如果在ViewModel中导入UIKit不是一种比在MVVM中使用UIImage picker的最佳方法更好的方法,那么我已经按照以下方法使用MVVM了
$("#deleteFormbtn").click(function(){
$.post ( "deleteRow.php", $("#editForm: input").serializeArray(), function(data){
alert(data);
});
});`
//要在下面的视图控制器实现中使用此方法,但是,我认为ViewModel在该实现中的位置在哪里,我知道这不是正确的方法,但我不知道什么是正确的解决方案:)
require_once ('sql/connectionstring/connectionstring.php');
$conn = SQLServerConnection();
if(isset($_POST['vendor'])){
$loc_sql = "SELECT TOP 1 loc_id FROM <table> WHERE loc_name = ?";
$parms = array($_GET['loc']);
$loc = sqlsrv_query($conn, $loc_sql, $parms) or die (print_r ( sqlsrv_errors(), true));
while ($q = sqlsrv_fetch_array($loc)){
$loc_id = $q["loc_id"];
}
$username = $_POST['username'];
$password = $_POST['password'];
$comments = $_POST['comments'];
$vendor_website = $_POST['website'];
$vendor = $_POST['vendor'];
$query = "UPDATE <table> SET <column> = '0' WHERE <column> = ?";
$parms = $username;
$result = sqlsrv_query($conn, $query, $parms) or die (print_r ( sqlsrv_errors(), true));
sqlsrv_close($conn);
}
header('Location: <location>);