Php use foreach post to add multiple data to mysql

Question

hello i have the following javascript which populates my fields and works fine

<script type="text/javascript">
var count = 0;
$(function(){
$('p#add_field2').click(function(){
    count += 1;
    $('#container').append(
            '<strong>Link #' + count + '</strong><br />'
            + '<input id="field_' + count + '" name="fields[]' + '" type="text" /><input     id="field1_' + count + '" name="idrima' + '" type="text" /><input id="field2_' + count + '" name="code_title' + '" type="text" /><input id="field3_' + count + '" name="name_title' + '" type="text" /><input id="field4_' + count + '" name="vathmos_title_a' + '" type="text" /><input id="field5_' + count + '" name="vathmos_title_d' + '" type="text" /><input id="field6_' + count + '" name="etos' + '" type="text" /><br />');

});
});
</script>

my problem is that when i use foreach i m able to add just the last field poppulation and not all. So i think foreach doesnt work cause adds only the last record. My code is

if ($_POST['fields']) {

    //get last inserted userid
    $inserted_user_id = $db->last_insert_id();

    //Loop through added fields
    foreach ( $_POST['fields'] as $key=>$value ) {

        //Insert into websites table
        $sql_website = sprintf("INSERT INTO spoudes (idrima,code_title) VALUES ('%s','%s')",
                    mysql_real_escape_string($_POST['idrima']),
                    mysql_real_escape_string($_POST['code_title']));
        $result_website = $db->query($sql_website);
        $inserted_website_id = $db->last_insert_id();


        //Insert into users_websites_link table
        $sql_users_website = sprintf("INSERT INTO spoudes_link (id_fakelos, id_spoudes) VALUES ('%s','%s')",
                               mysql_real_escape_string($inserted_user_id),
                               mysql_real_escape_string($inserted_website_id) );
        $result_users_website = $db->query($sql_users_website);


    }

Wing Lian · Accepted Answer · 2014-11-30 19:15:37Z

1

You should use <input name="idrima[]"> and <input name="code_title"> as each time you add a new field, it is overwriting that field on the post.

foreach ( $_POST['fields'] as $key=>$value ) {

    //Insert into websites table
    $sql_website = sprintf("INSERT INTO spoudes (idrima,code_title) VALUES ('%s','%s')",
                mysql_real_escape_string($_POST['idrima'][$key]),
                mysql_real_escape_string($_POST['code_title'][$key]));
    $result_website = $db->query($sql_website);
    $inserted_website_id = $db->last_insert_id();


    //Insert into users_websites_link table
    $sql_users_website = sprintf("INSERT INTO spoudes_link (id_fakelos, id_spoudes) VALUES ('%s','%s')",
                           mysql_real_escape_string($inserted_user_id),
                           mysql_real_escape_string($inserted_website_id) );
    $result_users_website = $db->query($sql_users_website);


}

edited Nov 30, 2014 at 19:15

answered Nov 30, 2014 at 18:43

Wing Lian

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2 Comments

teok1979 Over a year ago

it returns me error Warning: mysql_real_escape_string() expects parameter 1 to be string, array given in

Wing Lian Over a year ago

that is now a array and you'll have to access it via index as well within the foreach loop. try something like $_POST['idrima'][$key], $_POST['code_title'][$key]

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