Редактиране на 1 код

[niksdevil]

<?
$active=mysql_query("SELECT * FROM `pics` WHERE status='1' ORDER BY id DESC LIMIT 20") or die(mysql_error());
while($row = mysql_fetch_assoc($active)){   
//вадиш си инфото
?>
<input type="hidden" name="id" value="'.$row['id'].'"/><img src="<?php echo $row['link']; ?>" width="128" height="96">
<input type="submit" name="submit" value="Одобри" class="submit_big"/>    
<?php
}
if(isset($_POST['submit'])){

 $sql="UPDATE `pics` SET `status`='0' WHERE `id`='".$_POST['id']."'";

mysql_query($sql) or die(mysql_error());

echo '<meta http-equiv="refresh" content="0">';

}?>

Искам да направя от статус 1 да се преправя на статус 0, обаче след като натисна одобри не става идей :idea:

 

[Fakeheal]

<?php
$active=mysql_query("SELECT * FROM `pics` WHERE status='1' ORDER BY id DESC LIMIT 20") or die(mysql_error());
while($row = mysql_fetch_assoc($active)){   
//вадиш си инфото
?>
<input type="hidden" name="id" value="<?php echo $row['id'] ?>"/><img src="<?php echo $row['link']; ?>" width="128" height="96">
<input type="submit" name="submit" value="Одобри" class="submit_big"/>    
<?php
}
if(isset($_POST['submit'])){

 $sql="UPDATE `pics` SET `status`='0' WHERE `id`='".$_POST['id']."'";

mysql_query($sql) or die(mysql_error());

echo '<meta http-equiv="refresh" content="0">';

}
?>

 

[niksdevil]

Не става тъка

 

[Fakeheal]

Не става тъка

Сложил ли си input-ите в form?

<?php
$active=mysql_query("SELECT * FROM `pics` WHERE status='1' ORDER BY id DESC LIMIT 20") or die(mysql_error());
while($row = mysql_fetch_assoc($active)){   
//вадиш си инфото
?>
<form method="post">
<input type="hidden" name="id" value="<?php echo $row['id'] ?>"/><img src="<?php echo $row['link']; ?>" width="128" height="96">
<input type="submit" name="submit" value="Одобри" class="submit_big"/>    
</form>
<?php
}
if(isset($_POST['submit'])){

 $sql="UPDATE `pics` SET `status`='0' WHERE `id`='".$_POST['id']."'";

mysql_query($sql) or die(mysql_error());

echo '<meta http-equiv="refresh" content="0">';

}
?>