ERROR 500 when trying to insert excel data to mysql using PHP

Question

1.00/5 (1 vote)

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I'm trying to upload a file (xlsx) to mysql and I keep getting the same error when trying to submit. "127.0.0.1 is currently unable to handle this request". Any advices?

What I have tried:

<?php
	require('Classes/PHPExcel/IOFactory.php');
	
    $servername="localhost";
	$username="omri";
	$password="1234";
	$dbname="omri";
	
	if(isset($_POST['upload'])){
	
	$inputfilename= $_FILES['file']['tmp_name']	;
	$exceldata= array();
	
	$conn= mysqli_connect($servername, $username, $password, $dbname);
	
	if(!$conn){
	die("connection failed: ".mysqli_connect_error());	
	}	
	
	try{
		
	$inputfiletype= PHPExcel_IOFactory::identify($inputfilename);
	$objReader=	PHPExcel_IOFactory::createReader($inputfiletype);
	$objPHPExcel= $objReader ->load($inputfilename);	
		
	}	
		
	catch(Exception $e)
	{
		
	die('error loading file "'.pathinfo($inputfilename, PATHINFO_BASENAME).'": '.$e->getMessage());
		
	}	
	
	$sheet= $objPHPExcel->getSheet(0);	
	$highestrow= $sheet->getHighestRow();	
	$higestcolumn= $sheet->getHighestColumn();
		
	for($row=1; $row<=$highestrow; $row++)
	{
		
	$rowData= $sheet->rangeToArray('A'. $row .':'. $higestcolumn . $row, NULL, TRUE, FALSE );
	$sql= "INSERT INTO restricted (name, id, citizenship, expires) VALUES ('".$rowData[0][0]."','".$rowData[0][1]."','".$rowData[0][2]."','".$rowData[0][3]."')";
	
	if(mysql_query($conn,$sql)){
	$exceldata[]=$rowData[0];
	}
	
	else{
	echo "Error: " . $sql . "<br>" . mysql_error($conn);	
	}	
	}
		
	echo "<table border='1'>";
	foreach($exceldata as $index => $excelraw)
	{
		
	echo "<tr>";
	
	foreach($excelraw as $excelcolumn)
	{
		
	echo "<td>". $excelcolumn . "</td>";
		
	}	
		
	echo "</tr>";
		
	}	
		
	echo "</table>";	
	
	mysql_close($conn);	
	}
	?>
	
<html>
<head>
<meta charset="UTF-8">	
<title>Import Excel</title>
</head>

<body>
	
<form action="" method="post" enctype="multipart/form-data">
	
<input type="file" name="file">
<input type="submit" name="upload" value="upload">	
	
</form>	
	
</body>
</html>

Posted 27-Jul-18 2:43am

Member 13891176

Updated 4-Apr-23 22:32pm

v2

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Comments

Jochen Arndt 27-Jul-18 8:59am

Check the error log file of your web server. It should contain more information.

Richard Deeming 27-Jul-18 9:13am

Your code is vulnerable to SQL Injection[^]. NEVER use string concatenation to build a SQL query. ALWAYS use a parameterized query.

Everything you wanted to know about SQL injection (but were afraid to ask) | Troy Hunt[^]
How can I explain SQL injection without technical jargon? | Information Security Stack Exchange[^]
Query Parameterization Cheat Sheet | OWASP[^]

PHP: SQL Injection - Manual[^]
PHP: Prepared statements and stored procedures - Manual[^]

1 solution

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Patrice T · Answer 1 · 2018-07-27T03:53:00

PHP

$sql= "INSERT INTO restricted (name, id, citizenship, expires) VALUES ('".$rowData[0][0]."','".$rowData[0][1]."','".$rowData[0][2]."','".$rowData[0][3]."')";

Not a solution to your question, but another problem you have.
Never build an SQL query by concatenating strings. Sooner or later, you will do it with user inputs, and this opens door to a vulnerability named "SQL injection", it is dangerous for your database and error prone.
A single quote in a name and your program crash. If a user input a name like "Brian O'Conner" can crash your app, it is an SQL injection vulnerability, and the crash is the least of the problems, a malicious user input and it is promoted to SQL commands with all credentials.
SQL injection - Wikipedia[^]
SQL Injection[^]
SQL Injection Attacks by Example[^]
PHP: SQL Injection - Manual[^]
SQL Injection Prevention Cheat Sheet - OWASP[^]
How can I explain SQL injection without technical jargon? - Information Security Stack Exchange[^]
-----
Learn to indent properly your code, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.

PHP

<?php
require('Classes/PHPExcel/IOFactory.php');

$servername="localhost";
$username="omri";
$password="1234";
$dbname="omri";

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

	$inputfilename= $_FILES['file']['tmp_name']	;
	$exceldata= array();

	$conn= mysqli_connect($servername, $username, $password, $dbname);

	if(!$conn){
		die("connection failed: ".mysqli_connect_error());
	}

	try{

		$inputfiletype= PHPExcel_IOFactory::identify($inputfilename);
		$objReader=	PHPExcel_IOFactory::createReader($inputfiletype);
		$objPHPExcel= $objReader ->load($inputfilename);

	}

	catch(Exception $e)
	{

		die('error loading file "'.pathinfo($inputfilename, PATHINFO_BASENAME).'": '.$e->getMessage());

	}

	$sheet= $objPHPExcel->getSheet(0);
	$highestrow= $sheet->getHighestRow();
	$higestcolumn= $sheet->getHighestColumn();

	for($row=1; $row<=$highestrow; $row++)
	{

		$rowData= $sheet->rangeToArray('A'. $row .':'. $higestcolumn . $row, NULL, TRUE, FALSE );
		$sql= "INSERT INTO restricted (name, id, citizenship, expires) VALUES ('".$rowData[0][0]."','".$rowData[0][1]."','".$rowData[0][2]."','".$rowData[0][3]."')";

		if(mysql_query($conn,$sql)){
			$exceldata[]=$rowData[0];
		}

		else{
			echo "Error: " . $sql . "<br>" . mysql_error($conn);
		}
	}

	echo "<table border='1'>";
	foreach($exceldata as $index => $excelraw)
	{

		echo "<tr>";

		foreach($excelraw as $excelcolumn)
		{

			echo "<td>". $excelcolumn . "</td>";

		}

		echo "</tr>";

	}

	echo "</table>";

	mysql_close($conn);
}
?>

<html>
	<head>
		<meta charset="UTF-8">
		<title>Import Excel</title>
	</head>

	<body>

		<form action="" method="post" enctype="multipart/form-data">
			<input type="file" name="file">
			<input type="submit" name="upload" value="upload">
		</form>

	</body>
</html>

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