How to get the last inserted id in PHP

Question

Here I am trying to get the inserted id from MySQL database in the table I have product_id. After insert I want to get the latest inserted product_id and store it in array ['newid'][]

The insert query is going on pretty good, but I am not able to get the product_id in to the array. when I print the array I am getting NULL value.

 $link = mysqli_connect(db_host,db_user,db_password,db_name);    
    if (condition) {
    $sqlin = "INSERT INTO product_list (product_name, product_category, product_price,product_description,product_sharing_basis,product_co_owners,walden_product_price,product_referrence_URL,product_proposed_user_id,product_image_url,product_refurbish_factor,product_insurance_factor,product_life,product_size_category,product_publish_status) VALUES ('$product_name', '$product_category', '$product_price', '$product_description', '$share_basis', '$co_owners', '$walden_product_price', '$pro_url', '$proposed_by','files/uploaded_images/".$_FILES['file']['name']."', '$refurbishment_factor', '$insurance_factor', '$product_life', '$size_category','$approve')";          
    } else {
    $sqlin = "INSERT INTO product_list (product_name, product_category, product_price,product_description,product_sharing_basis,product_co_owners,walden_product_price,product_referrence_URL,product_proposed_user_id,product_image_url,product_refurbish_factor,product_insurance_factor,product_life,product_size_category) VALUES ('$product_name', '$product_category', '$product_price', '$product_description', '$share_basis', '$co_owners', '$walden_product_price', '$pro_url', '$proposed_by','files/uploaded_images/".$_FILES['file']['name']."', '$refurbishment_factor', '$insurance_factor', '$product_life', '$size_category')";
    }
    if(mysqli_query($link, $sqlin)){    
        $newisid = mysqli_insert_id($link);
        $_SESSION['newid'][] = $newisid;

How can I solve this?

How you print your array and have you start session on your page??? — Saty, Nov 02 '15 at 06:24
`"The insert query is going on quite good"` ... but your English grammar is going quite poorly. You meant to say `"going on quite well."` — Tim Biegeleisen, Nov 02 '15 at 06:24
@Saty yes i have started session. and i am using echo var_dump($_SESSION['newid']); to print. — CJAY, Nov 02 '15 at 06:25
Use just `$_SESSION['newid']= $newisid;` to store your last insert id — Saty, Nov 02 '15 at 06:27
@MithunRaikar can you please do something like mysqli_query() or die(mysqli_error($link)) and see what will happen. Additionally what's the response from mysqli_query and what's response from mysqli_insert_id()? — Nevio Vesić, Nov 02 '15 at 07:55
Try `echo mysqli_insert_id($link);` to see if you got the latest inserted id — nekiala, Nov 02 '15 at 09:00

score 1 · Answer 1 · answered Nov 02 '15 at 07:00

php.net : $mysqli->insert_id

    $mysqli = new mysqli("localhost", "my_user", "my_password", "world");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$mysqli->query("CREATE TABLE myCity LIKE City");

$query = "INSERT INTO myCity VALUES (NULL, 'Stuttgart', 'DEU', 'Stuttgart', 617000)";
$mysqli->query($query);

printf ("New Record has id %d.\n", $mysqli->insert_id);

Niranjan N Raju · Answer 2 · 2015-11-02T06:48:40.460

0

Try using mysql_insert_id() to get previously inserted id.

if (mysqli_query($conn, $sql)) {
    $last_id = mysqli_insert_id($conn);//previously insert id here.

} else {
    //error
}

See this and this

EDIT

You can also use LAST_INSERT_ID() for this. Check official mysql doc

edited Nov 02 '15 at 06:48

answered Nov 02 '15 at 06:32

Niranjan N Raju

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Mithun Raikar have already used `mysqli_insert_id` in his code – Suyog Nov 02 '15 at 06:36
What is coming if you are echoing the last insert id??@mithun – Niranjan N Raju Nov 02 '15 at 06:43

score 0 · Answer 3 · edited Nov 02 '15 at 07:15

Try this:

$link = mysqli_connect(db_host,db_user,db_password,db_name);

if (condition) {
    $sqlin = "INSERT INTO product_list (product_name, product_category, product_price,product_description,product_sharing_basis,product_co_owners,walden_product_price,product_referrence_URL,product_proposed_user_id,product_image_url,product_refurbish_factor,product_insurance_factor,product_life,product_size_category,product_publish_status) VALUES ('$product_name', '$product_category', '$product_price', '$product_description', '$share_basis', '$co_owners', '$walden_product_price', '$pro_url', '$proposed_by','files/uploaded_images/".$_FILES['file']['name']."', '$refurbishment_factor', '$insurance_factor', '$product_life', '$size_category','$approve')";          
} else {
    $sqlin = "INSERT INTO product_list (product_name, product_category, product_price,product_description,product_sharing_basis,product_co_owners,walden_product_price,product_referrence_URL,product_proposed_user_id,product_image_url,product_refurbish_factor,product_insurance_factor,product_life,product_size_category) VALUES ('$product_name', '$product_category', '$product_price', '$product_description', '$share_basis', '$co_owners', '$walden_product_price', '$pro_url', '$proposed_by','files/uploaded_images/".$_FILES['file']['name']."', '$refurbishment_factor', '$insurance_factor', '$product_life', '$size_category')";
}


if($link->query($sqlin)){
    $newisid = $link->insert_id;
    $_SESSION['newid'][] = $newisid;
}

Reference: http://php.net/manual/en/mysqli.insert-id.php

How to get the last inserted id in PHP

3 Answers3