Php I';我正在尝试使用';回声&x27;体内
当我在标记内运行php代码时,我希望“echo”在标记内回响,但它没有,而是打印我之前“echo”的内容 我已经修改了代码,修复了所有我知道的东西,但仍然不起作用 “dbh.inc.php”Php I';我正在尝试使用';回声&x27;体内,php,html,Php,Html,当我在标记内运行php代码时,我希望“echo”在标记内回响,但它没有,而是打印我之前“echo”的内容 我已经修改了代码,修复了所有我知道的东西,但仍然不起作用 “dbh.inc.php” $servername = "localhost"; $dBUsername = "root"; $dBPassword = ""; $dBName = "hayleyblog"; $conn = mysqli_connect($servername, $dBUsername, $dBPassword,
$servername = "localhost";
$dBUsername = "root";
$dBPassword = "";
$dBName = "hayleyblog";
$conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBName);
if (!$conn) {
die("Connection failed: ".mysqli_connect_error());
} else {
echo "<p>connected</p>";
}
$servername=“localhost”;
$dBUsername=“root”;
$dBPassword=“”;
$dBName=“hayleyblog”;
$conn=mysqli_connect($servername、$dBUsername、$dBPassword、$dBName);
如果(!$conn){
die(“连接失败:”.mysqli_connect_error());
}否则{
回声“已连接””;
}
我希望“connected”打印在body标记中,但它打印在html标记之前,检查连接时,应该将值存储在variable中,而不是echo中
$servername = "localhost";
$dBUsername = "root";
$dBPassword = "";
$dBName = "hayleyblog";
$conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBName);
if (!$conn) {
die("Connection failed: ".mysqli_connect_error());
} else {
$conn = "<p>connected</p>";
}
$servername=“localhost”;
$dBUsername=“root”;
$dBPassword=“”;
$dBName=“hayleyblog”;
$conn=mysqli_connect($servername、$dBUsername、$dBPassword、$dBName);
如果(!$conn){
die(“连接失败:”.mysqli_connect_error());
}否则{
$conn=“已连接””;
}
然后您可以如下所示进行回音。
<!DOCTYPE html>
<?php
require "dbh.inc.php";
?>
<html>
<head>
</head>
<body>
<?php
echo $conn;
?>
</body>
</html>
请理解include的概念,它可以帮助您在PHP中取得进步。 您将文件包括在其中,其工作方式如下 PHP include语句添加其他文件代码添加一个当前文件,将include语句附加到其中 包含的文件意味着您的代码最终存在
$servername = "localhost";
$dBUsername = "root";
$dBPassword = "";
$dBName = "hayleyblog";
$conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBName);
$connection_status="";
if (!$conn) {
die("Connection failed: ".mysqli_connect_error());
} else {
$connection_status = "<p>connected</p>";
}
<!DOCTYPE html>
<?php
require "dbh.inc.php";
?>
<html>
<head>
</head>
<body>
<?php
echo $connection_status;
?>
</body>
</html>
$servername=“localhost”;
$dBUsername=“root”;
$dBPassword=“”;
$dBName=“hayleyblog”;
$conn=mysqli_connect($servername、$dBUsername、$dBPassword、$dBName);
$connection_status=“”;
如果(!$conn){
die(“连接失败:”.mysqli_connect_error());
}否则{
$connection_status=“已连接””;
}
试试这个,我希望你能理解,如果有任何疑问,请在评论中问我,我可以向你详细解释。您应该深入理解PHP包含语句的概念。
谢谢change
echo“connected”代码>至$conn=“已连接””在您的else
中选择code>,然后选择echo$conn在你的HTMLHe中,他还需要echo$conn代码>我已经在我的系统中检查过了。它工作正常。可能你的数据库名称或用户名中缺少了什么。不,我明白“include'file.php'是什么意思,我自学了php,我13岁就学会了,刚刚开始学习,但是谢谢你的帮助,总是很高兴从别人的角度看问题
$servername = "localhost";
$dBUsername = "root";
$dBPassword = "";
$dBName = "hayleyblog";
$conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBName);
$connection_status="";
if (!$conn) {
die("Connection failed: ".mysqli_connect_error());
} else {
$connection_status = "<p>connected</p>";
}
<!DOCTYPE html>
<?php
require "dbh.inc.php";
?>
<html>
<head>
</head>
<body>
<?php
echo $connection_status;
?>
</body>
</html>