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Bug #13382 PHP - MySQL select statement
Submitted: 2001-09-21 16:27 UTC Modified: 2001-10-19 11:07 UTC
From: tranxuan dot thuyen at kpmg-ct dot com Assigned:
Status: Not a bug Package: Output Control
PHP Version: 4.0.6 OS: windows2000
Private report: No CVE-ID: None
 [2001-09-21 16:27 UTC] tranxuan dot thuyen at kpmg-ct dot com
Dear supporter,

My name is Thuyen Tran (KPMG CT Information) and I am working now with PHP and MySQL.
I just have a problem when I try to read and write the data from the database to the OUTFILE
using the following statement on the Unix web server:

$q_users = ("SELECT * INTO OUTFILE '$base_dir$csv_dir_name/$csv_file_users' FROM user_info");

but unfortunetly this doesn't work. The following error occurs:

Database error: Invalid SQL: SELECT * INTO OUTFILE '/space/web/kct/projects/testshop/export/outfile_users.csv' FROM user_info
MySQL Error: 1 (Can't create/write to file '/space/web/kct/projects/testshop/export/outfile_users.csv' (Errcode: 2))

It seem to me that the path '$base_dir$csv_dir_name'  
which is
"/space/web/kct/projects/testshop/export" as below can 
be seen, is not recognised (may be system drive??)
For the same code it works well for other web server. 
What I want to do is to put the *.csv file in a certain
specified directory.

I tried to open new file with success in the same directory as follows in order to see what happens:

$fp = fopen("./hallo.txt", "w+"); // only try to test 

and also 
$q_users = ("SELECT * INTO OUTFILE './$csv_file_users'  FROM user_info ");

all of these work well.

So please could you tell me what was wrong?

Your quick response will graetly be appreciated. Many thanks!


Here below is my code:

	//Create and make *.csv files to export
	//$base_dir 	= "/space/web/kct/projects/testshop/";
	$csv_dir_name 	= "export";
	$csv_file_orders	= "outfile_orders.csv";	
	$csv_file_users	= "outfile_users.csv";

	// open the directory with the path which is supplied as the sole parameter
    	chmod ("$base_dir", 0777);
	$dir = opendir(".");	
	$exist_flag = "false";	// the csv directory doesn't exist, need to create one.
	while ($dir_name=readdir($dir)) {	
		if ($dir_name == $csv_dir_name){
			$exist_flag = "true";			
	// create a new directory to store the *.csv file
	if ($exist_flag	== "false") {
		mkdir("./$csv_dir_name", "");
		echo ("New sub directory '$base_dir$csv_dir_name' is created!");

    	// change permission for the created directory
    	chmod ("./$csv_dir_name", 0777);
	// Set the current directory and open it
	chdir ("./$csv_dir_name");
	$dir = opendir(".");
	// Here just read all the 'entries' in it (both files and sub-folders)
	while ($file=readdir($dir)) {
		//echo("$file<BR>");	// only use to show all in the current directory
		// delete the csv file permanently if it exists.
		if ($file == $csv_file_users) {
			unlink ($csv_file_users);
		if ($file == $csv_file_orders) {
			unlink ($csv_file_orders);
    	 $fp = fopen("./hallo.txt", "w+"); // only try to test if it can, it works.

	// Get all information of USERS and store them in *.csv file
	$q_users = ("SELECT * INTO OUTFILE '$base_dir$csv_dir_name/$csv_file_users' FROM user_info");
    	//$q_users = ("SELECT * INTO OUTFILE './$csv_file_users'  FROM user_info "); it works.



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 [2001-10-19 11:07 UTC]
This is unlikely to be a bug in PHP. 
Check your permissions; probably MySQL has no permission to write there.

Ask further support questions on the appropriate mailinglist (see
 [2004-06-09 04:27 UTC] asdfsaf at asdfasf dot com
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