Dva input polja sa istom AJAX skriptom (problem sa ID-om)imam jedan problem sa kojim se mucim vec par dana. Naime nisam nesto strucan po pitanju Ajax-a, pa me zanima da li moze nekako da se modifikuje ovaj kod da radi. Kapiram da moze da se resi u dve linije koda, al sta cu kad ne znam :D
Koristim jednostavan kod za Ajax/JS live pretragu na sajtu i funkcionise odlicno, ali imam problem da ga pokrenem 2x na istoj strani. Prvi radi, drugi nece nikako. Shvatio sam da je problem u tome sto polja imaju isti ID. Pokusao sam da ga resim tako sto sam napravio novi .js fajl koji sam isto pozvao u .php ali normalno nece da radi :)
Dakle da pocnem:
U php pozivam <script type="text/javascript" src=http://www.sajt.rs/scripts.js"></script>
Sadrzaj scripts.js fajla je sledeci:
Code:
//Getting value from "ajax.php".
function fill(Value) {
//Assigning value to "search" div in "search.php" file.
$('#search').val(Value);
//Hiding "display" div in "search.php" file.
$('#display').hide();
}
$(document).ready(function() {
//On pressing a key on "Search box" in "search.php" file. This function will be called.
$("#search").keyup(function() {
//Assigning search box value to javascript variable named as "name".
var name = $('#search').val();
//Validating, if "name" is empty.
if (name == "") {
//Assigning empty value to "display" div in "search.php" file.
$("#display").html("");
}
//If name is not empty.
else {
//AJAX is called.
$.ajax({
//AJAX type is "Post".
type: "POST",
//Data will be sent to "ajax.php".
url: "check.php",
//Data, that will be sent to "ajax.php".
data: {
//Assigning value of "name" into "search" variable.
search: name
},
//If result found, this funtion will be called.
success: function(html) {
//Assigning result to "display" div in "search.php" file.
$("#display").html(html).show();
}
});
}
});
});
//Getting value from "ajax.php".
function fill(Value) {
//Assigning value to "search" div in "search.php" file.
$('#search').val(Value);
//Hiding "display" div in "search.php" file.
$('#display').hide();
}
$(document).ready(function() {
//On pressing a key on "Search box" in "search.php" file. This function will be called.
$("#search").keyup(function() {
//Assigning search box value to javascript variable named as "name".
var name = $('#search').val();
//Validating, if "name" is empty.
if (name == "") {
//Assigning empty value to "display" div in "search.php" file.
$("#display").html("");
}
//If name is not empty.
else {
//AJAX is called.
$.ajax({
//AJAX type is "Post".
type: "POST",
//Data will be sent to "ajax.php".
url: "check.php",
//Data, that will be sent to "ajax.php".
data: {
//Assigning value of "name" into "search" variable.
search: name
},
//If result found, this funtion will be called.
success: function(html) {
//Assigning result to "display" div in "search.php" file.
$("#display").html(html).show();
}
});
}
});
});
PHP forma u kojoj radi ovaj kod je sledeca:
Code:
<div class="col-xs-12 mb-20">
<div class="heading-part align-center">
<h3 class="sub-heading">Unesi</h3>
</div>
<div class="col-sm-12">
<input type="hidden" name="a" value="uredjaj" />
<input autocomplete="off" type="text" id="search" placeholder="Pretraži" name="id" value="" style="text-transform:uppercase" />
</div>
<div id="display"></div>
<button type="submit" name="submit" class="btn-color btn">Potvrdi</button>
</form>
</div>
<div class="col-xs-12 mb-20">
<div class="heading-part align-center">
<h3 class="sub-heading">Unesi</h3>
</div>
<div class="col-sm-12">
<input type="hidden" name="a" value="uredjaj" />
<input autocomplete="off" type="text" id="search" placeholder="Pretraži" name="id" value="" style="text-transform:uppercase" />
</div>
<div id="display"></div>
<button type="submit" name="submit" class="btn-color btn">Potvrdi</button>
</form>
</div>
Forma nije potpuno validna ali to nije problem ovde funkcionise kako treba. Zanima me kako sada da pozovem istu formu sa istom pretragom ali drugim ID-jem (da bi radilo), znaci da nekako modifikujem onaj prvi JS kod da kad unesem <div id="display2"></div>, znaci display2 da pozove istu formu. Kada je "display" ne radi jer ga ima 2x i to kapiram.
Hvala unapred!
