运行此代码时,它会返回错误,但仅返回 $sql
查询显示的错误代码的一部分。我没有说明为什么它不起作用。怪异的部分是INSERT
确实创建了具有完全相同代码的新记录。为什么这是?
我已经检查了root用户上的Pivileges。该用户有完整的访问权限,包括选择。
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "nny_gamerrum";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT email FROM booking;";
if ($conn->query($sql) === TRUE) {
echo 'Successfull';
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
echo $conn->error;
?>
返回屏幕:错误:从预订中选择电子邮件;
在这里您可以找到有关查询函数的文档。您有错误,因为此 $ conn->查询($ sql(=== true 语句给出false。
查询方法返回取决于您要插入哪种查询。为了成功选择,显示,显示,描述或解释方法返回对象,对于任何其他返回 true
尝试以下:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "nny_gamerrum";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT email FROM booking;";
if ($data = $conn->query($sql)) {
var_dump($data);
echo 'Successfull';
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
echo $conn->error;
?>