Ich versuche zu lernen, wie Sie ein Array verwenden, um mehrere Einträge in eine Datenbanktabelle einzufügen. Das war mein Versuch. Was mache ich falsch?Array in DB-Tabelle einfügen
$servername = "localhost";
$username = "#";
$password = "#";
$dbname = "hosts";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$client = array(
"1" => array("Jerry Garcia", "193.169.5.11"),
"2" => array("Bill Graham", "193.169.5.12"),
"3" => array("Arlo Guthrie", "193.169.5.13")
);
if(is_array($client) {
$DataArr = array();
foreach($client as $row) {
$fieldVal1 = mysqli_real_escape_string($client[$row][1]);
$fieldVal2 = mysqli_real_escape_string($client[$row][2]);
$fieldVal3 = mysqli_real_escape_string($client[$row][3]);
$DataArr[] = "('fieldVal1', 'fieldVal2', 'fieldVal3')";
}
$sql = "INSERT INTO ip_data (field1, field2, field3) values ";
$sql .= implode(',' , $DataArr);
mysqli_query($conn, $query);
}
Ich habe das versucht, aber es funktioniert immer noch nicht. Was vermisse ich?
$servername = "localhost";
$username = "#";
$password = "#";
$dbname = "hosts";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$client = array(
"0" => array("Jerry Garcia"),
"1" => array(""193.169.5.11"),
);
if(is_array($client)) {
$DataArr = array();
foreach($client as $row) {
$fieldVal1 = mysqli_real_escape_string($client[$row][0]);
$fieldVal2 = mysqli_real_escape_string($client[$row][1]);
$DataArr[] = "('$fieldVal1', '$fieldVal2')";
}
$sql = "INSERT INTO ip_data (field1, field2) values ";
$sql .= implode(',' , $DataArr);
mysqli_query($conn, $query);
}
Vielen Dank für Ihren Rat.
Nächster Versuch.
$servername = "localhost";
$username = "#";
$password = "#";
$dbname = "hosts";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$client = array(
"0" => array("name" => "Peter Maxx", "ip" => "193.169.5.16"),
"1" => array("name" => "Ravi Shankar", "ip" => "193.169.5.17")
);
if(is_array($client)) {
$DataArr = array();
foreach($client as $row) {
$DataArr[] = "('". mysqli_real_escape_string($conn, $row[0]) ."', '". mysqli_real_escape_string($conn, $row[1]) ."')";
}
$sql = "INSERT INTO ip_data (name, ip)
VALUES
('Peter Maxx', '193.169.5.16'),
('Ravi Shankar', '193.169.5.17')";
$sql .= implode(", " , $DataArr);
mysqli_query($conn, $sql);
}
Ich bekomme diese Fehlermeldungen.
PHP Notice: Undefined offset: 0 in php shell code on line 5
PHP Notice: Undefined offset: 1 in php shell code on line 5
PHP Notice: Undefined offset: 0 in php shell code on line 5
PHP Notice: Undefined offset: 1 in php shell code on line 5
Blick https://www.w3schools.com/sql/sql_insert.asp - Ihre Werte sollten in() –
@AlexeyShatrov sein - er beabsichtigt, zu , aber falsch. Siehe die Antwort. –