哪能如果else語句訪問功能的可調出身邊,你可以看到,當我宣佈opType == "Anand"
不是代碼將運行,但我使用後男人看到導致如何訪問函數的變量,如果外面聲明PHP
if (!(empty($_POST)) & isset($_POST)) {
$opType = $_POST["opType"];
if ($opType == "insertFuelEngineMap") {
//insert into fuel_engine_capacity_mapping setValuesForCreation(false);
$query = "INSERT INTO fuel_engine_capacity_mapping(cf_mapping_id,capacity_id) VALUES ('$cf_mapping_id','$capacity_id')";
$loginResult = mysqli_query($link, $query);
if (mysqli_affected_rows($link) > 0) {
$userdata = array('status' => '200', 'msg' => "insertlol into fuel_engine_capacity_mapping successfully", 'mapping_id' => $link->insert_id);
} else {
$userdata = array('status' => '404', 'msg' => " Cant fuel_engine_capacity_mapping " . mysqli_error($link));
}
}
if ($opType == "anand") {
function addition() {
$GLOBALS['z'] = $GLOBALS['x'] + $GLOBALS['y'];
}
}
}
addition();
echo $z;
請妥善格式化你的代碼。 –
你不認爲聲明'if/esle'的函數基礎是一個壞方法。你應該在課堂上創建一個功能並根據你的要求調用它 – urfusion
我實際上並不理解你在問什麼。 – RiggsFolly