php does not echo anything from ajax javascript?

Question

I cannot get my input to be echo when I input text in my search field. I do get the alert javascript but not the echo 2 to be display into the srv_search_mini_display field. Im thinking that the ajax cant access the find_Service.php

search.php

 <input id="service_search_box" class="form-control mr-sm-2" type="text" name="search" placeholder="Search Services..." oninput="searchService()">
 <div id="service_search_div" class="srv_search_mini_display d-none"></div>

<script type="application/javascript">

function searchService(){

    var srv_search = document.getElementById("service_search_box").value;

    var param = "service_search_box="+srv_search;

    if(srv_search.length > 0){

        $("#search_search_div").removeClass("d-none");

        var ajax = new XMLHttpRequest();

        ajax.onreadystatechange = function(){

            if(ajax.readyState === 4 && ajax.status === 200){

                if(ajax.responseText === ""){

                    document.getElementById("service_search_div").innerHTML = "";
                    $("#service_search_div").addClass("d-none");

                } else {

                    alert(param); document.getElementById("service_search_div").innerHTML = ajax.responseText;
                }

            }
        };

        ajax.open("POST",'find_Service.php',false);
        ajax.setRequestHeader("Content-type","application/x-www-form-urlencoded");
        ajax.send(param);

    } else{

        document.getElementById("service_search_div").innerHTML = "";
        $("#service_search_div").addClass("d-none");
    }

}
</script>

find_Service.php

 <?php

 require_once('database.php');
 $heidisql = pdo_con();

 echo 2; // does not even appear
  ?>

"echo 2; // does not even appear" Where should that appear? — user8034901
– user8034901, Commented Jan 18, 2019 at 9:09
inside the <div id="service_search_div" class="srv_search_mini_display d-none"></div> — Emanula Sohn
– Emanula Sohn, Commented Jan 18, 2019 at 9:14
document.getElementById("service_search_div").innerHTML = ajax.responseText; why this in the comment section. it should be in the next line to set the HTML. — Naveen K
– Naveen K, Commented Jan 18, 2019 at 9:16
Enable error_reporting in 'find_Service.php' and check. If you run the find_Service.php separately in browser is it echoing '2' have you checked that — Shibu
– Shibu, Commented Jan 18, 2019 at 9:58
Have you checked in browser console. Does it display any error related to javascript — Shibu
– Shibu, Commented Jan 18, 2019 at 10:05

Kyaw Kyaw Soe · Accepted Answer · 2019-01-18 10:08:01Z

1

First you have to make sure that your PHP file work, you can check it by open find_Service.php separately or like @ADyson said check it from inspect console in network tab.

Since I saw you using jQuery, I simplify your code a little bit by using jQuery's ajax and select element with $ sign.

I use fake API server to make it work here.

function searchService(){
  var srv_search = $('#service_search_box').val();
  var $srv_search_div = $('#service_search_div');

  if(srv_search.length > 0){
    $("#search_search_div").removeClass("d-none");

    $.post("https://jsonplaceholder.typicode.com/posts", //replace with "/find_Service.php"
    {
      service_search_box: srv_search,
    },
    function(data, status){
      if(status === "success" ){
        $srv_search_div.html(data.service_search_box); //replace with data
        $srv_search_div.addClass("d-none");
      }
      console.log(data);
    });

  }else{
    $srv_search_div.html("");
    $("#service_search_div").addClass("d-none");
  }
}

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>


<input id="service_search_box" class="form-control mr-sm-2" type="text" name="search" placeholder="Search Services..." oninput="searchService()">
 <div id="service_search_div" class="srv_search_mini_display d-none"></div>

answered Jan 18, 2019 at 10:08

Kyaw Kyaw Soe

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2 Comments

Emanula Sohn Over a year ago

ok i manage to get the 2 appear in console, but still nothing appear in the div

Kyaw Kyaw Soe Over a year ago

if you use my code change $srv_search_div.html(data.service_search_box); to $srv_search_div.html(data);

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