, so there shouldn't be an issue when inserting new records to your database table.

Additionally, you should also (later) do things like data sanitization and call wp_die() to end the AJAX request, and fix the issue as seen here.

answered Jan 15, 2021 at 10:15

Sally CJ

40.5k2 gold badges29 silver badges50 bronze badges

Yes, you are right. There is mistake in update query which I corrected. Now data updated but getting first row data. Screenshot - prnt.sc/wokuuk

vishal
– vishal

2021-01-15 12:44:37 +00:00
Commented Jan 15, 2021 at 12:44
1

Well, for that one, you need to show how you're outputting the data, i.e. show your form HTML and the relevant PHP code.

Sally CJ
– Sally CJ

2021-01-15 12:59:52 +00:00
Commented Jan 15, 2021 at 12:59
I added html code in above code. please have a look @Sally Cj.

vishal
– vishal

2021-01-15 13:07:20 +00:00
Commented Jan 15, 2021 at 13:07
1

Please add all PHP you used in that form - where/how you defined the $print, how you query your table (what PHP code you use), show any foreach/for loop, etc.

Sally CJ
– Sally CJ

2021-01-15 13:20:03 +00:00
Commented Jan 15, 2021 at 13:20

Add a comment |

vishal · Accepted Answer · 2021-01-15 14:46:40Z

I did wrong code in ajax and jquery- for update function in ajax and jQuery (Before)-

jQuery('.upd_btn').click(function(){
        var id = jQuery(this).attr('data-id');
        
        var question = jQuery('#question').val();
        var answer = jQuery('#answer').val();
        
        $.ajax({
            url: '<?php echo admin_url('admin-ajax.php');?>', 
            type: 'POST',
            data:{ 
              action: 'update_records', 
              update_record : id,
              update_question : question,
              update_answer : answer

            },
            success: function( data ){
                // alert("Records are successfully updated");
                // location.reload();
            }
         });
    });

After
What I did wrong was I got the same id for question('#question') and answer('#answer') column. So I got the first row data for any row Here is the answer

jQuery('.upd_btn').click(function(){
        var id = jQuery(this).attr('data-id');
        
        var question = jQuery(this).closest('table').find('.question').val();
        var answer = jQuery(this).closest('table').find('.answer').val();
        
        $.ajax({
            url: '<?php echo admin_url('admin-ajax.php');?>', 
            type: 'POST',
            data:{ 
              action : 'update_records', 
              update_record : id,
              update_question : question,
              update_answer : answer

            },
            success: function( data ){
                // alert("Records are successfully updated");
                
            }
         });
    });

So I changed id value and get class of each field and use closest function.

Yes, I did notice that (#question and #answer issue in your JS) before I posted my answer, but I was hoping to see more of your form HTML before commenting on that duplicate ID's. Anyway, glad that you figured it out. Happy coding! — Sally CJ
– Sally CJ, Commented Jan 15, 2021 at 15:15

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