Php 如何获取文本框的值并将其用作图像的文件名

有人能帮我吗？我正在尝试使用PHP将图像上载到特定目录，我想知道如何获取文本框的值并将其用作图像文件名。下面是我的代码

<?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["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["fileToUpload"]["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["fileToUpload"]["size"] > 500000) {
    echo "Sorry, your file is too large.";
    $uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
&& $imageFileType != "gif" ) {
    echo "Sorry, only JPG, JPEG, PNG & GIF 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["fileToUpload"]["tmp_name"], $target_file)) {
        echo "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.";
    } else {
        echo "Sorry, there was an error uploading your file.";
    }
}


?>

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如蒙答复，不胜感激。谢谢：）

PHP

move\u upload\u file（file，newloc）

将第二个选项作为源目标+文件名。比如：

move_uploaded_file(file, '/var/www/html/project/public'.$filename);

此处

$filename

是将其移动到服务器的文件名。您可以为文件指定任何名称，如时间戳值或用户指定的字符串

就你而言：

$filename = $_REQUEST['filename'];

---

目标文件名真正重要的是以下几行：

<?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);

---

谢谢你的评论，我只想获取textbox的值并将其设置为我的图像文件名如果我有两个textbox呢？
<form method="POST" action="target_process.php">
<input type="text" name="filename">
...(other form objects here)
</form>

<?php
$target_dir = "uploads/";
$file_parts = explode('.', $_FILES["fileToUpload"]["name"]);
$ext = $file_parts[count($file_parts) - 1];
$target_file = $target_dir . $_POST['filename'] . '.' . $ext;