php does not echo anything from ajax javascript?

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

  ?>

edited 20 hours ago

asked 21 hours ago

Emanula Sohn

275

"echo 2; // does not even appear" Where should that appear?

– kerbholz
21 hours ago

inside the <div id="service_search_div" class="srv_search_mini_display d-none"></div>

– Emanula Sohn
20 hours ago

1

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
20 hours ago

1

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

– Chris shi
20 hours ago

1

Have you checked in browser console. Does it display any error related to javascript

– Chris shi
20 hours ago

|
show 6 more comments

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

  ?>

edited 20 hours ago

asked 21 hours ago

Emanula Sohn

275

"echo 2; // does not even appear" Where should that appear?

– kerbholz
21 hours ago

inside the <div id="service_search_div" class="srv_search_mini_display d-none"></div>

– Emanula Sohn
20 hours ago

1

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
20 hours ago

1

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

– Chris shi
20 hours ago

1

Have you checked in browser console. Does it display any error related to javascript

– Chris shi
20 hours ago

|
show 6 more comments

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

  ?>

edited 20 hours ago

asked 21 hours ago

Emanula Sohn

275

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

  ?>

javascript php ajax

edited 20 hours ago

asked 21 hours ago

Emanula Sohn

275

edited 20 hours ago

asked 21 hours ago

Emanula Sohn

275

edited 20 hours ago

asked 21 hours ago

Emanula Sohn

275

asked 21 hours ago

Emanula Sohn

275

asked 21 hours ago

Emanula Sohn

275

"echo 2; // does not even appear" Where should that appear?

– kerbholz
21 hours ago

inside the <div id="service_search_div" class="srv_search_mini_display d-none"></div>

– Emanula Sohn
20 hours ago

1

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
20 hours ago

1

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

– Chris shi
20 hours ago

1

Have you checked in browser console. Does it display any error related to javascript

– Chris shi
20 hours ago

|
show 6 more comments

"echo 2; // does not even appear" Where should that appear?

– kerbholz
21 hours ago

inside the <div id="service_search_div" class="srv_search_mini_display d-none"></div>

– Emanula Sohn
20 hours ago

1

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
20 hours ago

1

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

– Chris shi
20 hours ago

1

Have you checked in browser console. Does it display any error related to javascript

– Chris shi
20 hours ago

"echo 2; // does not even appear" Where should that appear?

– kerbholz
21 hours ago

inside the <div id="service_search_div" class="srv_search_mini_display d-none"></div>

– Emanula Sohn
20 hours ago

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
20 hours ago

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

– Chris shi
20 hours ago

Have you checked in browser console. Does it display any error related to javascript

– Chris shi
20 hours ago

|
show 6 more comments

1 Answer
1

active

oldest

votes

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 20 hours ago

Kyaw Kyaw Soe

611211

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

– Emanula Sohn
19 hours ago

1

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

– Kyaw Kyaw Soe
19 hours ago

add a comment |

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1 Answer
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1 Answer
1

active

oldest

votes

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 20 hours ago

Kyaw Kyaw Soe

611211

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

– Emanula Sohn
19 hours ago

1

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

– Kyaw Kyaw Soe
19 hours ago

add a comment |

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 20 hours ago

Kyaw Kyaw Soe

611211

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

– Emanula Sohn
19 hours ago

1

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

– Kyaw Kyaw Soe
19 hours ago

add a comment |

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 20 hours ago

Kyaw Kyaw Soe

611211

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>

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>

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 20 hours ago

Kyaw Kyaw Soe

611211

answered 20 hours ago

Kyaw Kyaw Soe

611211

answered 20 hours ago

Kyaw Kyaw Soe

611211

answered 20 hours ago

Kyaw Kyaw Soe

611211

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

– Emanula Sohn
19 hours ago

1

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

– Kyaw Kyaw Soe
19 hours ago

add a comment |

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

– Emanula Sohn
19 hours ago

1

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

– Kyaw Kyaw Soe
19 hours ago

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

– Emanula Sohn
19 hours ago

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

– Kyaw Kyaw Soe
19 hours ago

add a comment |

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