Show data in PHP with AJAX

Question

I'm novice with Ajax syntax. So I create my load content with append but, I want to know if its possible with PHP? Say to ajax "Learn this php file to display data" and on my PHP file we find a while or foreach with the data of my json file.

I want to use it in PHP because I use this template in some file (this "append") was normally call in 10 files, so i don't want to copy and paste all code, I need to change something, I want to edit only one file.

This is my actually ajax file:

$.ajax(
    {
        url: 'http://localhost/concerts/json/getConcerts?date=previous&limit=10&offset=' + i++ + '',
        type: "get",
        dataType: 'json',
        success: function (data) {
            $.each(data, function (idx, elem) {

                $('#concertHome').append('<div>'+elem.bID9+'</div>')});

        },
        error: function () {
            alert('Erreur lors de la requête...');
        }
    });}

I understand English might not be your first language, so bare with me, but could you try to explain what you want to accomplish again (in a different way then above)? — Alex
– Alex, Commented Jun 5, 2018 at 0:25
Hi @Alex. I just dont find the word but now i think i find it, its templating like Handlebars.js — TiboK
– TiboK, Commented Jun 5, 2018 at 6:43

Masih IT · Accepted Answer · 2018-06-04 20:31:45Z

1

PHP scripts execute when you open a file and before everything else (It's server side language). Ajax calls exectue when the Javascript is completely loaded. (After php).

So you cant do that. they cant communicate with each other without someting like Ajax Call or Fetch etc.

answered Jun 4, 2018 at 20:31

Masih IT

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3 Comments

TiboK Over a year ago

Thanks for your reply. I find this example: itsolutionstuff.com/post/… but i dont undestand how its work..

Masih IT Over a year ago

What do you want to do? I think that's not what you want.

TiboK Over a year ago

Tis template (in my append), I call him in 8-9 ajax request (every time the same template, not the same data). I juste want to call him in all my ajax requet. I dont want on all my ajax request write the same content every time, if i need to make some modifications, i need to edit 8 files. Like a include in PHP

Carsten Massmann · Accepted Answer · 2018-06-05 06:22:35Z

It seems to me that you do not necessarily need to have the Ajax code portion in a PHP file. What you really require is some reusable chunk of code that you can apply in various places within the same page or, in your case, in different pages.

What you should take from the itsolutionstuff post is the idea of putting the Ajax call into a function which you can then call with varying arguments like:

function doAjax(params) 
  $.ajax(
  {
    url: 'http://localhost/concerts/json/getConcerts?date=previous&limit=10&offset=' 
        + (params.i++),
    type: "get",
    dataType: 'json',
    success: function (data) {
        $.each(data, function (idx, elem) {

            $(params.target).append('<div>'+elem[params.prop]+'</div>')});

    },
    error: function () {
        alert('Erreur lors de la requête...');
    }
  });}
}

You should put this code segment into a separate .js file which can then be included into all your pages with a <script src="ajaxcode.js"></script> directive. Each time you want to trigger the ajax event you need to call the function doAjax(actualParamaterObject). The actualParameterObject contains various properties to be used in the function, like - for example -

actualParameterObject={i:25, prop:'bID9', target:'#concertHome'};

Handing down the offset i inside this parameter object will have the desired side effect that it will be incremented within the function call. Handing it down as a direct argument would not work in this way.

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