我在创建一个允许用户收藏帖子的系统时遇到问题。我知道我在这里遗漏了一些东西,但我不确定是什么,也不知道如何修复它以使其正常工作。
mysql_select_db($database_connection, $connection);
$query_deltakelse = "SELECT brukernavn, postID FROM deltakelse";
$deltakelse = mysql_query($query_deltakelse, $connection) or die(mysql_error());
$row_deltakelse = mysql_fetch_assoc($deltakelse);
$totalRows_deltakelse = mysql_num_rows($deltakelse);
$username=$_SESSION['valid_user'];
if(in_array($_POST['id'], $row_deltakelse))
{
mysqli_query($connection,"DELETE FROM deltakelse WHERE postID='id'");
}
else
{
mysqli_query($connection,"INSERT INTO bruker (brukernavn, postID) VALUES ('$username', '$id')");
}
$(document).ready(function() {
$('.delta').on('click', null, function() {
var _this = $(this);
var post_id = _this.data('id');
$.ajax({
type : 'POST',
url : '/file.php',
dataType : 'json',
data : 'id='+ post_id,
complete : function(data) {
if(_this.text() == 'Fave this')
{
_this.html('Faved!');
}
else
{
_this.html('Fave this');
}
}
});
});
});
<a href="#" class="delta" data-id="<?php echo $post_id; ?>">Favorite</a>
最佳答案
您正在混合 mysql 和 mysqli 函数。你不能这样做。
这是MYSQL
mysql_select_db($database_connection, $connection);
$query_deltakelse = "SELECT brukernavn, postID FROM deltakelse";
$deltakelse = mysql_query($query_deltakelse, $connection) or die(mysql_error());
$row_deltakelse = mysql_fetch_assoc($deltakelse);
$totalRows_deltakelse = mysql_num_rows($deltakelse);
$username=$_SESSION['valid_user'];
此处您将 MySQLi 与 MySQL 资源一起使用。那是行不通的。
if(in_array($_POST['id'], $row_deltakelse))
{
mysqli_query($connection,"DELETE FROM deltakelse WHERE postID='id'");
}
else
{
mysqli_query($connection,"INSERT INTO bruker (brukernavn, postID) VALUES ('$username', '$id')");
}
也改变一下:
if(in_array($_POST['id'], $row_deltakelse))
{
mysql_query("DELETE FROM deltakelse WHERE postID='id'", $connection);
}
else
{
mysql_query("INSERT INTO bruker (brukernavn, postID) VALUES ('$username', '$id')", $connection);
}
关于php - 最喜欢的帖子脚本,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22645093/