对数据库中的数据进行操作

问题描述 投票:0回答:2

我有来自数据库的数据的页面:

<?php
$host = 'localhost';
$user = 'root';
$pass = '';
$base = 'zar';
try {

$db = new PDO('mysql:host='.$host.';dbname='.$base.';charset=utf8', $user, $pass, array(PDO::ATTR_EMULATE_PREPARES => false,PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION));

$name = 'name';
$pin = 'pin';
$ip = 'ip';
$id = 'id';

$statement = $db->query('SELECT id, ip, name, pin FROM devic');

foreach($statement as $wiersz)
{

?>
<div id = "<?php print($wiersz['id'])?>"
<img class= "obraz"  src="css/bulb_off.png" alt="...">
 <div class="relayBlock"><span class="relayTitle"><?php print($wiersz['name']) ?> </span>
    <button class="btn btn-block btn-lg btn-primary " value="on">Wł</button>
    <button class="btn btn-block btn-lg btn-danger " value="off">Wył</button>
</div>
</div>
<?php
}   
$statement->closeCursor();

} catch(PDOException $err) {
    exit('error: '.$err->getMessage());
}
?>

JS:

<script type="text/javascript">
        $(document).ready(function () {
            $('.btn').click(function() {
                var val = $(this).val();            
                $.ajax({
                    url: "try.php",
                    type: "POST",
                    data: {'myVar': val},                         
success : function(data) {
        if (data == 1) {

            $('.obraz').attr({src : "css/bulb_on.png"})
        }
        else{
            $('.obraz').attr({src : "css/bulb_off.png"})
    }
}
      });
            });
        });
    </script>

现在,如果我单击任何按钮,数据库中所有行的图片都会更改。我想这样做,在从数据库中单击特定行的按钮后,图片仅针对该行更改。我知道这是因为每张图片都有相同的图片类,但我可以用不同的方式吗?

php jquery mysql ajax
2个回答
1
投票
var idParent = getvalueId from the div parent
if (data == 1) {
     $('#' + idParent +' .obraz').attr({src : "css/bulb_on.png"})
} else{
     $('#' + idParent +' .obraz').attr({src : "css/bulb_off.png"})
}

jquery, selector for class within id


0
投票

我添加更改的代码,它的工作原理:JS:

<script type="text/javascript">
        $(document).ready(function () {
            $('.btn').click(function() {
                var val = $(this).val();
                var idParent = $(this).parent().attr("id");
                $.ajax({
                    url: "try.php",
                    type: "POST",
                    data: {'myVar': val},

success : function(data) {
        if (data == 1) {            
            $('#' + idParent +' .obraz').attr({src : "css/bulb_on.png"})
        }
        else{
            $('#' + idParent +' .obraz').attr({src : "css/bulb_off.png"})
    }
}
      });
            });
        });
    </script>

谢谢@RicardEspinàsLlovet的建议。

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