Друг начин на зареждане на данни в текстова кутия?

ivo75 · 7 Май 2013

Използвам 2 Jquery кода за зареждане на данни в combobox и textbox въпроса ми е имали други начини за зареждане на данни най-вече в textbox-а

ето и кодовете

1. Combobox

Код:

$(document).ready(function()
{
$(".model_id").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;

$.ajax
({
type: "POST",
url: "get_razmer.php",
data: dataString,
cache: false,
success: function(html)
{
$(".razmer_id").html(html);
} 
});

});
});

и PHP кода т.е. файла get_razmer.php

Код:

<?php require_once('../Connections/modeli.php'); ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<option selected="selected">--Избери размер--</option>
</body>
</html>
<?php
if(isset ($_POST['id']))
{
$id=$_POST['id'];
//$id=2;


$sql="SELECT DISTINCT razmeri.id, razmeri.razmer, stoki.razmer_id, stoki.model_id FROM razmeri INNER JOIN (modeli INNER JOIN stoki ON modeli.id = stoki.model_id) ON razmeri.id = stoki.razmer_id WHERE stoki.model_id='$id'";



//$sql="SELECT DISTINCT razmeri.razmer, razmeri.id FROM (stoki INNER JOIN razmeri ON stoki.razmer_id = razmeri.id) INNER JOIN modeli ON stoki.model_id = modeli.id WHERE stoki.model_id='$id'";
$result1=mysql_query($sql);
while($row=mysql_fetch_array($result1))
{
$razmer_id=$row['id'];
$razmer=$row['razmer'];
//echo $razmer_id;

echo '<option value="'.$razmer_id.'">'.$razmer.'</option>';
}
}
?>

2. textbox-a

Код:

function get_cena(){
        $.ajax({
            type: "POST",
            url: "get_cena.php",
            data:     "model=" + document.getElementById("model_id").value + 
                    "&razmer=" + document.getElementById("razmer_id").value,
            success: function(html){
				$("#response").html(html);
              // $("#cena").html(html);
            }
        });
        
        }

PHP кода т.е. файла get_cena.php

Код:

<?php require_once('../Connections/modeli.php'); ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
</body>
</html>
<?php
//echo $_POST['model'];
if ($_POST['model'])
{
$model=$_POST['model'];
//$id=2;
$razmer =$_POST['razmer'];

$sql="SELECT DISTINCT razmeri.id, razmeri.razmer, stoki.razmer_id, stoki.model_id, stoki.cena FROM razmeri INNER JOIN (modeli INNER JOIN stoki ON modeli.id = stoki.model_id) ON razmeri.id = stoki.razmer_id WHERE stoki.razmer_id='$razmer' AND stoki.model_id='$model'";



//$sql="SELECT DISTINCT razmeri.razmer, razmeri.id FROM (stoki INNER JOIN razmeri ON stoki.razmer_id = razmeri.id) INNER JOIN modeli ON stoki.model_id = modeli.id WHERE stoki.model_id='$id'";
$result1=mysql_query($sql);
$row=mysql_fetch_array($result1);
$cena=$row['cena']; ?>

<input type="text" name="cena" value="<?php echo $cena; ?>" size="10" class="cena" disabled="disabled" /> 
<?php
//echo $cena;
//echo $razmer_id;

}
?>

Друг начин на зареждане на данни в текстова кутия?

ivo75

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