我正在学习MySql,但我遇到了这个问题.以下代码应该只在db表中查询users表中的所有内容.但它返回此错误.Error: SELECT * FROM users这对我毫无帮助.我能够成功地将项目插入数据库,但我无法从中进行选择.我也试过$sql = "SELECT * FROM ama.users";我的数据库结构了
ama
|-users
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任何帮助将非常感激.
$conn = new mysqli($_ENV['OPENSHIFT_MYSQL_DB_HOST'],$_ENV['OPENSHIFT_MYSQL_DB_USERNAME'], $_ENV['OPENSHIFT_MYSQL_DB_PASSWORD'], 'ama');
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$username = "Doe";
$password = "johnexample";
$sql = "SELECT * FROM users";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
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从PHP手册:
mysqli :: query将成功返回对象并在失败时返回false.
所以你可以在不检查数据类型的情况下使用它(===):
if ($conn->query($sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
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为了更好地理解,您可以使用var_dump()并检查您得到的结果:
var_dump($conn->query($sql));
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