我如何从 link 中获取变量 "message" 并使其在数据库中更新?
How would I get the variable "message" from the link and make it update in the database?
嘿,所以我正在制作一个人们可以发送消息的站点,我想将其放入我的数据库中。但他们可以更新消息。这是 link 我正在使用 http://example.com/file.php?message=hi&referral=1 但在访问它时我收到此错误
0 Row updated, added hi to locker .
这是一个错误,因为没有更新任何行
<?php
define("MYSQL_HOST", "localhost");
define("MYSQL_PORT", "3306");
define("MYSQL_DB", "db");
define("MYSQL_TABLE", "table");
define("MYSQL_USER", "user");
define("MYSQL_PASS", "pass");
$mysqli = new mysqli(MYSQL_HOST, MYSQL_USER, MYSQL_PASS, MYSQL_DB);
if ($mysqli->connect_errno)
{
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
$referral = $_GET['refferal'];
$message = $_GET['message'];
if (!($stmt = $mysqli->prepare("UPDATE ".MYSQL_DB.".".MYSQL_TABLE." SET message=(?) WHERE referral=(?) ")))
{
echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
$stmt->bind_param('ds', $message, $referral);
if (!$stmt->execute())
{
echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
}
else
{
printf("%d Row updated, added ".$message." to locker ".$referral." .\n", mysqli_stmt_affected_rows($stmt));
}
?>
假设您有一个 url 作为 www.google.com
,并且您想从 url 参数传递某些数据属性。对于此示例,link 想要如下内容。
www.google.com?color=blue
您将执行以下操作以从 URL.
中检索值
// First lets define this variable so we do not get any undefined variable warnings in the error log.
$color = NULL;
// second would be lets validate that this get parameter is set and not empty.
if ( !empty($_GET['color']) ) {
// We will then define the REQUEST to a variable for later calling.
$color = $_REQUEST['color'];
}
然后你可以简单地 echo $color
它的值将是 blue
.
然后您可以简单地将您的 $color 变量应用到您的数据库语句。
最后,您的 refferal
REQUEST 中有一个拼写错误,它应该是 referral
。
嘿,所以我正在制作一个人们可以发送消息的站点,我想将其放入我的数据库中。但他们可以更新消息。这是 link 我正在使用 http://example.com/file.php?message=hi&referral=1 但在访问它时我收到此错误
0 Row updated, added hi to locker .
这是一个错误,因为没有更新任何行
<?php
define("MYSQL_HOST", "localhost");
define("MYSQL_PORT", "3306");
define("MYSQL_DB", "db");
define("MYSQL_TABLE", "table");
define("MYSQL_USER", "user");
define("MYSQL_PASS", "pass");
$mysqli = new mysqli(MYSQL_HOST, MYSQL_USER, MYSQL_PASS, MYSQL_DB);
if ($mysqli->connect_errno)
{
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
$referral = $_GET['refferal'];
$message = $_GET['message'];
if (!($stmt = $mysqli->prepare("UPDATE ".MYSQL_DB.".".MYSQL_TABLE." SET message=(?) WHERE referral=(?) ")))
{
echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
$stmt->bind_param('ds', $message, $referral);
if (!$stmt->execute())
{
echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
}
else
{
printf("%d Row updated, added ".$message." to locker ".$referral." .\n", mysqli_stmt_affected_rows($stmt));
}
?>
假设您有一个 url 作为 www.google.com
,并且您想从 url 参数传递某些数据属性。对于此示例,link 想要如下内容。
www.google.com?color=blue
您将执行以下操作以从 URL.
中检索值// First lets define this variable so we do not get any undefined variable warnings in the error log.
$color = NULL;
// second would be lets validate that this get parameter is set and not empty.
if ( !empty($_GET['color']) ) {
// We will then define the REQUEST to a variable for later calling.
$color = $_REQUEST['color'];
}
然后你可以简单地 echo $color
它的值将是 blue
.
然后您可以简单地将您的 $color 变量应用到您的数据库语句。
最后,您的 refferal
REQUEST 中有一个拼写错误,它应该是 referral
。