weixin_33725272 2020-04-02 20:50 采纳率: 0%
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I will begin with an apology: I'm bad with AJAX. I'm trying to learn, but this particular project is giving me an absolute headache. I've read through a lot of examples, but can't find anything that helps with this issue.

PROBLEM: Using an "old-fashioned" PHP/MySQL/HTML environment, I'm trying to send data from a simple HTML form using AJAX; send the ajax request to php for updating a record in a MySQL db. Unfortunately, my code just does not work. It seems that I am effectively sending data, but somewhere in all of this I am coming up short. NOTE: I've confirmed that there is no problem with my Database connection, I use "$db" when inputting my name, password, db_name as indicated in my "show_client.php" file.

MY GOAL: I'm hoping to: (a) Update the existing variable using this AJAX request (no refresh of page); AND (b) Display output in the div titled "case_activity_id2" after the form database row is updated and my ajax request has been submitted.

I have two relevant files: (1) index.php; and (2) show_client.php.


        <form id= "ajaxForm" action = ""  method = "POST" >
            <input type = "hidden" name='case_activity_id' id = 'case_activity_id' value = '<?php echo $case_activity_id?>'>
            <select name ='show_client_id' id = 'show_client_id' class="form-control">

  <td><input type="submit" value="send" id = "btnClick" class="btn btn-primary" /> </form></td>
  <td><div id = "case_activity_id2"></div></td>


    $("#btnClick").on("click", function(){ submitForm();});
function submitForm(){
    $(document).ready(function() {
            var case_activity_id = $("#case_activity_id").val();
            var show_client_id = $("#show_client_id").val();
           $.ajax( {
              type: "POST",
              dataType: 'json',
              data: {
              success:function(data) {

(2) show_client.php

require 'db/connect.php';

            $show_client_id = $_POST['show_client_id']; 
            $case_activity_id = $_POST['case_activity_id']; 

            $sql = "UPDATE case_activity
                    SET show_client_id = '$show_client_id'
                    WHERE case_activity_id = '$case_activity_id'";  
            mysqli_query($db, $sql);  

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