How To Use Ajax To Get Data From Db And Display On Page

I'm working in a project and I'm stuck here, I don't know why I can't get the list from my database\ Here is my JAVASCRIPT $(document).ready(function(){ $.ajax({ url:'d

Solution 1:

First, you will find it helpful to review these simple examples about AJAX. Do not just read them, copy them to your server and make them work. Change a few names or values -- see how they work.

AJAX request callback using jQuery

Next, here is a post that gives an overview to how PHP / web page / AJAX all work together. Take a few minutes and read it carefully. See if you can follow the logic. I bet the lightbulb will come on for you.

PHP - Secure member-only pages with a login system

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Make your code as standard as possible. Don't take any shortcuts. Use the full $.ajax() structure, not the shortcuts of $.post() or $.get() (these are both shortcut forms of $.ajax(). Don't skip anything. As you get better, you can start to take some shortcuts. But for now, make sure your AJAX code block looks like this:

var var_value = $('#someElement').val();

$.ajax({
    type: 'post',
     url: 'your_ajax_processor.php',
    data: 'post_var_name=' +var_value,
    success: function(dataz){
        if (dataz.length) alert(dataz);
        $('body').append(dataz);
    }
});

In your PHP, you will receive the value you posted in the $_POST array variable. If, in AJAX, you named your variable post_var_name (as we did in the example above), then that is how you access the contents:

$myVar = $_POST['post_var_name'];

When you are having trouble, a great idea is to put in some tests. (1) On the PHP side, comment out everything and at the top put in an echo command, like:

<?phpecho'I got here';
die();

Back on the web page, in the AJAX success function, just alert what you get:

success: function(d){
    alert(d);
}

At that point, you know two things:

1. Your AJAX -to- PHP communications are working, and

2. You see what value got passed over to PHP.

Then, you might do something like this

js/jQuery:

var var_value = $('#someElement').val();

$.ajax({
    type: 'post',
     url: 'your_ajax_processor.php',
    data: 'post_var_name=' +var_value,
    success: function(dataz){
        if (dataz.length) alert(dataz);
        $('body').append(dataz);
    }
});

PHP:

<?php$myVar = $_POST['post_var_name'];

    //Now you can do something with variable `$myVar`, such as:$out = '
        <div class="red-background"> '.$myVar.' </div>
    ';

    //This content is received in the AJAX code block using the variable name you specified: "dataz"echo$out; 

Solution 2:

is this in french? lol, luckily it's still all code. try this:

inside of your success(datos1) function replace datos1 with datos1.data. Ajax gives you the whole response, you just want the data from the response generally.

success:function(datos1){
        console.log(datos1);   
        console.log(datos1.data);   
        for(x = 0;x<=datos1.data.length;x++){
             //$("#PAIS").append("<option value='"+datos.data[x].id_pais+"'>"+datos.data[x].pais+"</option>");
              $("#REGION").append(new Option( datos.data[x].region, datos.data[x].id_region));
        }