PHP SimpleTableManager::insert示例

示例#1

文件： simple-createTableRecord.php 项目： hwit0308/AngularJS-CRUD-PHP

$data = array();
// array to pass back data
$tableName = FormSubmit::POST("table");
$crudType = FormSubmit::POST("crud-type");
/*
$tableName = FormSubmit::GET("table");
$crudType = FormSubmit::GET("crud-type");
*/
$tableManager = new SimpleTableManager();
$tableManager->Initialize($tableName);
$tableManager->debug = true;
//$tableManager::set("originalName", "POON");
if (isset($_POST["create"])) {
    foreach ($_POST["create"] as $key => $value) {
        if (isset($_POST["createDate"])) {
            if (!array_key_exists($key, $_POST["createDate"])) {
                $tableManager->{$key} = $value;
            }
        } else {
            $tableManager->{$key} = $value;
        }
    }
}
if (isset($_POST["createDate"])) {
    foreach ($_POST["createDate"] as $key => $value) {
        $tableManager->{$key} = $value;
    }
}
//echo json_encode($tableManager->_);
echo json_encode($tableManager->insert());