Question

我有一个简单的脚本，可以将CSV文件上传到MySQL数据库。它工作正常，但如果用户错过选择某个CSV文件上传并按下UPLOAD按钮，脚本会尝试在数据库中插入数千行。

我想知道如果用户已经选择了FILE，是否有办法检查befoere上传文件。

这是我的代码：

<?php 
include("conection.php");  //Connect to Database

    if (isset($_POST['submit'])) {

    if (is_uploaded_file($_FILES['filename']['tmp_name'])) {

    echo "<br><center><p>" . " ". $_FILES['filename']['name'] ." " . "</p></center>";
                                                            }

    //Import uploaded file to Database
    $handle = fopen($_FILES['filename']['tmp_name'], "r");
    $count = 0; //skip first line of the CSV file 
    while (($data = fgetcsv($handle, 1000, ";")) !== FALSE) {

    if($count) //skip first line of the CSV file 
    {
    $import="INSERT into PAX (name,surname,group) VALUES('$data[0]','$data[1]','$data[2]')";

    mysql_query($import) or die(mysql_error());
    }
    $count++;
    }

    fclose($handle);

    print "<br><center><p>UPLOADED </p></center> ";
    print "<br> ";
    print "<center><a href='index.php'><button>HOME</button</a></center> ";

    //view upload form
    }else {

    print "<center><br><p><b>UPLOAD</b> </p>\n";
    print "<center><br>Select the file for Upload \n";
    print "<form enctype='multipart/form-data' action='upload.php' method='post'>";
    print "<center><input size='50' type='file' name='filename'><br />\n";
    print "<br>";
    print "<input type='submit' name='submit' value='UPLOAD'></form>";
}

?>

Answer 1

很难用你的代码明确地说出它所呈现的＆amp;没有看到原始数据，但我认为只需设置检查$data的条件就足以防止您看到的问题：

所以这一行：

if ($count) //skip first line of the CSV file {

对此的更改：

if ($count && !empty($data)) //skip first line of the CSV file {

这是您的代码调整和缩进清理以便于阅读。编辑我已对其进行了编辑，因此现在有一个$SUCCESS_SUBMIT变量。这样，如果提交成功＆amp;您可以true或false采取行动。

  include("conection.php");  //Connect to Database

  $SUCCESS_SUBMIT = false;
  if (isset($_POST['submit'])) {
    if (is_uploaded_file($_FILES['filename']['tmp_name'])) {
      echo "<br><center><p>" . " ". $_FILES['filename']['name'] ." " . "</p></center>";
    }       

    //Import uploaded file to Database
    $handle = fopen($_FILES['filename']['tmp_name'], "r");

    $count = 0; //skip first line of the CSV file 

    while (($data = fgetcsv($handle, 1000, ";")) !== FALSE) {
      if ($count && !empty($data)) //skip first line of the CSV file {
        $import = "INSERT into PAX (name,surname,group) VALUES('$data[0]','$data[1]','$data[2]')";
        mysql_query($import) or die(mysql_error());
        $SUCCESS_SUBMIT = true;
      }
      $count++;
    }
    fclose($handle);
  }

  if ($SUCCESS_SUBMIT) {
    print "<br><center><p>UPLOADED </p></center> ";
    print "<br> ";
    print "<center><a href='index.php'><button>HOME</button</a></center> ";
  }
  else {
    print "<center><br><p><b>UPLOAD</b> </p>\n";
    print "<center><br>Select the file for Upload \n";
    print "<form enctype='multipart/form-data' action='upload.php' method='post'>";
    print "<center><input size='50' type='file' name='filename'><br />\n";
    print "<br>";
    print "<input type='submit' name='submit' value='UPLOAD'></form>";
  }

Answer 2

“有一种方法可以显示用户是否未选择任何csv文件给出消息文件”请选择文件“或类似的东西”

如果没有选择文件，你可以使用javascript来发送消息：

print "<form enctype='multipart/form-data' action='upload.php' 
  onsubmit='return validateUpload();' name='formUpload'
method='post'>";
print "<center><input size='50' type='file' name='filename'><br />\n";
print "<br>";
print "<input type='submit' name='submit' value='UPLOAD'></form>";
}

?>

<script type="text/javascript" > 
  function validateUpload()
  {
    var fName = document.forms["formUpload"]["filename"].value;
    if (fName==null || fName=="")
    {
      alert("Please select file !");
      return false;
    }
 }
</script>

如果用户选择了CSV文件，如何在将CSV文件上传到MySQL数据库之前进行检查

2 个答案: