【发布时间】:2014-12-20 09:14:15
【问题描述】:
我想将 CSV 文件中的数据插入 MySQL 表。为此,我现在使用以下代码,但是当我上传文件时,我的浏览器变得没有响应。几次之后,会弹出一个显示重启 Firefox 或退出 Firefox。我只是想知道,我在给定代码中的错误在哪里?
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<html>
<head>
<title></title>
<meta name="" content="">
</head>
<body>
<form method="POST" enctype="multipart/form-data">
<input type="file" name="imageup" /><input type="submit" name="submit" value="Upload"/>
</form>
</body>
</html>
<?php
function generateRandomString($length = 10){
$characters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
$randomString = '';
for($i = 0; $i < $length; $i++){
$randomString .= $characters[rand(0, strlen($characters) - 1)];
}
return $randomString;
}
if(isset($_POST['submit'])){
$t= generateRandomString();
$path = 'csv/';
$image = $_FILES["imageup"]["name"];
// $tmp = explode(".",$image);
$type = end($tmp);
$file = array("csv");
$csv_file = $path.$image;
if(in_array(strtolower($type), $file)){
if(move_uploaded_file($_FILES["imageup"]["tmp_name"], $path.$image)){
readfile($_FILES['imageup']['tmp_name']);
$open = fopen($_FILES['imageup']['tmp_name'], 'r');
$theData = fgets($open);
$i = 0;
while(!feof($open)){
$csv_data[] = fgets($open, 1024);
$csv_array = explode(",", $csv_data[$i]);
$insert_csv = array();
$insert_csv['ID'] = $csv_array[0];
$insert_csv['firstname'] = $csv_array[1];
$insert_csv['lastname'] = $csv_array[2];
$insert_csv['email'] = $csv_array[3];
$isql = "INSERT INTO `myguests`(`id`, `firstname`, `lastname`, `email`) VALUES ('','".$insert_csv['firstname']."','".$insert_csv['lastname']."','".$insert_csv['email']."')";
$run = mysqli_query($con, $isql);
$i++;
}
fclose($open);
echo "File upload successfully";
mysqli_close($con);
}
}
else{
echo "Not valid file formate";
}
}
?>
【问题讨论】:
-
从 CLI 而不是浏览器运行它怎么样?
标签: php html mysql firefox csv