PHP插入到两个不同的表中
我需要知道是否可以在两个数据库表中插入一些数据 我有以下代码,用于验证用户是否已登录:PHP插入到两个不同的表中,php,mysql,database,insert,Php,Mysql,Database,Insert,我需要知道是否可以在两个数据库表中插入一些数据 我有以下代码,用于验证用户是否已登录: <?php // Start the session (pretty important!) session_start(); // Establish a link to the database $dbLink = mysql_connect('localhost', 'DBUSER', 'PW'); if (!$dbLink) die('Can\'t establish a connection
<?php
// Start the session (pretty important!)
session_start();
// Establish a link to the database
$dbLink = mysql_connect('localhost', 'DBUSER', 'PW');
if (!$dbLink) die('Can\'t establish a connection to the database: ' . mysql_error());
$dbSelected = mysql_select_db('DBNAME', $dbLink);
if (!$dbSelected) die ('We\'re connected, but can\'t use the table: ' . mysql_error());
$isUserLoggedIn = false;
$query = 'SELECT * FROM users WHERE session_id = "' . session_id() . '" LIMIT 1';
$userResult = mysql_query($query);
if(mysql_num_rows($userResult) == 1){
$_SESSION['user'] = mysql_fetch_assoc($userResult);
$isUserLoggedIn = true;
}else{
if(basename($_SERVER['PHP_SELF']) != 'index.php'){
header('Location: index.php');
exit;
}
}
?>
HTML表单不是必需的,因为您知道它的外观
现在我有了下面的代码,它将一些数据插入到数据库表中。它的数据库表与上面的代码不同
这就是php代码:
<?php
include ('adresa-site.php');
if(isset($_POST['add']))
{
$dbhost = 'localhost';
$dbuser = 'DBUSER';
$dbpass = 'PW';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Eroare de conexiune: ' . mysql_error());
}
if(! get_magic_quotes_gpc() )
{
$link = $web_site.''.$_POST["stire"].'?idstire='.$_POST["idstire"];
$poza = addslashes ($_POST['poza']);
$nume = addslashes ($_POST['nume']);
$prenume = addslashes ($_POST['prenume']);
$varsta = addslashes ($_POST['varsta']);
$localitatea = addslashes ($_POST['localitatea']);
$numep = addslashes ($_POST['numep']);
$idstire = addslashes ($_POST['idstire']);
$stire = addslashes ($_POST['stire']);
$data = gmDate('Y-m-d H:i:s');
}
else
{
$link = $web_site."".$_POST['stire']."?idstire=".$_POST['idstire'];
$poza = htmlentities($_POST['poza'], ENT_QUOTES | ENT_HTML5);
$nume = htmlentities($_POST['nume'], ENT_QUOTES | ENT_HTML5);
$prenume = htmlentities($_POST['prenume'], ENT_QUOTES | ENT_HTML5);
$varsta = $_POST["varsta"];
$localitatea = htmlentities($_POST['localitatea'], ENT_QUOTES | ENT_HTML5);
$numep = htmlentities($_POST['numep'], ENT_QUOTES | ENT_HTML5);
$idstire = htmlentities($_POST['idstire'], ENT_QUOTES | ENT_HTML5);
$stire = $_POST["stire"];
$data = gmDate('Y-m-d H:i:s');
}
$sql = "INSERT INTO stiri2".
"(link, poza, nume, prenume, varsta, localitatea, numep, idstire, stire, accesari, data) ".
"VALUES('$link','$poza','$nume','$prenume','$varsta','$localitatea','$numep','$idstire','$stire','$accesari', NOW())";
mysql_select_db('DBNAME');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Nu s-au putut adauga datele: ' . mysql_error());
}
echo "
<p class='text-center'>Copiaza si trimite link-ul de mai jos, prietenului tau</p>
<textarea rows='2' cols='55'>http://site.ro/" . $_POST['stire'] . "?idstire=" . $_POST['idstire'] . "</textarea>
<br />
<p class='text-center'>Acceseaza <a target='_blank'href='http://site.ro/" . $_POST['stire'] . "?idstire=" . $_POST['idstire'] . "'>ACEST</a> link pentru previzualizare</p>
";
mysql_close($conn);
}
?>
现在,来自users数据库表的用户希望在stiri2数据库表中插入新信息
我必须从users数据库表中获取会话id,并使用插入的信息更新users和stiri2数据库表
我将创建一个小地图:
用户已登录,其会话id为328234917238914
现在他想插入一些新信息,这些信息将存储在stiri2数据库表中
每个用户都会添加他们的信息,因此我必须采取措施避免与插入的信息重叠
我需要知道的是,如何为当前用户创建insert。您想要的可能是这样的:mysql\u insert\u id 你要做的是:一个用户注册,你得到这个id,我们把它作为你的用户id 您可以在会话中存储此用户标识
如果用户登录,您将查找此id并在其他查询中使用它。为什么需要将会话id保存在数据库中?您可以将信息与用户id关联起来。这不需要任何会话id,但我需要知道如何为每个用户进行插入!没人能帮我解决那个问题?太复杂了。。对我来说是的,但我想不是对你们!这并不复杂,但您需要阅读stackoverflow帮助页面和。之后,试着自己解决这个问题。在那之后,发布一个更简洁、更通俗的问题。好吧,如果你这样做,你可能会在发布问题之前找到答案。。我会努力解决自己的问题。我希望如果有人能帮助我,因为我是php.Ty的新手,请发表评论。我将从php.net阅读整个页面。另外,我发现了一些有趣的东西。。Insert into stiri2从id='CURENT id'…是正确id的用户中选择*。@MuzicaVeche您应该在检查用户凭据时检索您的用户id我尝试过:$sql=Insert into stiri。链接、poza、nume、prenume、varsta、localitatea、idstire、stire、accesari、数据。值“$link”、“$poza”、“$nume”、“$prenume”、“$varsta”、“$localitatea”、“$idstire”、“$stire”、“$accesari”,现在。从id='.$\会话['id']的用户中选择*限制1,但是不起作用。。你有别的解决办法吗?
<?php
include ('adresa-site.php');
if(isset($_POST['add']))
{
$dbhost = 'localhost';
$dbuser = 'DBUSER';
$dbpass = 'PW';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Eroare de conexiune: ' . mysql_error());
}
if(! get_magic_quotes_gpc() )
{
$link = $web_site.''.$_POST["stire"].'?idstire='.$_POST["idstire"];
$poza = addslashes ($_POST['poza']);
$nume = addslashes ($_POST['nume']);
$prenume = addslashes ($_POST['prenume']);
$varsta = addslashes ($_POST['varsta']);
$localitatea = addslashes ($_POST['localitatea']);
$numep = addslashes ($_POST['numep']);
$idstire = addslashes ($_POST['idstire']);
$stire = addslashes ($_POST['stire']);
$data = gmDate('Y-m-d H:i:s');
}
else
{
$link = $web_site."".$_POST['stire']."?idstire=".$_POST['idstire'];
$poza = htmlentities($_POST['poza'], ENT_QUOTES | ENT_HTML5);
$nume = htmlentities($_POST['nume'], ENT_QUOTES | ENT_HTML5);
$prenume = htmlentities($_POST['prenume'], ENT_QUOTES | ENT_HTML5);
$varsta = $_POST["varsta"];
$localitatea = htmlentities($_POST['localitatea'], ENT_QUOTES | ENT_HTML5);
$numep = htmlentities($_POST['numep'], ENT_QUOTES | ENT_HTML5);
$idstire = htmlentities($_POST['idstire'], ENT_QUOTES | ENT_HTML5);
$stire = $_POST["stire"];
$data = gmDate('Y-m-d H:i:s');
}
$sql = "INSERT INTO stiri2".
"(link, poza, nume, prenume, varsta, localitatea, numep, idstire, stire, accesari, data) ".
"VALUES('$link','$poza','$nume','$prenume','$varsta','$localitatea','$numep','$idstire','$stire','$accesari', NOW())";
mysql_select_db('DBNAME');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Nu s-au putut adauga datele: ' . mysql_error());
}
echo "
<p class='text-center'>Copiaza si trimite link-ul de mai jos, prietenului tau</p>
<textarea rows='2' cols='55'>http://site.ro/" . $_POST['stire'] . "?idstire=" . $_POST['idstire'] . "</textarea>
<br />
<p class='text-center'>Acceseaza <a target='_blank'href='http://site.ro/" . $_POST['stire'] . "?idstire=" . $_POST['idstire'] . "'>ACEST</a> link pentru previzualizare</p>
";
mysql_close($conn);
}
?>
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| id | link | poza | nume | prenume | varsta | localitatea | numep | idstire | stire | data | accesari |
--------------------------------------------------------------------------------------------------------