British-based php programmer wanted

Hello,

I'm looking to partner with someone in a project. I am happy to invest some money, but this will be a project that, hopefully, will grow significantly and, therefore, should be self-financing. Initially, I am not considering this for an income stream, so I am pleased to reinvest all profits during a period of development and commercialization; In essence, you can take a large part of the income to help improve the project.

It is a fintech project. I am fully qualified, authorized and insured with a successful business in the Financial Services industry (operations since 2013) and would make sure that this company is authorized by the FCA. I would like to start a business division where everything is online. There are similar companies in the sector, so I know it works, but I have a unique twist and I can go a step further and, in fact, have a basic model. to show that it can work.

My problem is that I lack the skills to develop the site beyond basic html and limited php and, having invested in another IT project recently, I've noticed that IT developers have a way of getting every penny of a job if you're not 100% sure of everything This is the reason why my preference is oriented to work with someone who could think about the problems of my plans in advance instead of waiting until the problem arises and then charging more to correct it. If you work with someone who has invested in the company, then you are interested in avoiding any problems.

I am happy to have an initial conversation over the phone and finally meet. However, due to the rules and regulations with data protection and regulatory requirements, this position is specifically for someone based in the United Kingdom.
SEMrush

Thank you.

php – How to know if there is already a file for not writing about it?

I am a beginner, what I am doing is a form that is saved in a .txt document, the problem is that when you save it is not known if it already exists with that name or not and if you put the same name it is written above the one that already existed , can you help me please?

``

PHP makes an example of what is different between return and return ()

I have seen the return of PHP (value); vs return value;

You should never use parentheses around your return variable when you return by reference, as this will not work. You can only return variables by reference, not the result of a declaration. If you use return ($ a); then it is not returning a variable, but the result of the expression ($ a) (which is, of course, the value of $ a).

But I want to know how to illustrate that. returns $ a Y return ($ a) They are different?

javascript – Doubt with implementation of Tag Manager in php code

I have the question of how to implement a datalayer in condigo php. Someone could advise me please.

Example datalayer:

dataLayer = [{
    'seccion' : ‘Personas’, //Dinamico
    'sub-seccion' : ‘Seguro de Gastos Medico’, //Dinamico
    'tipo-de-seguro' : ‘Planes Nacionales’, //Dinamico
    'nombre-del-seguro' : ‘Linea Azul Flexible’  //Dinamico
}]

merge 2 tables in duplicate update in php

Hi, I have 2 tables that have the same structure. I want to update t1 with the values ​​of t2.
I used this query but it does not work
$ QRY = "INSERT IN mutashabeha c (
SELECT * FROM mutashabehat t
ON (c.SurahNumber = t.SurahNumber AND c.AyahNumber = t.AyahNumber AND c.SurahNumber2 = t.SurahNumber2 AND c.AyahNumber2 = t.AyahNumber2) UPDATE
SimilarParts = t.SimilarParts,
DifferentParts1 = t.DifferentParts1,
DifferentParts2 = t.DifferentParts2,
Markers = t.Markers,
Comments = t.Comments,
PoeticVerses = t.PoeticVerses
);
";

How to obtain a variable from a From :: text php without loading Page

I wanted to ask how I can get the value of the Form :: text and save it in a php variable when I click on the button

THANK YOU

{!! Form :: text ('message', '', array ('class' => 'form-control', 'id' => 'message')) !!} Start

php – admin-ajax return 400 – can not find a custom function

I am adding a function that should be called when a button is clicked, I have a js file with the following jquery code and the ajax call:

jQuery (document) .ready (function () {

function getUrlParameter (sParam) {

var sPageURL = window.location.search.substring (1);

var sURLVariables = sPageURL.split (& # 39; && # 39;);

for (var i = 0; i <sURLVariables.length; i ++) {
var sParameterName = sURLVariables[i].split (& # 39; = & # 39;);
if (sParameterName[0] == sParam) {
return sParameterName[1];
}
}
console.log (& url function entered & # 39;);
}

jQuery (& # 39; # upload_btn & # 39;). click on (function (e) {

e.preventDefault ();

var flag = true;
var postId = getUrlParameter (& # 39; preview_id & # 39;);
var files = jQuery (& # 39; # file_tool & # 39;). prop (& # 39; files & # 39;);
console.log (files);

var dataS = {{
& # 39 ;: & # 39; upload_button & # 39 ;,
& # 39; preview_id & # 39 ;: postId,
& # 39; files & # 39 ;: files,
& # 39; set flag
}
jQuery.ajax ({
type: & # 39; post & # 39 ;,
url: diario.upload,
processData: false,
Content type: false,
data: data
success: function (answer) {
if (response.type == "success") {
console.log (& # 39; jquery works & # 39;);
} else console.log (answer);
}
});

});

});

When I click on the button the console.log shows the obj files, so that at least the onClick works, but just after it appears. jquery.js: 4 POST custom / wp-admin / admin-ajax.php 400. This is how you see my functions.php:

add_action (& # 39; wp_enqueue_scripts & # 39 ;, & # 39; enqueue_onClick_upload & # 39;);

// custom_js_enqueuer
function enqueue_onClick_upload () {

wp_register_script (& # 39; onClick_upload & # 39 ;, WP_CONTENT_URL. & # 39; / themes / microjobengine / onClick_upload.js & # 39 ;, array (& # 39; jquery & # 39;));
wp_localize_script (& # 39; onClick_upload & # 39 ;, & # 39; diary & # 39 ;, array (& # 39; upload & # 39; => admin_url (& # 39; admin-ajax.php & # 39;)));

wp_enqueue_script (& # 39; jquery & # 39;);
wp_enqueue_script (& # 39; onClick_upload & # 39;);

}


add_action (& # 39; wp_ajax_upload_button & # 39 ;, & # 39; upload_button & # 39;);
add_action (& # 39; wp_ajax_nopriv_upload_button & # 39 ;, & # 39; upload_button & # 39;);

// button up function
upload_button () {function

$ postID = $ _POST['preview_id'];

yes ($ _POST['set']) {
yes ($ _ POST['files']['size']    === 0)
echo "";

}
$ result['type'] == & # 39; success & # 39 ;;
echo json_encode ($ result);
wp_die ();
}

I have no idea why he does not find the action, everything is done as wp says he should, I hope someone can help out, thanks.

PHP7 I use a mysql query in php and it gives me a different value than if I do it in phpMyAdmin

I have this query =

SELECT
- tags_in_messages.id AS 'EM_id',
- tags_in_messages.id_message AS 'EM_id_message',
- tags_in_messages.id_tag ​​AS 'EM_id_tag',
 messages.id AS 'message_id',
 messages.title AS 'message_title',
 messages.content AS 'message_content',
 messages.createAt AS 'message_createAt',
 messages.screador AS 'message_creator',
 messages.url_img AS 'message_url_img',
 Tags.id AS 'label_id',
 tags.name AS 'label_name'
DESDE
tags_in_messages,
messages,
labels
WHERE
tags_in_messages.id_message = messages.id
AND
tags_in_messajes.id_tag ​​= tags.id



ORDER BY
messages.id;

In HeidiSQL(a program like MySQL Workbench) gives me this result =
enter the description of the image here

But if I do it with PHP look what I get, exactly with the same query:

$ sql =
"SELECT
- tags_in_messages.id AS 'EM_id',
- tags_in_messages.id_message AS 'EM_id_message',
- tags_in_messages.id_tag ​​AS 'EM_id_tag',
 messages.id AS 'message_id',
 messages.title AS 'message_title',
 messages.content AS 'message_content',
 messages.createAt AS 'message_createAt',
 messages.screador AS 'message_creator',
 messages.url_img AS 'message_url_img',
 Tags.id AS 'label_id',
 tags.name AS 'label_name'
DESDE
tags_in_messages,
messages,
labels
WHERE
tags_in_messages.id_message = messages.id
AND
tags_in_messajes.id_tag ​​= tags.id



ORDER BY
messages.id;

";

if ($ result = $ connection-> query ($ sql)) {
    var_dump ($ result);
    while ($ row = mysqli_fetch_array ($ result)) {
        echo 'MESSAGE_ID ='. $ row['mensaje_id']. "<br";
        echo 'MESSAGE_TITLE ='. $ row['mensaje_titulo']. "<br";
        echo 'MESSAGE_CONTINUED ='. $ row['mensaje_contenido']. "<br";
        echo 'MESSAGE_CREATEAT ='. $ row['mensaje_createAt']. "<br";
        echo 'MESSAGE_CREATER ='. $ row['mensaje_creador']. "<br";
        echo 'MESSAGE_URL_IMG ='. $ row['mensaje_url_img']. "<br";
        echo 'LABEL_ID ='. $ row['etiqueta_id']. "<br";
        echo 'EITQUETA_NOMBRE ='. $ row['etiqueta_nombre']. "<br";
    }
}

I really do not understand how I can get a result so different between one and another. Excuse my ignorance about it, but I have not managed to fix that.

Services code .psd to HTML5 / CSS3 / JAVASCRIPT receptive also PHP / MYSQL

Embed

HTML:

BBCode:

Link image:

Problem with special characters, question mark php?

I have a problem, the accents appear a question mark.

(in the event that this is not the case,
but now interrogation points appear when I put an accent.
I have
apache2 with default_charset = iso-8859-1
and my file is iso-8859-1
and I already tried all the php functions, nothing solves: iconv (), mb_decoding, utf8_encode, etc.
impossivelllll. does not disappear ??? n? n?