我有這個與MySQL非常奇怪的問題。我可以插入並創建表格。但是我不能選擇什麼,當我嘗試它不顯示任何錯誤,它只是回聲我行「錯誤:」無法從表中選擇,但我可以插入並創建?
getDiff("validTable");
function getDiff($regNr) {
global $servername, $username, $password, $dbname;
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//todo
$sql = "SELECT * FROM $regNr ORDER BY id LIMIT 1";
$result = $conn->query($sql);
if ($result === TRUE) {
echo "done";
} else {
$error = $result->error;
echo "Error: " . $error;
}
$conn->close();
}
但是當我插入到數據庫使用此
$query = "CREATE TABLE IF NOT EXISTS $regNr (`id` MEDIUMINT NOT NULL AUTO_INCREMENT, `mail` INT NOT NULL , `price` INT NOT NULL , `views` INT NOT NULL , `the_date` DATE NOT NULL , `time` VARCHAR(5) NOT NULL, PRIMARY KEY (id))";
runQuery($query);
function runQuery($todo) {
global $servername, $username, $password, $dbname;
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//todo
$sql = $todo;
if ($conn->query($sql) === TRUE) {
echo "done";
} else {
echo "Error: " . $conn->error;
}
$conn->close();
}
它的工作原理很好,很漂亮。我做了什麼錯誤?!這真讓我抓狂!
謝謝!還有更多謝謝fetch_array()!一直工作太久,看不到明顯的! –