我创建了一个简单的php脚本,用于使用wamp将数据上传到MySQL服务器。 我的脚本 -Wamp服务器显示任何实际的语法错误宽度错误
<?php
$server = "127.22.9.0";
$user = "root";
$password = "";
$db = "complaints";
$conn = new mysqli($server, $user, $password, $db);
if ($conn->connect_error) { // This is that line 9
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$sql = "INSERT INTO complaints (email, title, details) VALUES ('".$_POST['email']."', '".$_POST['title']."', '".$_POST['dtls']."')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
?>
当我执行此WAMP的服务器上,它显示了一个错误
Parse error: syntax error, unexpected '{' in C:\wamp64\www\index.php on line 9
我已经彻底搜查了错误,但一无所获。
任何人都可以帮助我吗?
编辑:我解决我的问题,但在一个不寻常的方式 -
我重新输入我的代码,并去掉了一些不必要的空间和改变,如果语句的结构。现在,我的代码工作 -
$server = "127.22.9.0";
$user = "root";
$password = "";
$db = "complaints";
$conn = new mysqli($server, $user, $password, $db);
nl2br($_POST['dtls']);
if($conn->connect_error)
{
die("Error: ".$conn->connect_error);
}
$sql = "INSERT INTO complaints (email, title, details) VALUES ('".$_POST['email']."', '".$_POST['title']."', '".$_POST['dtls']."')";
if($conn->query($sql))
{
echo "<h1>Success!! Your complaint has been successfully registered!!</h1>";
}
else
{
echo "Error: " . $sql . "<br>" . $conn->error;
}
这些更改是相当不必要的,但它的工作原理。
$ conn-> connect_error显示错误,因为这是mysqli的方法使用$ conn-> connect_error() –
我补充说。现在它显示错误'死''语法错误,意想不到的'死'(T_EXIT)在C:\ wamp64 \ www \ index.php在线10' –
你使用哪个php版本? – PassionInfinite