Pass values from JQUERY to PHP then output to INPUT in FORM

Question

i need to get a value from the a form then send it to php using jquery then output the result a dropdown select menu

get the value of using jquery

 <input id="search" name="search" type="text">

send it to php and perform a query

  <select  id="farmertype" name="farmertype" >

         <option value="" > - PLEASE SELECT FARM -</option>

         ////   output here as options                                      
  </select>

my php file farm.php

<?php
include_once("../init.php");
$q = ($_POST["search"]);

$db->query("SELECT * FROM farmers ");
  while ($line = $db->fetchNextObject()) {

      $idno = $line->idno;
      echo "<option value='$idno'>$idno</option>";

   }
 }

?>

the jquery part is so messy this is where i really need help

$("#search").click(function() {
    search = $(this).attr('#search');
    $.ajax({
        type: 'GET',
        url: 'farm.php',
        data: "#search=" + search,

    });
});

saw a question just like this earlier but from another account and was deleted. — Funk Forty Niner
– Funk Forty Niner, Commented Oct 29, 2015 at 13:59
@Fred-ii- yet this one managed to get 2 upvotes and 2 favourite!! — Steve
– Steve, Commented Oct 29, 2015 at 13:59
@Steve and they think they can truly hide ;-) hah - I guess some just can't take the pressue. Hm... upvotes,... from who's "account"? — Funk Forty Niner
– Funk Forty Niner, Commented Oct 29, 2015 at 14:00

Jrey · Accepted Answer · 2015-10-29 14:03:46Z

1

try this, it will help you.

JQuery:

$("#search").click(function() {
    search = $(this).val();
    $.ajax({
        type: 'POST',
        url: 'farm.php',
        data: {searchValue:search},
        success:function(result) {
            console.log(result);
        }
    });
});

PHP:

<?php
    include_once("../init.php");
    $q = ($_POST["searchValue"]);
    $db->query("SELECT * FROM farmers");
    $result = [];
    while ($line = $db->fetchNextObject()) {
        $idno = $line->idno;
        $result = "<option value='$idno'>$idno</option>";
    }
    print_r($result);
?>

what is the purpose of your variable $q?

answered Oct 29, 2015 at 14:03

Jrey

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2 Comments

thumbi njoroge Over a year ago

this one works. how do i output the result to a particular id in the form

Jrey Over a year ago

$('#id').text(result[array_index]) if you need to replace a text or $('#id').val(result[array_index]) if you need to replace a value

fico7489 · Accepted Answer · 2015-10-29 13:59:42Z

0

Your jquery can be like :

$("#search").click(function() {
    search = $('#search').val();
    $.ajax({
        type: 'GET',
        url: 'farm.php',
        data: {search : search},
        success: function(html){
            alert(html);
        }

    });
});

edited Oct 29, 2015 at 13:59

answered Oct 29, 2015 at 13:54

fico7489

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1 Comment

Steve Over a year ago

$(this).attr('#search'); Whats that?? Also, the php file specifies a POST request (though it never uses it).

Ankit rana · Accepted Answer · 2015-10-29 14:13:26Z

0

$("#search").click(function() { /* I think you should use keyUp or use click on a button, nobody clicks an input box */
var search = $(this).val();
$.ajax({
    method: 'POST', // 
    url: 'farm.php',
    data: {'search' : search},
    success: function(data){
        alert(data);
    }

});

});

edited Oct 29, 2015 at 14:13

answered Oct 29, 2015 at 14:05

Ankit rana

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1 Comment

Steve Over a year ago

the php file specifies a POST request (though it never uses it)

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