I am trying to upload a image to MySQL databases using php5 script. And I am receiving an notice error.
Error, query failed
UploadImage.php
<?php
session_start();
?>
<HTML>
<HEAD>
<TITLE> Image Upload</TITLE>
</HEAD>
<BODY>
<FORM NAME="f1" METHOD="POST" ACTION="uploadImage2.php" ENCTYPE="multipart/form-data">
<table>
<tr><td> Image Upload Page </td></tr>
<tr><td> <input type="file" name="imgfile"/></td></tr>
<tr><td> <input type="submit" name="submit" value="Save"/> </td></tr>
</table>
</FORM>
</BODY>
</HTML>UploadImage2.php
<?php
include "dbconfig.php";
$dbconn = mysql_connect($dbhost, $dbusr, $dbpass) or die("Error Occurred-".mysql_error());
mysql_select_db($dbname, $dbconn) or die("Unable to select database");
if(isset($_REQUEST['submit']) && $_FILES['imgfile']['size'] > 0)
{
$fileName = mysql_real_escape_string($_FILES['imgfile']['name']); // image file name
$tmpName = $_FILES['imgfile']['tmp_name']; // name of the temporary stored file name
$fileSize = mysql_real_escape_string($_FILES['imgfile']['size']); // size of the uploaded file
$fileType = mysql_real_escape_string($_FILES['imgfile']['type']); //
$fp = fopen($tmpName, 'r'); // open a file handle of the temporary file
$imgContent = fread($fp, filesize($tmpName)); // read the temp file
$imgContent = mysql_real_escape_string($imgContent);
fclose($fp); // close the file handle
$query = "INSERT INTO img_tbl (img_name, img_type, img_size, img_data )
VALUES ('$fileName', '$fileType', '$fileSize', '$imgContent')";
mysql_query($query) or die('Error, query failed'.mysql_errno($dbconn) . ": " . mysql_error($dbconn) . "
");
$imgid = mysql_insert_id(); // autoincrement id of the uploaded entry
//mysql_close($dbconn);
echo "<br>Image successfully uploaded to database<br>";
echo "<a href=\"uploadImage2_viewimage.php?id=$imgid\">View Image</a>";
}else die("You have not selected any image");
?>I have upload an image file but still have error on it.
But now I have counter another error for view Image.
<?php
// get the file with the id from database
include "dbconfig.php";
$dbconn = mysql_connect($dbhost, $dbusr, $dbpass) or die("Error Occurred-".mysql_error());
mysql_select_db($dbname, $dbconn) or die("Unable to select database");
if(isset($_REQUEST['id']))
{
$id = $_REQUEST ['id'];
$query = "SELECT img_name, img_type, img_size, img_data FROM img_tbl WHERE id = ‘$id’";
$result = mysql_query($query) or die(mysql_error());
list($name, $type, $size, $content) = mysql_fetch_array($result);
header("Content-length: $size");
header("Content-type: $type");
print $content;
mysql_close($dbconn);
}
?>The error code:
Notice: Undefined variable: id� in C:\xampp\htdocs\sandbox\Testing\uploadImage2_viewimage.php on line 12 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '�' at line 1
Please advise...
</div>
