Question

可能会再次询问，但我找不到我的＆＃34;没有选择数据库的解决方案＆＃34;问题。基本上，我想将CSV内容上传到页面，然后将它们插入到MYSQL数据库中。以下代码在将内容显示到网页之前一直很好。我无法将详细信息存入数据库。我得到＆＃34;没有选择数据库＆＃34; ......任何人都引导我完成

＆＃13;

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Upload page</title>
<style type="text/css">
body {
	background: #E3F4FC;
	font: normal 14px/30px Helvetica, Arial, sans-serif;
	color: #2b2b2b;
}
a {
	color:#898989;
	font-size:14px;
	font-weight:bold;
	text-decoration:none;
}
a:hover {
	color:#CC0033;
}

h1 {
	font: bold 14px Helvetica, Arial, sans-serif;
	color: #CC0033;
}
h2 {
	font: bold 14px Helvetica, Arial, sans-serif;
	color: #898989;
}
#container {
	background: #CCC;
	margin: 100px auto;
	width: 945px;
}
#form 			{padding: 20px 150px;}
#form input     {margin-bottom: 20px;}
</style>

</head>
<body>
<div id="container">
<div id="form">

<?php

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "btccredentials";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}



//Upload File
if (isset($_POST['submit'])) {
	if (is_uploaded_file($_FILES['filename']['tmp_name'])) {
		echo "<h1>" . "File ". $_FILES['filename']['name'] ." uploaded successfully." . "</h1>";
		echo "<h2>Displaying contents:</h2>";
		readfile($_FILES['filename']['tmp_name']);
	}

	//Import uploaded file to Database
	$handle = fopen($_FILES['filename']['tmp_name'], "r");

	while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
		$import = "INSERT into btcdetails(mobilenumber,amount) values('$data[0]','$data[1]'),$dbname";
      
		mysqli_query($conn,$import);
	}

	fclose($handle);

	print "Import done";

	//view upload form
}else {

	print "Upload new csv by browsing to file and clicking on Upload<br />\n";

	print "<form enctype='multipart/form-data' action='uploadcsv.php' method='post'>";

	print "File name to import:<br />\n";

	print "<input size='50' type='file' name='filename'><br />\n";

	print "<input type='submit' name='submit' value='Upload'></form>";

}

?>

</div>
</div>
</body>
</html>

＆＃13;

Answer 1

问题是你混淆了两个功能。一个是弃用的，另一个是好的，但我建议你使用PDO http://php.net/manual/en/book.pdo.php进行安全的数据库操作。

现在您的问题出在下面的代码中。

new mysqli($servername, $username, $password, $dbname); // Your connection 

mysql_query($import) or die(mysql_error()); // Your function

使用以下代码更改它。

mysqli_query($conn,$import) or die(mysql_error());

请记住，对于所有实例都使用mysqli，因为您已使用mysqli作为连接。

如何在PHP中将csv数据插入到mysql数据库中

1 个答案: