php mysql如何使用mysql_insert_id（）能够同时插入2个表中？

Question

I am using php with mysql database to do some insert. I have 2 tables.

• user
• status where the second table has as foreign key user_id to relate between these 2 tables

my problem is that when i insert into status table the user_id field do change and take 0 no matter what is the user_id.

so how to fix this problem ???

this is the code of login.php

<?php
//array for JSON response
$response = array();

// check for required fields
if(empty($_POST['user']) || empty($_POST['password'])){
       $response["success"] = 0;
       $response["message"] = "enter Both Fields";
        // echoing JSON response
        die (json_encode($response));


}

else if (isset($_POST['user']) && isset($_POST['password']) ) {

    $user = $_POST['user'];
    $password = $_POST['password'];


    // include db connect class
    require_once '/db_connect.php';

    // connecting to db
    $db = new DB_CONNECT();

    // mysql inserting a new row
    $sql = mysql_query("Select user, password from users where user='$user'")or die(mysql_error());
    $count_query = mysql_num_rows($sql);
    if($count_query >0){
     $response["success"] = 1;
        $response["message"] = "correct Informations";

   // echoing JSON response
        echo json_encode($response);
    }
     else{
        $response["success"] = 0;
        $response["message"] = "Wrong User Or Pass";

        // echoing JSON response
        echo json_encode($response);
    }
}
else {
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);
}   
?>

status.php

<?php

/*
 * Following code will create a new product row
 * All product details are read from HTTP Post Request
 */

// array for JSON response
$response = array();
if(empty($_POST['status'])){
       $response["success"] = 0;
       $response["message"] = "You must Write something";
        // echoing JSON response
        die (json_encode($response));
}

// check for required fields
else if (isset($_POST['status'])) {

    $status = $_POST['status'];

    // include db connect class
    require_once __DIR__ . '/db_connect.php';

    // connecting to db
    $db = new DB_CONNECT();



    // mysql inserting a new row
    $result = mysql_query("INSERT INTO status(status, user_id) VALUES('$status' , '$last_insert_id')") or die(mysql_error);
    $last_insert_id = mysql_insert_id();

    // check if row inserted or not
    if ($result) {
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = "Your Status has been saved.";

        // echoing JSON response
        die (json_encode($response));
    } else {
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";

        // echoing JSON response
        die (json_encode($response));
    }

}

?>

i am using php with android so their is no html form

Answer 1

You can't. mysql_insert_id() only applies to the LAST insert performed. If you're doing two inserts, and call insert_id() after the second one, the first ID is lost.

There is no way around this.

You must have something like:

INSERT INTO foo ....
$fooid = mysql_insert_id();
INSERT INTO bar .... foo_id=$fooid
$barid = mysql_insert_id();

Given that your code actually seems to be split into multiple pages, it's even worse. mysql_insert_id() only applies to the CURRENT connection to the database. Once your first script exits, the connection is closed and the insert_id is lost.

The next script will get a NEW connection, and have its own completely separate insert_id system going.

For chaining multiple pages together like this, you'll have to retrieve/pass the insert ID around yourself, eg

page1:

INSERT ...
$_SESSION['page1_id'] = mysql_insert_id();

page2:

$last_id = $_SESSION['page1_id'];
INSERT ..... id=$last_id