Question

我有一个程序可以获取用户输入并将其更改为文件或允许用户上传文件。我从用户那里获取上传的文件时遇到问题。现在，我已将其硬编码为我上传的名为sample.fasta的示例文件。我希望能够获取用户上传的文件的名称，然后使用该文件的名称调用我的程序。

我将发布有关此问题的所有相关代码。

此页面名为blast.php

<?php
if(isset($_POST['submit2'])){
            //echo "submit2";
    //      echo $_FILES['uploadedfile']['name'];

     //declare variables to what the user defines them as
            $db = $_POST['database'];
            $evalue = $_POST['evalue'];
            $sequence = $_POST['BlastSearch'];
            $hits = $_POST['hits'];
            $userid = $_SESSION['uid'];

            //insert the values into the database
            $mysqli->query("INSERT INTO `Job` (`uid`, `input`, `status`, `start_time`, `finish_time`) VALUES ('1', 'used a file', 'running' , NOW(), NOW())");

            $mysqli->query("INSERT INTO `BLAST`(`db_name`, `evalue`, `job_id`) VALUES ('" . $db . "','" . $evalue . "', '".$mysqli->insert_id."')") or die(mysqli_error($mysqli));

            //need to change the name of sample.fasta to whatever file uploaded
            exec('/students/groups/cs4380sp15grp4/blast/blast-2.2.26/bin/blastall -p blastp -d db -i /students/groups/cs4380sp15grp4/public_html/home/uploads/sample.fasta -m '.$evalue.' -o outputFILE -v '.$hits.' -b '.$hits);
?>
          <form enctype="multipart/form-data" action="upload.php" method="POST" class="form-inline">
                            <input type="file" name="fileToUpload" id="fileToUpload" class="form-control"/>
                            <input type="submit" value="upload" name="upload" class="form-control"/>
                            <input type="reset" value="reset" name="reset" class="form-control"/>
            </form>

此文件名为upload.php，用于上传文件的表单。

 <?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$FileType = pathinfo($target_file,PATHINFO_EXTENSION);

// Allow certain file formats
if($FileType != "fasta" ) {
    echo "Sorry, only fasta 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.";
}
}
header('Location: http://babbage.cs.missouri.edu/~cs4380sp15grp4/home/blast.php');
?>

所以基本上在我的exec函数中而不是读取sample.fasta，我需要读取用户上传的文件...

Answer 1

您只需将表单发回到包含表单的页面，然后将上传的文件名和位置用于执行您需要执行的操作。

为此，您可以检查您的帖子值是否设置在PHP文件的开头并进行处理，如果它们就像您在upload.php文件中所做的那样。

Answer 2

将upload.php最后一行更改为：

header('Location: http://babbage.cs.missouri.edu/~cs4380sp15grp4/home/blast.php?file='.urlencode($target_file));

并检查$_GET中的blast.php：

if(isset($_GET['file'])){
    $fastaFile = $_GET['file']
....

如何获取添加到目录的最后一个文件的名称

2 个答案: