我有这个代码，我已经写（upload.php），然而，当它上传，我似乎不能保存一个新的文件与一个-我也试图找到一种方法来做到这一点，但似乎不能让它工作.
[上传. php]

<?php
/* Get the name of the uploaded file */
$filename = $_FILES['file']['name'];
/* Choose where to save the uploaded file */
$location = "upload/".$filename;
/* Save the uploaded file to the local filesystem */
if ( move_uploaded_file($_FILES['file']['tmp_name'], $location) ) { 
  echo 'Success'; 
} else { 
  echo 'Failure'; 
}
?>

下面是[index.php]的一部分：

<div class="chat-message" id="chat-area">
    <div class="message-content-grid">
        <div class="add-file">
            <label for="fileupload">
                <i class="fa-solid fa-circle-plus"></i>
            </label>
            <input id="fileupload" type="file" name="fileupload" onchange="uploadFile()"/>
        </div>
        <div class="message-area">
            <form id="send-message-area">
                <textarea id="sendie" maxlength ="300" placeholder="Message"></textarea>
            </form>
        </div>
    <div>
</div>
<script>document.getElementById('chat-area').scrollTop = document.getElementById('chat-area').scrollHeight;</script>
<script>
    async function uploadFile() {
        let formData = new FormData(); 
        formData.append("file", fileupload.files[0]);
        await fetch('upload.php', {
            method: "POST", 
            body: formData
        }); 
        alert('The file has been uploaded successfully.');
    }
</script>

因此，当您单击上传图标时：

<label for="fileupload">
    <i class="fa-solid fa-circle-plus"></i>
</label>
<input id="fileupload" type="file" name="fileupload" onchange="uploadFile()"/>

它将运行javascript：

<script>
async function uploadFile() {
    let formData = new FormData(); 
    formData.append("file", fileupload.files[0]);
    await fetch('upload.php', {
        method: "POST", 
        body: formData
    }); 
    alert('The file has been uploaded successfully.');
}
</script>

这将实习，运行php脚本在[upload.php].

<?php
/* Get the name of the uploaded file */
$filename = $_FILES['file']['name'];
/* Choose where to save the uploaded file */
$location = "upload/".$filename;
/* Save the uploaded file to the local filesystem */
if ( move_uploaded_file($_FILES['file']['tmp_name'], $location) ) { 
  echo 'Success'; 
} else { 
  echo 'Failure'; 
}
?>

我想要的是检查是否有一个文件具有相同的名称，如果有，添加一个1或2或3取决于该文件是否也存在（file1.txt，file2.txt，file3.txt...）任何人有什么想法？Thx