我只是在浏览一些示例网页,因为一段时间以来我没有接触过数据库提交的表单,而且我注意到mysql_ *不再被广泛使用,并已被mysqli_ *取代。我有一个简单的表单(可以在需要时发布,但我认为问题不在于表单内)。我的数据库连接代码如下,
<?php
//addslash() for sanitization
ExtendedAddslash($_POST);
//correct form name's for data to be inserted
$name = $_POST['name'];
$specName = $_POST['specName'];
$weight = $_POST['weight'];
$weightUnits = $_POST['weight-units'];
$description = $_POST['description'];
$conn = mysqli_connect("localhost","root","CENSORED","testDB");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL :" . mysqli_conect_error();
}
else {
echo "Connection was OK!\n";
}
$result = mysqli_query($conn,"INSERT into 'testDB.users' (name, specName, description, weight, weightUnits) VALUES ('$name','$specName','$description', '$weight', '$weightUnits')");
if ($result) {
echo "Success\n";
}
else {
echo "Error\n";
}
mysqli_close($conn);
?>
过去,这很简单,但是我更改了一些语法以匹配mysqli标准。连接测试始终会导致“连接正常”,但是我似乎总是会收到“错误”。问题是$result
变量吗?还是我只是没有正确执行此查询。