如果输入为空，如何从运行停止一行代码

Question

我有这行代码，我希望它在 "beg_img" 为空时不要运行，我从 w3schools 获得代码并更改了一些小东西，我对此很陌生，但我想我必须使用某种 og if/else isset?

<?php
$target_dir = "../../img/images/";
$target_file = $target_dir . basename($_FILES["beg_img"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
    $check = getimagesize($_FILES["beg_img"]["tmp_name"]);
    if($check !== false) {
        echo "File is an image - " . $check["mime"] . ".";
        $uploadOk = 1;
    } else {
        echo "File is not an image.";
        $uploadOk = 0;
    }
}

// Check if file already exists
if (file_exists($target_file)) {
    echo "Sorry, file already exists.";
    $uploadOk = 0;
}
// Check file size
if ($_FILES["beg_img"]["size"] > 500000) {
    echo "Sorry, your file is too large.";
    $uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg") {
    echo "Sorry, only JPG, JPEG & PNG files are allowed.";
    $uploadOk = 0;
}
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
    echo "Sorry, your file was not uploaded.";
// if everything is ok, try to upload file
} else {
    if (move_uploaded_file($_FILES["beg_img"]["tmp_name"], $target_file)) {
        echo "The file ". basename( $_FILES["beg_img"]["name"]). " has been uploaded.";
    } else {
        echo "Sorry, there was an error uploading your file.";
    }
}

$beg_i_img = basename($_FILES["beg_img"]["name"]);

?>

Answer 1

用 "if":

包围您的整个代码

if (($_FILES["beg_img"]) != NULL) && (!empty($_FILES["beg_img"])) { 
   //yourcode 
}

你也可以直接检查$_FILES["beg_img"]是否不存在

if (isset($_FILES["beg_img"])){
   //yourcode
}

Answer 2

或者您可以在脚本的开头尝试这样做：

// If no file was uploaded
if($_FILES['beg_img']['error'] == 4){
    return;
}

并且最好在脚本的开头验证一些东西，比如你想要的东西，而不是用 if 包装你的所有代码，因为再包装一个，然后再包装一个，再包装一个......如果我们可以避免，代码会更好:)

您应该查看 return，也许 continue 会对您的脚本有所帮助 :)

Answer 3

你可以试试这个：

if (isset($_FILES["beg_img"])){
}

我在我的脚本中使用它并且工作正常

Answer 4

Return false 当 $_FILES["beg_img"] 为空时，如：

if (empty($_FILES["beg_img"]))
{
    return false;
}

如果输入为空，如何从运行停止一行代码

How to stop a line of code from running, if a input is empty

php

file-upload

input

如果输入为空，如何从 运行 停止一行代码

How to stop a line of code from running, if a input is empty

php

file-upload

input

如果输入为空，如何从运行停止一行代码