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in Technique[技术] by (71.8m points)

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

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 data

<tr>
                     <td><?php echo $q ?></td>

                     <td>
                        <input type="text" name="question" class="question"  value="<?php echo $print->question ?>" ></td>
                     <td>
                     <input type="text" name="answer" class="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>
                </tr>
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 ) )
          ); 
}

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

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1 Reply

0 votes
by (71.8m points)

The problem might be that your loop which is creating the list for displaying on admin page is causing multiple elements to have the same id, when id attributes must be unique within the DOM. To fix this use common classes on the elements within the loop, then DOM traversal to find them when the button is clicked.

Change the code block for display like this:

$mysql = Database::DBCon()->prepare('SELECT * FROM table_name');
$mysql->execute();
while ($fetch = $mysql->fetch(PDO::FETCH_ASSOC)){
    $html = '<tr>
    <td><input type="text" value="'. $fetch['question'] .'" class="question"></td>
    <td><input type="text" value="'. $fetch['answer'] .'" class="answer"></td>
    <td>
        <button type="button" class="btn btn-danger insert-btn" data-id="'. $fetch['ID'] .'">Insert</button>
        <button type="button" class="btn btn-warning update-btn" data-id="'. $fetch['ID'] .'">Update</button></td>
    </tr>';
    //echo $html;
}

Then change implementation of function to something like this:

$(document).ready(function() {
    $(document).on('click', '.update-btn', function() {
        var $row = $(this).closest('tr');
        var id = $(this).data('id');
        var question= $row.find('.question').val();
        var answer = $row.find('.answer').val();

        // ajax request here...
   });
})

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