Insert array into db table - php

Question

I'm trying to learn how to use an array to insert multiple entries into a database table. This was my attempt. What am I doing wrong?
$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);
}
I tried this but it still doesn't work. What am I missing?
$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);
}
Thanks for your advice.
Next try.
$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);
}
I get these error messages.
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

Answer 1

It is unclear what your purpose is since the SQL-query in your question specify 3 fields (field1, field2, field3) to be inserted into your table, but you only have 2 values in your clients array. If you want to insert multiple rows in one single query, lets say for the "name" and "ip" value in your clients array, you can do this:
if(is_array($client)) {
$DataArr = array();
foreach($client as $row) {
//CREATE ARRAY WITH name AND ip VALUES FOR EACH USER...
$DataArr[] = "('". mysqli_real_escape_string($conn, $row[0]) ."', '". mysqli_real_escape_string($conn, $row[1]) ."')";
}
$sql = "INSERT INTO ip_data (name, ip) VALUES ";
$sql .= implode(", " , $DataArr);
mysqli_query($conn, $sql);
}
The $sql variable will contain the following query compliant with the syntax for a multiple insert:
INSERT INTO ip_data (name, ip)
VALUES ('Jerry Garcia', '193.169.5.11'),
('Bill Graham', '193.169.5.12'),
('Arlo Guthrie', '193.169.5.13')
Note that you have $query instead of $slqin your question. It should be: mysqli_query($conn, $sql);.
Another thing: Using mysqli_real_escape_string() the procedural way instead of the object oriented way, as in your example, requires to pass a link identifier to the connection as parameter: mysqli_real_escape_string($conn, $row[0])
UPDATE:
In your latest attempt you have changed the array to an associative array, so this should do it:
$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, $row["name"]) ."', '". mysqli_real_escape_string($conn, $row["ip"]) ."')";
}
$sql = "INSERT INTO ip_data (name, ip) VALUES ";
$sql .= implode(", " , $DataArr);
mysqli_query($conn, $sql);
}

Answer 2

Please change append to $DataArr[] to this one:
$DataArr[] = "('$fieldVal1', '$fieldVal2', '$fieldVal3')";

Answer 3

Correct
From
if(is_array($client) {
To
if(is_array($client)){
And your array $client will have only 2 elements in each iteration
From
# You will get PHP Warning: Illegal offset type
# since $row is already array
$fieldVal1 = mysqli_real_escape_string($client[$row][1]);
$fieldVal2 = mysqli_real_escape_string($client[$row][2]);
$fieldVal3 = mysqli_real_escape_string($client[$row][3]);
to
// For first iteration
// $row[0] = "Jerry Garcia"
// $row[1] = "193.169.5.11"
$fieldVal1 = mysqli_real_escape_string($conn, $row[0]);
$fieldVal2 = mysqli_real_escape_string($conn, $row[1]);
// $row[3] does not exists so comment it and set $fieldVal3 some data
// $row[3] not exists in your array and even $row[2]
// or add one value to your $client array and access using $row[2]
// $fieldVal3 = mysqli_real_escape_string($row[3]);
$fieldVal3 ='somedata';
and Finally
From
$DataArr[] = "('fieldVal1', 'fieldVal2', 'fieldVal3')";
To
$DataArr[] = "('$fieldVal1', '$fieldVal2', '$fieldVal3')";