我有这个问题:在下面的代码中,我制作了一个带有条件的表单,其中如果"profileid"在数组friends中,则打印按钮"add to friends",否则打印按钮"remove to friend斯",但第二个条件不起作用,则不打印任何内容。当我第一次加载页面时,如果数组中已经有"friend id",则会有按钮"add tofriends。
这是我的代码:
<?php
$userid = $_SESSION['userid'];
$profileid = $_SESSION['profileID'];
$compressed_friends=mysql_query("SELECT friends FROM users WHERE id LIKE '$userid'");
$friends = explode (',',$compressed_friends);
if(isset($_POST['addFriends']))
{
$compressed_friends=$profileid.','.$compressed_friends;
mysql_query("UPDATE users SET friends='$compressed_friends' WHERE id='$userid'");
}
elseif(isset($_POST['removeFriends']))
{
array_filter($friends,$profileid);
$compressed_friends=implode(',', $friends);
mysql_query("UPDATE users SET friends='$compressed_friends' WHERE id='$userid'");
}
else
{
?>
<form role="form" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<?php
if(!in_array($profileid, $friends))
{
echo' <button type="submit" name="addFriends" class="btn btn-primary col-lg-3">Add to friends</button>';
}
elseif(in_array($profileid, $friends))
{
echo '<button type="submit" name="removeFriends" class="btn btn-danger col-lg-3">Remove to Friends</button>';
}
?>
</form>
<?php } ?>
让我试着回答,因为我是PHP初学者。我发现回声有问题。你曾经报道过"."
echo' <button type="submit" name="addFriends" class="btn btn-primary col-lg-3">Add to friends</button>';
你可以试试这个,也许它能解决你的问题。
echo "<button type='submit' name='addFriends' class='btn btn-primary col-lg-3'>Add to friends</button>";
谢谢!