Supplement Genie

@codersantosh if I chose to use the image as a normal image this then does not display correctly as a header image which I am trying to use the block for.

Is there a better block to use as a header image block that might do what I am trying to achieve?

@wpmudev-support9 I sorted the problem. I hade removed jquery.js from the exception list as it was shown as render-blocking.

But it appears that it does not work correctly if combined and minified so when I re0added it, the forms worked fine.

Brilliant, the problem appears to be sorted ??

My current Exception list is:

wp-includes/js/dist/, wp-includes/js/tinymce/, plugins/gutentor/assets/js/google-map-loader.min.js, plugins/gutentor/assets/js/google-map-loader.js

I have also noticed that forminator – contact form doesnt load either and seems to be given an element style of display:none;

ah… Got it ??

@optimizingmatters

I added

plugins/gutentor/assets/js/google-map-loader.min.js, plugins/gutentor/assets/js/google-map-loader.js

To the exceptions, sorry if that is what you meant

@optimizingmatters I tried that but it had no effect, the maps block still does not show.

If I add that to the exceptions, would that not then leave all gutentor blocks from being cached and as a result negate the need for this plugin?

I am thinking, from looking that it has something to do with one of these files:

https://github.com/wp-invoice/wp-invoice/blob/latest/lib/class_template_functions.php

https://github.com/wp-invoice/wp-invoice/blob/latest/lib/class_functions.php

https://github.com/wp-invoice/wp-invoice/blob/latest/lib/class_ajax.php

https://github.com/wp-invoice/wp-invoice/blob/latest/lib/ui/wp-invoice_page_wpi_page_manage_invoice.php

https://github.com/wp-invoice/wp-invoice/blob/latest/lib/ui/metabox/wp-invoice_page_wpi_page_manage_invoice.php

@a2hostingrj I think the PDF will be the easy part.

However, I need to add a new field to the admin end of the plug in so that I can insert the PO number. Currently When I add an invoice, there is no option for this

I have mainly done front end development so not sure how to add new fields to a plugin?

@macmanx thank you for your reply.

I have tried doing the above. I have even set my website back to yesterdays settings. the website worked fine earlier today and assumed that if I regressed to yestersay that it should be fine but it done nothing

@joyously @bcworkz Thank you for the help. The form is intended to be used internally and only logged-in users within the company will have access so im hoping that security won’t be an issue.

Can I ask, I have tried a number of different things but apparently none of them being the right thing.

based on the code, I have… After form submission, I am keen for the form to be cleared however as it stands, once submitted, all of the form fields remain populated.

The jQuery/Ajax, passes all the details over for PHP to process so I am thinking that PHP will be the better place to put the code…

I have tried adding at the end:

echo:"<script></script>"

and putting different code in after the

if (mysqli_query($conn, $sql)) {
        echo "New record created successfully";
    } 

as well as after the ajax:

success: function() {
                    alert('Message sent');
                    document.getElementById('dataForm').reset();
                }

Is there a better way to do this that might actually work? lol

Update…

Solved the problem..

success: function() {
                    alert('Message sent');
                },
                complete: function(){
                    $('#dataForm').each(function(){
                        this.reset();   //Here form fields will be cleared.
                    });
                }

The bit that I cant get to clear is the <select> areas, they remain populated. however if I fill in the other areas and then press submit, although they have values in them, the script doesn’t recognsise that theyre there so I need to clear the values myself and re-enter

• This reply was modified 5 years, 2 months ago by Supplement Genie.
• This reply was modified 5 years, 2 months ago by Supplement Genie.
• This reply was modified 5 years, 2 months ago by Supplement Genie.
• This reply was modified 5 years, 2 months ago by Supplement Genie.

@joyously
Simple fix, thank you, thats it working now ??

In all honesty in the few replies youve given me you have been a massive help… You have helped me to also see that when I am debugging I spend too much time looking at the code and what it is saying as above, my code seemed to have all the right info etc… I dont spend enough time looking at the other details such as punctuation which can have a massive effect on the code.

I looked at that over and over and never noticed the missing ,

Thank you again and sorry for my negligence ?? lol

@joyously @bcworkz yeah the mysqli was misspelt but I amended that earleir, not that it made any difference lol.

I now have error messages showing…

Have done a bit of tinkering and the error message I am getting:

Error: INSERT INTOadd_job(id,adddate,addcontact,adda,addb,addincome,addpayment,addsubbie,addclient`) VALUES (NULL ‘2019-09-06’, ‘test’, ‘test’, ‘test’, ‘test’, ‘test’, ‘test’, ”)
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ”2019-09-06′, ‘test’, ‘test’, ‘test’, ‘test’, ‘test’, ‘test’, ”)’ at line 2`

I have changed to code to:

JS

    jQuery(document).ready(function () {
    $("#addsubmit").click(function(e) {
        e.preventDefault();
        e.stopPropagation();

        var date = document.getElementById("adddate").value;
        var contact = document.getElementById("addcontact").value;
        var a = document.getElementById("adda).value;
        var b = document.getElementById("addb").value;
        var income = document.getElementById("addincome").value;
        var payment = document.getElementById("adddate").value;
        var subbie = document.getElementById("addsubbie").value;
        var client = document.getElementById("addclient").value;

        // Returns successful data submission message when the entered information is stored in database.
        var dataForm = 'date1=' + date + '&contact1=' + contact + '&a1=' + a + '&b1=' + b + '&income1=' + income + '&payment1=' + payment + '&subbie1=' + subbie + '&client1=' + client;
        if (date == '' || contact == '' || a == '' || b == '' || income == '' || payment == '' || subbie == '' || client == '') {
        alert("Please Fill All Fields");
        } 
        else {
        // AJAX code to submit form.
            jQuery.ajax({
                type: "POST",
                url: 'wp-content/themes/EazyFreight/database.php',
                data: dataForm,
                dataType:"json", 
                success:function(strMessage) {
                    $("#message").text(strMessage);
                    $("#dataForm")[0].reset();
                }
            });
            return false;  
        }
    });

PHP

<?php

    //Register variables
    $adddate = $_POST['date1'];
    $addcontact = $_POST['contact1'];
    $adda = $_POST['a1'];
    $addb = $_POST['b1'];
    $addincome = $_POST['income1'];
    $addpayment = $_POST['payment1'];
    $addsubbie = $_POST['subbie1'];
    $addclient = $POST['client1'];

    //connect with Database
    if (!empty($adddate) || !empty($addcontact) || !empty($adda) || !empty($addb) || !empty($addincome) || !empty($addpayment) || !empty($addsubbie) || !empty($addclient)){
        $host_name = 'xxx.hosting-data.io';
        $database = 'xxx';
        $user_name = 'xxx';
        $password = 'xxx';
        $conn = mysqli_connect($host_name, $user_name, $password, $database);
    }
    // Check connection

    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }

    $sql = "INSERT INTO <code>add_job</code>(<code>id</code>, <code>adddate</code>, <code>addcontact</code>, <code>adda</code>, <code>addb</code>, <code>addincome</code>, <code>addpayment</code>, <code>addsubbie</code>, <code>addclient</code>)
    VALUES (NULL '$adddate', '$addcontact', '$adda', '$addb', '$addincome', '$addpayment', '$addsubbie', '$addclient')";
    
    if (mysqli_query($conn, $sql)) {
        echo "New record created successfully";
    } 
    else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
    mysqli_close($conn);
    die();

Have you any idea what might cause the error? or the best way to deb

@joyously @bcworkz yeah the mysqli was misspelt but I amended that earleir, not that it made any difference lol.

no the way it was shown in the question was wrong, I tried to edit but was too late. I had added ” to the terms in the brackets on the lower one but all it done was add grey areas lol… And this is one big grey area for me

I feel as though I have used the correct code I cannot see anything that I should change.
But I am unsure how to determine what is causing the error.

When I check console and params, it gives all the right parameters with the correct titles.

But the php doesnt seem to take them and insert them into the table. I am unsure how to debug this as the echo statements I have added dont seem to work which makes it harder to determine where the error lies

@bcworkz
Hopefully you or someone can help with this final thing. I have tried ammending based on what I have seen online but with no joy.

I have the ajax sending to the correct file, everything seems to be going great, however, the database table is not being populated…

The code I am using for the table is as before:

    <div class="addjob tab-pane show active" id="home" role="tabpanel" aria-labelledby="home-tab">
        <h2>Add Job</h2>
        <form id="dataForm" name="dataform" method="POST" action="">
	    <span>
	        <label>Job Date:</label>
		<input class="add-date" type="date" placeholder="Placeholder Text" name="adddate">
	    </span>
	    <span>
	        <label>Client Name:</label>
		<select name="addclient" class="limitedNumbChosen add-client" multiple="false">
		    <option value="1">ClientA</option>
		    <option value="2">ClientB</option>
		</select>
		<div id="add-client" class="fas fa-plus"> Add New Client</div>
	    </span>
            <span>
	        <label>Contact Name:</label>
		<input class="add-contact" type="text" placeholder="John Doe"name="addcontact">
	    </span>
	    <span>	
	        <label>a:</label>
		<input id="add_a" class="add-from" type="text" placeholder="Postcode from" name="adda">
	    </span>
	    <span>
	        <label>b:</label>
		<input class="add-b" type="text" placeholder="Postcode to" name="addb">
            </span>
	    <span>
	        <label>Who Done Job:</label>
		<select name="addsubbie" class="limitedNumbSelect2 add-subbie" multiple="true">
		    <option value="1">Dave</option>
		</select>
		<div id="add-subbie" class="fas fa-plus"> Add Subbie </div>
	    </span>
	    <span>
	        <label>Income (￡):</label>
		<input class="add-income" type="number" placeholder="￡120" name="addincome">
	    </span>
	    <span>
	        <label>Driver Payment (￡):</label>
		<input class="add-payment" type="number" placeholder="￡100" name="addpayment">
	    </span>
	    <span>
	        <button id="submit" type="submit" value="submit">Submit</button>
	    </span>
	</form>
    </div>

The PHP used in database.php (which the ajax calls)

    <?php

    function so56917978_upload_callback() {  

        //Register variables

        $adddate = $_POST['adddate'];
        $addcontact = $_POST['addcontact'];
        $adda = $_POST['adda '];
        $addb = $_POST['addb'];
        $addincome = $_POST['addincome'];
        $addpayment = $_POST['adddate'];
        $addsubbie = $_POST['addsubbie '];
        $addclient = $POST['addclient'];
    
        //connect with Database

        $host_name = 'xxx.hosting-data.io';
        $database = 'xxx';
        $user_name = 'xxx';
        $password = 'xxx';
        $conn = mysqil_connect($host_name, $user_name, $password, $databse);

        // Check connection
        if (!$conn) {
            die("Connection failed: " . mysqli_connect_error());
        }

        $sql = "INSERT INTO add_job (id, adddate, addcontact, adda, addb, addincome, addpayment, addcompany, addsubbie, addclient)
                VALUES (NULL, '$adddate', '$addcontact', '$adda', '$addb', '$addincome', '$addpayment', '$addsubbie', '$addclient')";

        if (mysqli_query($conn, $sql)) {
            echo "New record created successfully";
        } else {
            echo "Error: " . $sql . "<br>" . mysqli_error($conn);
        }

        mysqli_close($conn);
    }

    ?>

The code seems to execute ok, when i check in firefox, the Params section is as expected so the correct info is being sent with the correct names…
no response…

However, timings shows…

Blocked:
1 ms
DNS resolution:
0 ms
Connecting:
0 ms
TLS setup:
0 ms
Sending:
0 ms
Waiting:
84 ms
Receiving:
0 ms

Not sure if this means anything. I have changed the database and am now using the same database that the CMS is installed on.

Should it make a difference wether i use INSERT as shown above:

    $sql = "INSERT INTO add_job (id, adddate, addcontacct, adda, addb, addincome, addpayment, addcompany, addsubbie, addclient)
                VALUES (NULL, '$adddate', '$addcontact', '$adda', '$addb', '$addincome', '$addpayment', '$addsubbie', '$addclient')";

or like

$sql = "INSERT INTOadd_job(id,adddate,addcontact,adda,addb,addincome,addpayment,adddsubbie,addclient`)
VALUES (NULL, ‘$adddate’, ‘$addcontact’, ‘$adda’, ‘$addb’, ‘$addincome’, ‘$addpayment’, ‘$addsubbie’, ‘$addclient’)”;’

• This reply was modified 5 years, 2 months ago by Supplement Genie.