"如果目录已存在则退出脚本"

Question

我知道这听起来像一个愚蠢的问题，但我无法解决这个问题。 基本上我只想在目录中的文件夹已经存在的情况下退出脚本，但我得到的只是带有回显“la cartella esiste”的空白页面，我猜这是因为退出功能。 我想要的是将消息显示在页面本身上，就像显示的其他错误一样，但不继续使用代码。 你们对如何进行有什么建议吗？

<?php

    
    require('db.php');

    if (isset($_POST['upload'])) {
        $batch_number = stripslashes($_REQUEST['batch_number']);
        $batch_number = mysqli_real_escape_string($con, $batch_number);
        $product_name = stripslashes($_REQUEST['product_name']);
        $product_name = mysqli_real_escape_string($con, $product_name);
        $vial_size = stripslashes($_REQUEST['vial_size']);
        $vial_size = mysqli_real_escape_string($con, $vial_size);
        $Sterile = $_POST['Sterile'];
        $Macchina = $_POST['Macchina'];
        $location = "immagini/$batch_number/";
        $statusMsg = $errorMsg = $insertValuesSQL = $errorUpload = $errorUploadType = ''; 
        $allowTypes = array('jpg','png','jpeg', 'bmp');
        $img_name = array_filter($_FILES['image']['name']);
        $select = mysqli_query($con, "SELECT * FROM info_flaconi WHERE batch_number = '". $_REQUEST['batch_number']."'");
        if(file_exists($location) && is_dir($location)) {
            exit ("la cartella esiste");
            
        } else{
            mkdir("immagini/$batch_number/", 0777, true);
            echo "Cartella creata.";        

        }
                   
        if(!empty($img_name)){
            foreach($_FILES['image']['tmp_name'] as $key=>$val){
                //file upload path
                $fileName= $_FILES['image']['name'][$key];
                $fileName_tmp= $_FILES['image']['tmp_name'][$key];
                $targetPath = $location .$fileName;
                $ext=strtolower(pathinfo($targetPath, PATHINFO_EXTENSION));
                $uploadDate = date('Y-m-d H:i:s');
                $uploadOk = 1;
                
                // Check whether file type is valid 
                
                if(in_array($ext, $allowTypes)){ 
                // Upload file to server 
                    if(move_uploaded_file($fileName_tmp, $targetPath)){
                        $sqlVal = $fileName;
                    }  else{
                        $response = array(
                            "status" => "alert-danger",
                            "message" => "File coud not be uploaded.");
                        }
                }else{
                    $response = array(
                    "status" => "alert-danger",
                    "message" => "Only .jpg, .jpeg, .png and bmp file formats allowed.");
                }
                if(!empty($sqlVal)){                   
                    $query    = "INSERT INTO `info_flaconi` (batch_number, product_name,  vial_size, vial_image, uploaded_on, Sterile_Area, Macchina)
                    VALUES ('$batch_number', '$product_name', '$vial_size', '$sqlVal', '$uploadDate', '$Sterile', '$Macchina')";
                    $result   = mysqli_query($con, $query);
                    if($result){
                        $response = array(
                            "status" => "alert-success",
                            "message" => "Immagini caricate correttamente.");    
                    
                    }else{
                        $response = array(
                            "status" => "alert-danger",
                            "message" => "Files coudn't be uploaded due to database error.");
                    }  
                }
            }
        }else {
            // Error
            $response = array(
            "status" => "alert-danger",
            "message" => "Per favore seleziona le immagini da caricare.");
           
        }
    }                                       
    ?>  

Answer 1

看起来您需要为其他代码（您尚未向我们展示）设置$response<\/code>数组以显示消息。 一种方法是将您向我们展示的代码包装在do { ... } while (false)<\/code>块中（其中“...”是您在上面显示的所有代码），并在您当前调用exit()<\/code>的位置exit()<\/code> ，将其更改为：

$response = array(
  "status" => "alert-danger",
  "message" => "la cartella esiste"
);
break;