submit in html on sql query

Posted by user1644661 on Stack Overflow See other posts from Stack Overflow or by user1644661
Published on 2012-09-03T21:36:17Z Indexed on 2012/09/03 21:37 UTC
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i need to run this sql query , which give me a list of Id and Dates i want to click each result and take with me the Id value to the next form i wrote this query above but i see in the debager that the hidden ID get his value but not pass to the next form i think i have a problem with the submit() . where should i put him ?

thanks anat

function ShowAllCarts($user_email) {

$connB = new ProductDAO();
$connB->Connect();
$pro_query = "SELECT * FROM Cart WHERE `Email`='$user_email';";
$db_result = $connB->ExecSQL($pro_query);

$html_result = '<div data-role="content"> <ul data-role="listview" data-theme="b"> ';
$html_result .= '<form action="PreviouscartProduct.php" method="POST"/>';

while($row_array = $db_result->fetch_array(MYSQLI_ASSOC))
{
    $Id= $row_array['Id'];
    $Date= $row_array['Date'];

    //$html_result //
    $html_result .="<li><a href='PreviouscartProduct.php'>Cart number: $Id from Date: $Date><input type='hidden' name='Id' value'<?=$Id?>'</input></a></li>'";
    $html_result .= '<a onclick="this.form.submit();" </a>;
}

    $html_result .= ' </ul> </div>';
    $html_result .= '</form>';  
$connB->Disconnect();
return $html_result;

}

//display all carts $func_result = ShowAllCarts($Email);

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