Question

我有一个脚本，它接受一个文本文件并将其插入到db。我已经为xls，xlsx，csv做了exacly脚本，并且工作正常，但是使用.txt文件，我在读取文件时出错。这是我的代码

//html form that uploads the file
<html>
    <body>
    <form action="" method="post" enctype="multipart/form-data">
        <label for="file">Filename:</label>
        <input type="file" name="file" id="file" /><br />
        <input type="submit" name="submit" value="Submit" />
    </form>
    </body>
</html>

//this is my php script
if(isset($_POST['submit'])){

        move_uploaded_file($_FILES["file"]["tmp_name"],$_FILES["file"]["name"]);
        $file = $_FILES["file"]["name"];
        $document = fopen($file,'r');
        $contents = fread($document, filesize($file));
        fclose($document);
        $contents = str_replace("\r\n","|",$contents);
        $contents = str_replace("\"","",$contents);
        $contents = explode("|",$contents);
        $contents = array_chunk($contents, 10);

        $count = 0;
        for($i = 1; $i < sizeof($contents);$i++){
            $categoria = $contents[$i][0];
            $ragsoc    = $contents[$i][1];
            $telefono  = $contents[$i][2];
            $via       = $contents[$i][3];
            $comune    = $contents[$i][4];
            $cap       = $contents[$i][5];
            $pro       = $contents[$i][6];
            $reg       = $contents[$i][7];
            $gas       = $contents[$i][8];
            $scadenza  = convert_date($contents[$i][9]);
            $query = "insert into tabela set
                categoria='".mysqli_real_escape_string($categoria)."',
                ragsoc='".mysqli_real_escape_string($ragsoc)."',
                telefono='".mysqli_real_escape_string($telefono)."',
                via='".mysqli_real_escape_string($via)."',
                comune='".mysqli_real_escape_string($comune)."',
                cap='".mysqli_real_escape_string($cap)."',
                pro='".mysqli_real_escape_string($pro)."',
                reg='".mysqli_real_escape_string($reg)."',
                gas='".mysqli_real_escape_string($gas)."',
                scadenza='".mysqli_real_escape_string($scadenza)."',
                data_ins='$data'";
            $result = mysqli_query($query);
            if(!$result){
                echo mysqli_error();
            }
            $count++;
            //echo $query;exit;
        }

        echo "Total records :".$count;
    }

问题是我无法读取文件。如果我把文件来源像这样

$file = "myfile.txt";

脚本将正常工作，但当我指向这样的脚本

$file = $_FILES["file"]["name"];

我得到一个emty数组（）

上传文本文件无效

0 个答案: