stevieontario

Ahh, thanks Pika. okay, I put in an error statement and got the following fatal mysql error: Odd: I checked username and password and can't see anything wrong, plus I gave user2 -- i.e., myself -- all privileges. I'm checking with my provider, maybe this is at their end.

Ahh, thanks Pika. okay, I put in an error statement and got the following fatal mysql error: Odd: I checked username and password and can't see anything wrong, plus I gave user2 -- i.e., myself -- all privileges. I'm checking with my provider, maybe this is at their end.

Afternoon Freaks, PHP Version 5.2.15 mysql server 5.1.48-log MySQL client version: 5.1.48 I loaded the following code and got no errors but also no data appears in db2.table3. <?php $db_hostname = "localhost"; $db_name = "db1"; $db_username = "user1"; $db_password = "pw1"; $cxn = mysql_connect($db_hostname,$db_username,$db_password) or die ("Could not connect: " . mysql_error()); mysql_select_db($db_name); $db_name2 = "db2"; $db_username2 = "user2"; $db_password2 = "pw2"; $cxn2 = mysql_connect($db_hostname,$db_username2,$db_password2) or die ("Could not connect: " . mysql_error()); mysql_select_db($db_name2); $sql1 = "SELECT sourceid as sourceid2, unitname, fuel, output, capability, capabilityfactor, generated, version as hour FROM table1, table2 where table1.sourceid=table2.id and fuel like 'gasoline%'"; $result = mysql_query($sql1, $cxn) or die("<b>A fatal MySQL error occured</b>.\n<br />Query: " . $sql1 . "<br />\nError: (" . mysql_errno() . ") " . mysql_error()); while ( $row = mysql_fetch_assoc($result)) { extract($row); $sql2 = "insert into db2.table3 (sourceid2, unitname, fuel, output, capability, capabilityfactor, generated, hour) values ('$sourceid2', '$unitname', '$fuel', '$output', '$capability', '$capabilityfactor', '$generated', '$hour')"; $result1 = mysql_query($sql2, $cxn2); } ?> I pre-tested it in xampp on my pc, and it worked fine. I suspect it has something to do with the fact that I'm testing it on a live server, but I'm stumped -- no errors are displayed either on the screen or on my server error log. Any ideas? Thanks in advance!

Arbitter, many thanks -- that worked. I should have indicated I am a wet-behind-the-ears novice when it comes to this. This is all a result of my old computer crashing and me having to rebuild the application from memory and old files. One further question: is there any way I could have discovered this on my own, by going through the files in the xampp/phpmyadmin folder?

Hello everyone, I created a username and password for phpMyAdmin, and now my own phpMyAdmin won't accept my own password! Here's the error I get: Following someone else's advice, I deleted the files "user.frm", "user.MYD", and "user.MYI" from the folder xampp/mysql/data/mysql then put new ones back in. Any advice would be greatly appreciated! Thanks,

As always, thanks for your time. I did the "show create table performance2" query in phpMyAdmin, and just copied the results. "Binary_file.dat" appears in each field in the perfid column when I click on the performance2 table in phpMyAdmin, in Browse mode.

Thanks Mchl, and sorry for the delay (computer problems related to office move). Here's the table structure: performance2 CREATE TABLE `performance2` ( `perfid` int(125) ... Hope this helps, Steve

whoops sorry, perfid is my auto incremented primary key column for the table performance2. Since it's auto-incremented, I don't include it in the insert.

Afternoon Freaks, I have been getting a weird INSERT query result. I put data into a table using the following query: $cquery = "insert into performance2 (sourceId, column1, column2, column3, column4) VALUES ('$lastSid', '$value1', '$value2', '$value3', '$value4')"; mysql_query($cquery, $cxn); The table performance2 contains an id field (perfid) which is an auto-incremented, indexed, primary key. When I run that query, everything works -- except that all the perfid rows are filled with "binary_file.dat". I've never seen this before, and I've tried googling the problem and still can't find out what's going on. Anybody have any ideas? Thanks!

thanks PFM. This is a mystery. The variable $object doesn't appear anywhere else in "Functions1.php" or in the main file. Also, the function object2array is not called anywhere else, and the code I displayed in the main file is all the code there is that is relevant to this function (from there on down it's just a bunch of if and foreach loops, which all produce the right output). Here's another interesting thing: I finally figured out how to do the $insert2sourceinfo2 query properly, and the variable $generated_date, which is based on $rr, gives the right output! I still get the same warning, but it produces the right output.

var_dump($output) shows the structure of the xml file "xmlfilename.xml": SimpleXMLElement Object ( [@attributes] => Array ( [docID] => etc. The file is fairly large and complex, so I won't show the whole thing

okay thanks, that was my next question: is there a way to migrate a thread to another forum. Now, another elementary question (and please pardon my inexperience): how exactly do I bump a thread?

sorry, I'm not trying to waste anyone's time. I noticed my earlier thread had gone to page 2 of the previous forum, which gives me to believe it will fall off the radar. Also I felt I had not explained it properly initially; plus I realized I should have posted it to this one in the first place.

I should have also pointed you to line 20 in Functions1.php: $var = get_object_vars($object);

Hi everyone, I have stared at the following problem for so long that I may be missing something very elementary; I was hoping somebody might spot it. I wrote a piece of code which includes a call to a function someone else wrote. They are in two different files: "File.php" (which I wrote) and "Functions1.php." The latter, Functions1.php, contains the fuction the other guy wrote. When I launch File.php to the browser, I get the following message: Here is the pertinent part of File.php: require_once ("Functions1.php"); ////////////////////////////////////////////////////////// $lastxml = basename('c:/program files/xampp/htdocs/dev/xmlfilename.xml'); $output= simplexml_load_file($lastxml); $rr = object2array($output); $fdate = getdatafromfilename($lastxml); $fversion = getdatafromfilename($lastxml,1); $generated_date = trim($rr['IMODocHeader']['CreatedAt']); $generated_date = str_replace("T"," ",$generated_date); /////////// Database Settings ////////////// $db_hostname = "localhost"; $db_name = "db1"; $db_username = "jacksmith"; $db_password = "password1"; $cxn = mysql_connect($db_hostname, $db_username, $db_password); ///////////////////////////////////////////// $insert2sourceinfo2 = "insert into sourceinfo2 (xmlname, filedate, version, generated) VALUES ('$lastxml', '$fdate', '$fversion', '$generated_date')"; mysql_query($insert2sourceinfo2, $cxn); $lastSid = mysql_insert_id(); And here's the code for the function "object2array" in the file Functions1.php: // converts the xml file into a php object; function object2array($object) { $return = NULL; if(is_array($object)) { foreach($object as $key => $value) $return[$key] = object2array($value); } else { $var = get_object_vars($object); if($var) { foreach($var as $key => $value) $return[$key] = object2array($value); } else return strval($object); } return $return; } The variable $lastxml outputs properly when I execute print_r($lastxml), and the variables $fdate and $fversion also output properly. Any ideas? Thanks!

yes, an array of xml elements, which I am able to loop through successfully (that code is below the sql query in the main program). Just to narrow down the problem, I put echo statements inside the looping scripts, and lauched the script to my browser. The echo statements return the expected output. I turn $lastxml (which is an xml file) into an object using simplexml_load_file, then the function object2array turns that object into an array. Or at least it's supposed to.

team, thanks again. I should have mentioned that I tested $lastxml using echo var_dump($lastxml); and it printed fine, so I think the problem resides elsewhere. I also echoed $fdate and $fversion (the function getdatafromfilename also resides in "functions1.php"), and that was successful too.

thanks teamatomic, That's just it -- the code that I displayed is all the code. $object appears only in the function object2array. I have never touched it before (someone else wrote it), and it worked previously.

Morning Freaks, I'm trying to recreate code which includes a call to a function in an include file. The original code worked, but the recreated code is giving me problems. I get the following error message Also, the database insert query doesn't put anything into the database. I haven't altered the include file at all. This leads me to believe that my problem is with the main file. Here's the main file code: require_once ("Functions1.php"); ////////////////////////////////////////////////////////// $lastxml = basename('c:/program files/xampp/htdocs/dev/xmlfilename.xml'); $output= simplexml_load_file($lastxml); $rr = object2array($output); $fdate = getdatafromfilename($lastxml); $fversion = getdatafromfilename($lastxml,1); $generated_date = trim($rr['IMODocHeader']['CreatedAt']); $generated_date = str_replace("T"," ",$generated_date); /////////// Database Settings ////////////// $db_hostname = "localhost"; $db_name = "newstats"; $db_username = "steveaplin"; $db_password = "nafta"; $cxn = mysql_connect($db_hostname, $db_username, $db_password); ///////////////////////////////////////////// $insert2sourceinfo2 = "insert into sourceinfo2 (xmlname, filedate, version, generated) VALUES ('$lastxml', '$fdate', '$fversion', '$generated_date')"; mysql_query($insert2sourceinfo2, $cxn); $lastSid = mysql_insert_id(); And here's the include file code: // converts the xml file into a php object; function object2array($object) { $return = NULL; if(is_array($object)) { foreach($object as $key => $value) $return[$key] = object2array($value); } else { $var = get_object_vars($object); if($var) { foreach($var as $key => $value) $return[$key] = object2array($value); } else return strval($object); } return $return; } Line 20 is the following: $var = get_object_vars($object); Anybody have any ideas? Thanks in advance!

I'm playing whack-a-mole here. I uninstalled xampp, and reinstalled it only to get (1) a warning that and (2) a fatal error: Yikes! Any ideas on what's going on?

Jay, many thanks. I'll try that doc you recommended. However, in attempting to work through the problem outlined in my post, I ran into the problem indicated in the Subject. i.e., I went and changed the local password (in phpMyAdmin), and now I can't even load the local page in my browser without getting the following error message: I have googled the heck out of this, and keep running into suggestions that I reset privileges via the command line prompt. It is at that point that I realize I'm in way over my head: I don't even know how to access mysql from the command line prompt. In fact I'm not even sure what the command line prompt is!

Afternoon Freaks, I'm having a weird experience trying to insert data to a Mysql db using php. I'm getting the following message: Playing in a xampp sandbox and editing/launching code from DreamWeaver CS3 PHP v. 5.2.9 MySQL API 5.0.51a My code: $db_hostname = "localhost"; $db_name = "mydb"; $db_username = "myname"; $db_password = "mypassword"; $cxn = mysql_connect($db_hostname, $db_username, $db_password); /////////////////////////////////////////////////////////////////////////////////////////////////////// $insert2sourceinfo2 = "insert into sourceinfo2 (filename, filedate, version, generated) VALUES ('$filename', '$fdate', '$fversion', '$generated_date')"; mysql_query($insert2sourceinfo2, $cxn); $lastSid = mysql_insert_id(); I set up xampp on a machine I don't normally use it on (the one I normally use is kaput for the time being), and as far as I can remember all the settings (db host, db name, etc.) are exactly the same. If anyone has any ideas where the little gremlin is, I'd sure appreciate it! Thanks,

cags, thanks very much, that was exactly the problem. Everything is good now. I love this forum! thanks again

Afternoon PHPFreaks, I'm having a difficult time displaying the results of a mysql query. Here's the code: $sql = "SELECT fuel, sum( output ), avg(capabilityfactor) FROM performance\n CROSS JOIN\n (SELECT MAX(id) AS sourceid FROM sourceinfo) AS sq\n USING (sourceid)\n GROUP BY fuel\n ORDER BY sum( output ) DESC"; $result = mysql_query($sql, $cxn) or die ("fuggoff, cahsuh muhfuh"); while ( $row = mysql_fetch_assoc($result)) { extract($row); echo "<pre>$fuel: $output&#09;$capabilityfactor</pre>"; } mysql_close($cxn); The browser displays only $fuel (five rows) in the correct order, but not $output or $capabilityfactor. When I do the query in phpMyAdmin, it runs just fine. This leads me to suspect I am using mysql_fetch_assoc() or extract() incorrectly, but I can't figure out what I'm doing wrong. Any suggestions would be much appreciated! Thanks in advance

ahh... beautiful! thanks so much, this was driving me nuts