Question

我有一个关于PHP表单检查发布文本和图像的问题。我如何处理这两种数据的表单检查？

$latErr = $lngErr = $photoErr = "";
$lat = $lng = $photo = "";

//what if i am not using form action-PHP_SELF, currently i am use anotherfile.php, what do i use below?
if ($_SERVER["REQUEST_METHOD"] == "POST") {

if (empty($_POST["inputfield1"])) {
 $latErr = "Latitude is required. Enable your browser geolocation.";
} else {
 $lat = test_input($_POST["inputfield1"]);
}

if (empty($_POST["inputfield2"])) {
 $lngErr = "Longitude is required. Enable your browser geolocation.";
} else {
 $lng = test_input($_POST["inputfield2"]);
}

//can i check for image like this?
if (empty($_POST["photo"])) {
 $photoErr = "Please select your image.";
} else {
 $photoErr = test_input($_POST["photo"]);
}
}

Answer 1

您需要使用输入类型file，然后检查$_FILES是否包含任何上传的图片：

if (empty($_FILES['photo']['name'])) {
    // error - no photo selected
}
if (empty($_POST['inputfield1'])) {
    // error - no text input
}


// ...
// ... form ...
<form method="POST" enctype="multipart/form-data">
    <input type="file" name="photo" />
    <input type="text" name="inputfield1" />
// ...

Answer 2

如果有人在没有发送图片的情况下使用cURL，您的代码就会崩溃，因为它正在尝试查看是否有通过POST发送的“照片”。你应该尝试做这样的事情：

if( $_FILES['photo']['error'] == 4 )
 $photoErr = "No file was uploaded";

Answer 3

if(isset($_FILES['input_field_name']['name'])){
    //File Exists
}

if(isset($_POST['input_fiel_text'])){
    //Text field exists
}

Answer 4

您可以获得有关图像的更多信息以获得良好的控制

        $name = $_FILES['filename']['name'];
        $dir = pathinfo($name);
        $extension = $dir['extension'];
        $extensionsAutorisees = array("png","gif","bmb");
        $to_dir = dirname(__FILE__)."/img/";
        $new_name = $name.".".$extension;

        if (!(in_array($extension, $extensionsAutorisees))) {
             .... error with extension alow
        } else {    
             ..... your action
        }

Answer 5

我喜欢Mark M的回答，他是对的，但我想将上传图片的验证添加到：

public static function isValidUpload() {

    if ($_FILES['photo']['error'] !== UPLOAD_ERR_OK) {
        return false;
    }

    $info = getimagesize($_FILES['photo']['tmp_name']);

    return !(($info === FALSE) || (($info[2] !== IMAGETYPE_GIF) && ($info[2] !== IMAGETYPE_JPEG) && ($info[2] !== IMAGETYPE_PNG)));
}

不要检查扩展名，因为我可以编写PHP文件并将其另存为evil.png。在这种情况下，尽管扩展名是png，但这将是一个潜在危险的php。

PHP表单检查文本和图像

5 个答案: