我有一个应用程序,我的用户可以创建类似文章的内容。现在,我不想限制每个用户写的文章数量。
我只是不知道该怎么做 - 我可以想象我应该使用:mysql_count_num_rows或类似的东西。
在我的数据库中,我有一个名为:opslag的表,其中我有userid_bywho和user_bywho。我想我应该计算表中找到用户ID的次数,如果它高于5,则用户不应该能够创建该文章。
我在SESSION中有用户ID。
我希望你能帮我知道如何计算ID的显示次数。
if(isset($_POST['title']) && isset($_POST['daystart']) && isset($_POST['address']) && isset($_POST['phone']) && isset($_POST['email']) && isset($_POST['website']) && isset($_POST['content']))
{
if($_POST['title'] === ''){
$errMsg = "Du skal udfylde titel";
}
elseif($_POST['daystart'] === ''){
$errMsg = "Du skal udfylde start dag";
}
elseif($_POST['address'] === ''){
$errMsg = "Du skal udfylde adresse";
}
elseif($_POST['phone'] === ''){
$errMsg = "Du skal udfylde telefon";
}
elseif($_POST['email'] === ''){
$errMsg = "Du skal udfylde email";
}
elseif($_POST['website'] === ''){
$errMsg = "Du skal udfylde website";
}
elseif($_POST['content'] === ''){
$errMsg = "Du skal udfylde beskrivelse";
} else {
$sql = mysql_query("INSERT INTO opslag (userid_bywho, user_bywho, title, category, link, daystart, address, phone, email, website, content)VALUES('$userid_bywho', '$user_bywho', '$title', '$category', '$link', '$daystart', '$address', '$phone', '$email', '$website', '$content')")or die(mysql_error());
$add_success = "Dit opslag er oprettet.";
}
}
答案 0 :(得分:4)
$user_id=$_SESSION['user_id'];
$num=10; //default he can't create the article
if (!is_numeric($user_id)) die('SQL INJECTION!');
$sql="SELECT userid_bywho, count(*) as num FROM opslag WHERE userid_bywho=$user_id GROUP BY userid_bywho";
$result = mysql_query($sql);
if (!$result) {
echo "can't run query";
} else {
$row = mysql_fetch_row($result);
$num = $row[1];
}
if ($num>=5 ){
$errMsg = "You have more than 5 articles";
} elseif(isset($_POST['title']) && isset($_POST['daystart']) && isset($_POST['address']) && isset($_POST['phone']) && isset($_POST['email']) && isset($_POST['website']) && isset($_POST['content']))
{
if($_POST['title'] === ''){
$errMsg = "Du skal udfylde titel";
}
elseif($_POST['daystart'] === ''){
$errMsg = "Du skal udfylde start dag";
}
elseif($_POST['address'] === ''){
$errMsg = "Du skal udfylde adresse";
}
elseif($_POST['phone'] === ''){
$errMsg = "Du skal udfylde telefon";
}
elseif($_POST['email'] === ''){
$errMsg = "Du skal udfylde email";
}
elseif($_POST['website'] === ''){
$errMsg = "Du skal udfylde website";
}
elseif($_POST['content'] === ''){
$errMsg = "Du skal udfylde beskrivelse";
} else {
$sql = mysql_query("INSERT INTO opslag (userid_bywho, user_bywho, title, category, link, daystart, address, phone, email, website, content)VALUES('$userid_bywho', '$user_bywho', '$title', '$category', '$link', '$daystart', '$address', '$phone', '$email', '$website', '$content')")or die(mysql_error());
$add_success = "Dit opslag er oprettet.";
}
}
答案 1 :(得分:0)
SELECT COUNT(*)FROM opslag WHERE userid = _bywho = USERID
然后获取结果
if ($fetch > 5 ) {
echo 'oops you posted more than 5!'
}
答案 2 :(得分:0)
不要使用mysql_count_num_rows,假设ID是数字,请使用SQL:
$sql = "select count(*) from table
where {$_SESSION['ID']} in ( userid_bywho , user_bywhouser_id)";
注意:注意SQL注入,转义它,准备并执行句子。