Question

我有一个网站，用户可以使用标题和链接提交帖子。我目前正在努力，如果没有提交链接，它将默认为自己的评论。（这将是/comments.php?comments=$postsId）

这里是我认为是问题的代码部分，当我删除if else语句时，它完美地运行

function addhttp($url) {
    if (!preg_match("~^(?:f|ht)tps?://~i", $url)) {
        $url = "http://" . $url;
    }
    return $url;
};

if(empty($link)){
    $postsID = mysqli_query($conn, 'SELECT `id` FROM `data` ORDER BY id DESC LIMIT 0, 1')or die(mysqli_error($conn));
    $row = mysqli_fetch_array($postsID)
    $link = 'http://linkdump.rf.gd/comments.php?comments='.$row['id'].'';
    echo "Self post"
}else{
    $link = addhttp($link);
    echo "Link post";
};
$subtitle = strip_tags ( $title );
mysqli_query($conn,"INSERT INTO `data`(`title`,`link`, `nsfw`) VALUES ('$subtitle', '$link', '$nsfw')") or die(mysqli_error($conn));

但是，当用户被重定向到页面时，它只是空白...

这是页面上的完整代码：

$title = $_POST['title'];
$link = $_POST['link'];
$nsfw = $_POST['nsfw'];
include 'functions/db.php';
$conn = mysqli_connect("$url","$user","$pass","$db", null, '/tmp/mysql.sock');
$title = mysqli_real_escape_string($conn, $title);
$link = mysqli_real_escape_string($conn, $link);
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
echo 'Connected successfully<br><br>';

function addhttp($url) {
    if (!preg_match("~^(?:f|ht)tps?://~i", $url)) {
        $url = "http://" . $url;
    }
    return $url;
};

if(empty($link)){
    $postsID = mysqli_query($conn, 'SELECT `id` FROM `data` ORDER BY id DESC LIMIT 0, 1')or die(mysqli_error($conn));
    $row = mysqli_fetch_array($postsID)
    $link = 'http://linkdump.rf.gd/comments.php?comments='.$row['id'].'';
    echo "Self post"
}else{
    $link = addhttp($link);
    echo "Link post";
};
$subtitle = strip_tags ( $title );
mysqli_query($conn,"INSERT INTO `data`(`title`,`link`, `nsfw`) VALUES ('$subtitle', '$link', '$nsfw')") or die(mysqli_error($conn));
?>

我觉得我错过了一些简单的事情......

Answer 1

尝试替换那些';'

    if(empty($link)){
    $postsID = mysqli_query($conn, 'SELECT `id` FROM `data` ORDER BY id DESC LIMIT 0, 1')or die(mysqli_error($conn));
    $row = mysqli_fetch_array($postsID);
    $link = 'http://linkdump.rf.gd/comments.php?comments='.$row['id'].'';
    echo "Self post";
}else{
    $link = addhttp($link);
    echo "Link post";
}

告诉我们是否有帮助

PHP提交是空白的，没有回声

1 个答案: