我已经完成了这项工作,但我确信有更好的方法来完成这项工作。但是,我已经搜索了很多个小时,但没有找到我想要做的事情的确切答案。基本上从URL获取变量usrID,我需要在MySQL中搜索该用户的相应信息。后来我想使用我的页面(更好的网站)上的不同字段来个性化体验。
<?php
$servername = "localhost";
$username = "authorized-user";
$password = "secret";
$dbname = "agentDB";
$usrID = "001";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM agentInfo WHERE usrID = '$usrID'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$Lname = $row["Lname"];
$Fname = $row["Fname"];
$tl = $row["tl"];
}
}
mysqli_close($conn);
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Load MySQL Data into Corresponding PHP Variables</title>
</head>
<body>
here is the body<br>
My name is: <?php echo $Fname; ?> <?php echo $Lname; ?><?php echo $tl; ?>
</body>
</html>
最佳答案
您可以创建一个变量来存储全名,然后在其上存储“tl”,如下所示:
$user_info = $Lname . ", " . $Fname . ": " . $tl;
然后:
<?php echo $user_info; ?>
无论您在哪里需要该信息。
如果您想最大限度地减少分配的变量数量,您可以将其包装在函数中并返回所需的数据字段:
function fetchUserData(userData) {
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM agentInfo WHERE usrID = '$usrID'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$userData = $row[userData];
}
}
return $userData;
}
mysqli_close($conn);
你可以这样获取指定的数据:
<?php echo fetchUserData("Fname"); ?>
关于php - 将 MySQL 数据加载到相应的 PHP 变量中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50477901/