Display content of a file

Judit1983

I want to display the content of a file which is stored in a database. I put a read button next to each column and if you click on this button you should be able to read the content of the stored file on a new page.

Here's the code.

<?php

if (isset($_POST['Read']))
{
mysql_connect("localhost","root","") or die ("Connection failed");
mysql_select_db("txt_archive") or die ("Connection failed");

$sql2 = "SELECT location FROM files";
$myData = mysql_query ($sql2,$con) or die ("Fájl nem található.");

while ($result = $myData->fetch_assoc()){

$loc = $result["location"];

$content = file_get_contents($loc);

print ($content);

}
};

But got this error: Fatal error: Call to a member function fetch_assoc() on a non-object in C:\xampp\htdocs\txtupload\read.php on line 11

bradgrafelman

[man]mysql_query/man doesn't return an object.

Furthermore, you shouldn't even be using the entire [man]mysql[/man] extension at all - it's outdated and superseded by newer extensions such as [man]MySQLi[/man] (or [man]PDO[/man]). See the manual page [man]mysqlinfo.api.choosing[/man] for more info.

Judit1983

I place here the index.php file as well: (Some messages are in Hungarian, but it's unimportant)

<?php
ini_set("memory_limit","128M");
ini_set("post_max_size","64M");
ini_set("upload_max-filesize","64M");
if(isset($POST['ok']))
{
if($FILES['fajl']['name']!="")
{
if($FILES['fajl']['type']=="text/plain")
{
if($FILES['fajl']['error']==0)
{
if(!is_dir("uploads"))
{
mkdir("uploads");
}
$filename=basename($FILES['fajl']['name']);
$title=$POST['title'];
move_uploaded_file($_FILES['fajl']['tmp_name'],"uploads/".$filename);
$location="uploads/".$filename;
print $location;
if(file_exists("uploads/".$filename))
{
$error="Sikeres fájlfeltöltés!";
$con = mysql_connect("localhost","root","");
if (!$con)
{
die ("Kapcsolódás nem sikerült a kapcsolódás ehhez: " . mysql_error());
}
mysql_real_escape_string($filename);
mysql_real_escape_string($title);
strip_tags($filename);
strip_tags($title);
mysql_select_db("txt_archive",$con);
$sql = mysql_query("INSERT INTO files (id,location,title) VALUES ('','$location','$title')");

			}
			else
			{
				$error="A Fájl feltöltése sikertelen!";
			}
		}
		else
		{
			$error="Hiba a feltöltés során: ".$_FILES['fajl']['error'];
		}
	}
	else
		{
			$error="Hibás fájlformátum - csak TXT megengedett";
		}
}
else
{
	$error="Nem adott meg fájlt!";
}

}
echo "Korábban feltöltött fileok:";
echo "<br />";
$con = mysql_connect("localhost","root","");
if (!$con){
die ("Nem sikerült kapcsolódni a következő adatbázishoz:" . mysql_error());
}
mysql_select_db("txt_archive",$con);

$sql="SELECT * FROM files";
$myData = mysql_query ($sql,$con);
echo"<table border=1>
<tr>
<th>Cím</th>
<th>Olvasás</th>
</tr>";

while($record = mysql_fetch_array($myData))
{
echo "<form action=read.php method=post>";
echo "<tr>";
echo "<td>" . $record['title'] . "</td>";
echo "<td>" . "<input type=submit name=Olvas value=Olvas>" . "</form>" . "</td>";
echo "</tr>";
echo "</form>";
}
echo "</table>";
mysql_close ($con);
?>
</body>
</head>
</html>
<!DOCTYPE html>
<html>
<head></head>
<body>
<?php
if(isset($error))
{
print($error);
}
?>
<form method="post" enctype="multipart/form-data">
Kérem töltsön fel egy TXT fájlt.<input type="file" name="fajl" /> <br />
Adjon meg hozzá egy nevet: <input type="text" name="title" id="title"/>
<input type="hidden" name="MAX_FILE_SIZE" value="2MB">
<br />
<input type="reset" name="reset" value="Törlés" />
<input type="submit" name="ok" value="Feltöltés">
</form>
</body>
</html>

bradgrafelman

What's unimportant is actually everything you posted. What is important is both of the issues I mentioned in my previous reply.

Judit1983

I managed to display the file content, the problem is that to display on the content of the file, where I clicked on the Read button next to it. Now it lists all the content of all the files in the database. Here's my code:

<?php

if (isset($_POST['Olvas']))
{
mysql_connect("localhost","root","") or die ("Kapcsolódás nem sikerült");
mysql_select_db("txt_archive") or die ("Kapcsolódás nem sikerült");

$sql2=("SELECT location FROM files");
$myData = mysql_query ($sql2) or die ("Fájl nem található.");

while ($result = mysql_fetch_assoc($myData)) {
$loc = $result["location"];

    $content = file_get_contents($loc);
    print ($content);
    }

};

Weedpacket

When posting code, please use the markup tags described in the FAQs to mark up your code and make it readable.

$sql2=("SELECT location FROM files");

You don't say anything about which file you want to display (there's no WHERE clause), so you retrieve all of them.

Judit1983

And how can I say to PHP that I want display the content of that file, that is in the same row with the Read button?

Judit1983

How to change the following code not to retrieve the content of all the uploaded files, but only the content of one life (which is in the same row with the Read button)

Here's the code when I create the button:

while($record = mysql_fetch_array($myData))
{
  echo "<form action=read.php method=post>";
  echo "<tr>";
  echo "<td>" . $record['title'] . "</td>";
  echo "<td>" . "<input type=submit name=Read value=" . $record['id']">" . "</td>";
  echo "<td>" . "<input type=hidden name=hidden value=" . $record['id'] . " </td>";
  echo "</tr>";
  echo "</form>";
}

and when I make the mysql query

$sql2="SELECT location FROM files WHERE id="$record['id']";

Judit1983

I also got this error:

Warning: imagepng() [function.imagepng]: gd-png error: compression level must be 0 through 9 in C:\xampp\htdocs\imageconvert\function.php on line 65