basically this a code for uploading images and text info into a mysql
database, however, when I execute this code, an error occured as:
Undefined variable: action in \phpmySQL\add.php on line 2
what will be the problem?
<?php
if ($action == "upload") {
// ok, let's get the uploaded data and insert it into the db now
include "open_db.inc";
if (isset($binFile) && $binFile != "none") {
$data = addslashes(fread(fopen($binFile, "r"), filesize($binFile)));
$strDescription = addslashes(nl2br($txtDescription));
$sql = "INSERT INTO tbl_Files ";
$sql .= "(description, bin_data, filename, filesize, filetype) ";
$sql .= "VALUES ('$strDescription', '$data', ";
$sql .= "'$binFile_name', '$binFile_size', '$binFile_type')";
$result = mysql_query($sql, $db);
mysql_free_result($result); // it's always nice to clean up!
echo "Thank you. The new file was successfully added to our
database.<br><br>";
echo "<a href='main.php'>Continue</a>";
}
mysql_close();
} else {
?>
<HTML>
<BODY>
<FORM METHOD="post" ACTION="add.php" ENCTYPE="multipart/form-data">
<INPUT TYPE="hidden" NAME="MAX_FILE_SIZE" VALUE="1000000">
<INPUT TYPE="hidden" NAME="action" VALUE="upload">
<TABLE BORDER="1">
<TR>
<TD>Description: </TD>
<TD><TEXTAREA NAME="txtDescription" ROWS="10" COLS="50"></TEXTAREA></TD>
</TR>
<TR>
<TD>File: </TD>
<TD><INPUT TYPE="file" NAME="binFile"></TD>
</TR>
<TR>
<TD COLSPAN="2"><INPUT TYPE="submit" VALUE="Upload"></TD>
</TR>
</TABLE>
</FORM>
</BODY>
</HTML>
<?php
}
?>
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