Bug #13382 Updated: PHP - MySQL select statement

ID: 13382 Updated by: sander Reported By: tranxuan.thuyen@kpmg-ct.com Old Status: Open Status: Bogus Bug Type: Output Control Operating System: windows2000 PHP Version: 4.0.6 New Comment: 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 http://www.php.net/support.php). Previous Comments: ------------------------------------------------------------------------ [2001-09-21 16:27:59] tranxuan.thuyen@kpmg-ct.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! Regards, Thuyen Here below is my code: <?php //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){ $exist_flag = "true"; break; } } // create a new directory to store the *.csv file if ($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. $db->query($q_users); $db->next_record(); ?> ------------------------------------------------------------------------ Edit this bug report at http://bugs.php.net/?id=13382&edit=1