Unable to set PHP variables as values for input field in form

Question

I have a PHP file which SELECT's all from the row found based on an SQL query. When I put the echo in a div, I get all information, but when I try to echo it into an input box in a form, it does not shows.

What am I doing wrong?

Please also note that I am aware that I am (most likely) making a lot of mistakes when it comes to security practices or programming standards, but this whole thing (PHPDesktop > https://github.com/cztomczak/phpdesktop) will get packed into an EXE file which will run locally only (no need for an online connection as the SQLite3 DB gets packed in with the EXE), and I am still figuring out how to program this in the first place, so efficient and tidy coding are not high on my list yet ;-)

DO_CUSTEDIT.PHP

$custName = $_POST['custName'];

$query = "SELECT * FROM `CustDB` WHERE CustName LIKE '%$custName%'";

$result = $db->query($query);

while ($row = $result->fetchArray()) {
      $custID = $row['CustID'];
      ......;
}

if (!$result) {
   echo $db->lastErrorMsg();
   $db->close();
   exit;
} else {
   echo $custID;
   echo ......;
   $db->close();
   exit;
}

EDITCUST.PHP / Javascipt

    <script>
      $(document).ready(function() {
        $("#subeditcust").on('click',function(e) {
          e.preventDefault();
          $.ajax( {
            url: "lib/do_editcust.php",
            method: "post",
            data: $("form").serialize(),
            dataType: "text",
            success: function(strMessage) {
              if (strMessage == "Customer updated successfully") {
                $("#message").text(strMessage);
                $("#neweditform").get(0).reset();
                } else {
                  $("#message").text(strMessage);
                }
              }
          });
        });
      });
    </script>

EDITCUST.PHP / HTML

      <form id="editcustform" name="editcustform" action="" method="post">
        <div class="row">
          <div class="column-half" style="background-color:#fff">
            <table>
              <tr>
                <td>
                  <a>Customer ID</a>
                </td>
                <td>
                  <div class="inputb">
                    <input type="text" name="custID" value="<?php echo (isset($custID)) ? $custID: ''; ?>" readonly/>
                  </div>
                </td>
              </tr>
            </table>
          </div>
        </div>
      <div class="row">
        <table style="table-layout:fixed">
          <td style="background-color: rgb(215,215,215); padding:0 10px;">
            <button id="subeditcust" type="submit" class="mainbtn">Create</button>
          </td>
          <td style="background-color: rgb(215,215,215); padding:0 10px;">
            <button id="reseteditcust" type="reset" class="mainbtn">Reset</button>
          </td>
        </table>
      </div>
    </form>

Furqan Ansari · Accepted Answer · 2019-10-23 16:34:48Z

1

<input type="text" name="custID" value="<?php echo (isset($custID) ? $custID: ''); ?>" readonly/>

Replace this line with yours it will work. IA

answered Oct 23, 2019 at 16:34

Furqan Ansari

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2 Comments

Stephen R Over a year ago

In PHP 7: <?=$custID ?? ''?>

Roy Monden Over a year ago

Unfortunately, still doesn't pass through the variable.

Volkkerine · Accepted Answer · 2019-10-23 14:52:27Z

0

It seems you have two different files, right? DO_CUSTEDIT.PHP and EDITCUST.PHP

The variables are being created on DO_CUSTEDIT.PHP and the when you are creating the HTML code those variables ($custID) are not setted.

Is one file including or requiring the other?

answered Oct 23, 2019 at 14:52

Volkkerine

864 bronze badges

1 Comment

Roy Monden Over a year ago

Correct, these are indeed two seperate files, with the DO_CUSTEDIT.php file sitting in a subfolder 'Lib' (hence the lib/DO_CUSTEDIT.php). The AJAX script inside EDITCUST.php calls the DO_CUSTEDIT.php file for execution, and the $custID variable is created in DO_CUSTEDIT.php. To check the output of the script, I added an ECHO, which is then caught by the AJAX script in EDITCUST.php (the 'success' operator) and shown as the 'strMessage' variable in EDITCUST.php. The variable shows the output correctly, so I know they communicate correctly, but I cannot call that variable in an input field.

Nilesh · Accepted Answer · 2019-10-23 17:02:04Z

0

EDITCUST.PHP is your first page from you are submitting form to DO_CUSTEDIT.PHP

When you land on EDITCUST.PHP variable $custID is not created
When the form is submitted through ajax, the ajax returns the data inform of object or array depending on how you are echoing the data from DO_CUSTEDIT.PHP I would recommend to use json_encode() php function to return inform of array
To debug the value by logging the data in console (though function console.log())
After returning the value from ajax, you have to populate the value in form through jquery something like $(input[name="custID"]).val( data.custID )

edited Oct 23, 2019 at 17:02

answered Oct 23, 2019 at 15:55

Nilesh

777 bronze badges

4 Comments

Roy Monden Over a year ago

I understand your reasoning, but am unsure where to insert the new code. The json_encode() i suspect goes in the last line of DO_CUSTEDIT.php like 'echo json_encode($custID);', but where do I put in the '$(input[name="custID"]).val( data.custID )'? Does that go under the success operator in tje AJAX script, or as value under the '<input type="text" name="custID" value="code here"/>'?

Nilesh Over a year ago

As an Example, in DO_CUSTEDIT.PHP you save your results in an array $results and then echo them as json_encode($return, JSON_PRETTY_PRINT) which you will take as input into your Javascript, inside $.ajax call as .done(function ajaxDone(data) { console.log(data); }) and inside this .done function you can run $(input[name="custID"]).val( data.custID )

Roy Monden Over a year ago

So now I added '$results = array($custID, $custName);' and 'echo json_encode($results);' to my DO_EDITCUST.php. Then, to check the output, I call the array to a temporary line '<p id="message"</p>' in my EDITCUST.php using my above mentioned SUCCESS operator 'success: function(strMessage) {$("#message").text(strMessage);}' I get the output [1,"BlaBla"], which I think means that the array is correctly parsed through? But the .done function completely confuses me I am afraid.

Nilesh Over a year ago

If you are confused and afraid, nothing can be done. Just try the code and make it work. That's how atleast I do

Roy Monden · Accepted Answer · 2019-10-24 11:34:59Z

I managed to figure it out!

DO_CUSTEDIT.php / PHP

$results = array($custID, $custName);
echo json_encode($results, JSON_PRETTY_PRINT);

EDITCUST.php / HTML

<input type="text" id="custID" name="custID" value="" readonly/>
<input type="text" id="custName" name="custName" value="" readonly/>

EDITCUST.php / JS

    <script>
      $(document).ready(function() {
        $("#subeditcust").on('click',function(e) {
          e.preventDefault();
          $.ajax( {
            url: "lib/do_editcust.php",
            method: "post",
            data: $("form").serialize(),
            dataType: 'json',
            success: function(data){
              document.getElementById('custID').value = data[0];
              document.getElementById('custName').value = data[1];
            }
          });
        });
      });
    </script>

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