foreach loop returns nothing - php

I am trying to pull user data from a Cart66 table I have and put it into a shortcode in wordpress. $account is an integer pulled from session data. The code below returns nothing.
$account =Cart66Session::get(Cart66AccountId);
global $wpdb;
$fname=$wpdb->get_results("SELECT * FROM 'vfp_cart66_accounts' WHERE id = '$account', ARRAY_N");
foreach ($fname AS $row)
{
echo $row;
}
This returns "Array"
return $fname;

Ok firstly, maybe I am the only one who saw this, and it could be the source of your entire problem, but you have a misplaced double quote, at the end of your SQL line, which should live at the end of the actual SQL string, not after the requested return type:
// at the end of this line you have: '$account', ARRAY_N");
// this should be changed to: '$account'", ARRAY_N);
$fname=$wpdb->get_results("SELECT * FROM 'vfp_cart66_accounts' WHERE id = '$account', ARRAY_N");
Even the first person who answered the question did not correct you, so I am assuming he didn't see it either. Secondly, using single quotes (') to escape a table name is invalid. If it is quoted at all, use backticks (`). Single quotes indicate a string, not an database, table, or field, all three of which should only be quoted with backticks (except on utility queries like SHOW). Use this instead:
select * from `vfp_cart66_accounts` where id = '$account'
Thirdly, as your commenters point out, you could be vulnerable to SQL Injection. Make sure to use the tools that WP gives you, and do this, or similar, instead:
$fname = $wpdb->get_results(
$wpdb->prepare(
'select * from `vfp_cart66_accounts` where id = %d',
$account
),
ARRAY_N
);
Lastly, you are requesting an array from the DB, but you are trying to echo it as if it were a scalar value. This explains why printing the value of $row yields "Array". When you convert an array() to a string, by default, you get "Array", since arrays can be complex data that may not be beautifully converted to a string. As a correction of this, you can do one of two things.
First, if you need the entire resulting array that represents the entire row of the table, then you can simply change your echo code to this:
foreach ($fname as $row) {
// print the fname of the row
echo $row['fname'];
// do the other stuff you need to do with $row
...
}
OR, if you simply need the fname field out of that table, for the given id, you could use a different $wpdb function, called $wpdb->get_var(), which gets one specific field from the first entry of the resulting data from the database, coupled with some minor SQL changes:
// use the get_var() function instead
$fname = $wpdb->get_var(
$wpdb->prepare(
// 1) change the 'fields' of your sql to only get the `fname` field
// 2) also add limit 1, to reduce load by only asking for one row
// NOTE: #2 is optional really, because WP does this for you when using get_var,
// but is good practice to only ask for what you need. so do it
'select fname from `vfp_cart66_accounts` where id = %d limit 1',
$account
),
ARRAY_N
);
echo $fname; // print the value of field fname from vfp_cart66_accounts for id $account
Now. I don't have specific knowledge of Cart66. That being said, if the above changes to PHP, WordPress, and SQL syntax do not yield results, then you are probably having one of the following other problems instead:
there is a different PHP error somewhere in the code, causing this to never run
this code is never called, and thus it is never executed
you misspelled the table name, which is causing an SQL error
the table exists, but does not have a field named id
both table and field exist, but there are no entries in the table
some other random thing that is not coming to mind
DEBUG #1
For #1, you could try turning on error_reporting() and display_errors early in the code execution. In a normal, run of the mill PHP script you could add the following two lines somewhere early in the code:
error_reporting(E_ALL);
ini_set('display_errors', 1);
However, you are using WordPress, so you will need to do something like this in your wp-config.php file:
// find the line that looks like this and comment it out
// define('WP_DEBUG', false);
// add these two lines directly below it
define('WP_DEBUG', true);
ini_set('display_errors', 1);
DEBUG #2
Make sure your code is running. Don't be afraid to throw a die() statement directly above it, to make sure it is running. Something like this:
// add a die() before everything
die('I am running. Awesome!');
// revised code
$account = Cart66Session::get(Cart66AccountId);
global $wpdb;
$fname = $wpdb->get_var(
$wpdb->prepare(
'select fname from `vfp_cart66_accounts` where id = %d limit 1',
$account
),
ARRAY_N
);
echo $fname;
DEBUG #3
To debug #3, you need either access to a commandline tool for MySQL or some type of GUI interface like phpMyAdmin, so that you can run a query directly from the database. Here is the query you should run:
show tables like 'vfp_cart66_%';
This is an example of one of the only places in SQL that you should ever quote a table name in single quotes. Running this will yield a list of all the tables that start with vfp_cart66_. If you get no results, then your table name is wrong. If your results do not include vfp_cart66_accounts, then your table name is wrong. If you see vfp_cart66_accounts, you are good to go.
DEBUG #4
This one will need to be run directly from the DB or through something like phpMyAdmin also. You are trying to make sure you have the correct field name. The way you do that is:
show create table `vfp_cart66_accounts`;
Assumedly, the field you are calling id would be the auto_incremented field in the table. Thus you are looking for a line, similar to this one:
`id` bigint(20) NOT NULL AUTO_INCREMENT,
Make sure that the line that has AUTO_INCREMENT on it, begins with:
`id`
If it does not, and the name is something else other than id, then you probably have the wrong field name.
DEBUG #5
Make sure you actually have data to display. From your mysql console or phpMyAdmin, run:
select * from `vfp_cart66_accounts` limit 1;
If you bet any results, then you have data, and you are good.
DEBUG #3 - #5 (alternate methods)
Another option you have is to dump the $wpdb object, directly after you run the query, because it contains the last error you received from MySQL. You can do this like so:
$fname = $wpdb->get_var(
$wpdb->prepare(
'select fname from `vfp_cart66_accounts` where id = %d limit 1',
$account
),
ARRAY_N
);
// dump a readable version of the $wpdb object
echo '<pre>';
print_r($wpdb);
die('</pre>');
Often times, reading the MySQL error message helps narrow down the problem in your SQL syntax.
DEBUG #6
If none of this has helped at all, then you will need to use your experience to trackdown a random bug in either your plugins or theme, what could literally be anything. You may as well not even dig in core WP code because, while it does have a couple minor bugs unrelated to your problem, which are getting repaired as we speak, it is one of the most stable CMS platforms out there. It is used by more of the top 10 million sites on the internet than any other CMS, for a good reason. It works, it is up-to-date, and most of all, it is stable.
I really hope you found this helpful or at least learned something from it. Hopefully others find it useful as well.

$fname=$wpdb->get_results(
"SELECT * FROM `vfp_cart66_accounts` WHERE id = '$account'",
ARRAY_N"
);

Related

PHP mySQL variable access

Say I do something like this:
//get unit id
$query = "SELECT id FROM units WHERE unit_name = '".$unit."'";
$id = mysqli_query($con, $query);
$unit_id = 0;
while ($row = mysqli_fetch_array($id))
{
$unit_id = $row['id'];
}
why is $unit_id not changed outside of the while loop?
What happens is this: I have a selection dropdown with a list of units and when on is clicked that php code is fired (along with other code in the file) and it makes a hidden input field with the id in it. I unhide the id and find that the id is not correct. What is displays, rather, is, say I click the first option, 1001, second option, 1002, third, 1003, etc. These ids do not correspond to my database at all, though the units begin at 1001 in the database. Because of all of that I assumed that my $unit_id just wasn't getting read properly and that somehow PHP didn't let one access the variable outside of a while loop in that way. I see now that assumption was premature. Thanks.
Two possible explanations:
Your query is failing but you have error reporting disabled (nor are you outputting/logging MySQL errors).
There is no unit_name with that name.
If you don't have error reporting enabled by default, try putting:
ini_set('display_errors',1);
error_reporting(E_ALL);
at the top of your script (in a dev enviornment, error reporting should be enabled by default, by the way). Also, you can try using:
$id = mysqli_query($con, $query) or trigger_error(mysqli_error($con));
to view any MySQL errors that may have occurred.
Actually it changes.
Two cases for this..
Maybe your $row['id'] value is also 0.
[or]
Your query returning (0 results) and it is not entering the loop.
What happens is this: I have a selection dropdown with a list of units and when on is clicked that php code is fired (along with other code in the file) and it makes a hidden input field with the id in it. I unhide the id and find that the id is not correct. What is displays, rather, is, say I click the first option, 1001, second option, 1002, third, 1003, etc. These ids do not correspond to my database at all, though the units begin at 1001 in the database. Because of all of that I assumed that my $unit_id just wasn't getting read properly and that somehow PHP didn't let one access the variable outside of a while loop in that way. I see now that assumption was premature. Thanks.

Unable to subtract a table value via variable

I can not get an SQL update statement to subtract a variable from a table value. Here is my code:
$_SESSION_Job101=mysql_fetch_array(mysql_query("SELECT * FROM job_101 WHERE job_101.username='$_SESSION_User'"));
mysql_query("UPDATE characters SET currenergy=currenergy-$_SESSION_Job101['ecost'] WHERE username='$_SESSION_User'");
$_SESSION_Job101 is a perfectly valid result, as I pull from it on another page; I even pull the 'ecost' on said page. I also update currenergy this way in another script, except I use the number 1 instead of the variable. So I've narrowed it down to that variable.
It wouldn't matter that $_SESSION_Job101 is the result from a second table (job_101), and that query is updating to the table characters, would it?
We don't have enough information, but since you don't perform ANY error handling or validation that SQL resultset is returned, it could be an error caused by issues such as:
no rows returned in first query
some other parsing issue not directly evident
I would propose that you use temporary strings and echo the actual SQL queries.
Continue by actually testing them with MYSQL (through workbench, queryviewer, or console) in order to see where and what the error is.
Also, it's not recommended to skip error checking and try to combine so many lines/steps into 2 lines.
Imagine the first query does not return any results for example...
Debugging:
$query1 = "SELECT * FROM job_101 WHERE job_101.username='$_SESSION_User'";
echo $query1."<br/>";
$_SESSION_Job101=mysql_fetch_array(mysql_query($query1 ));
$query2 = "UPDATE characters SET currenergy=currenergy-$_SESSION_Job101['ecost'] WHERE username='$_SESSION_User'";
echo $query2."<br/>";
mysql_query($query2);
Update
Based on your comment I suggest you try the following two options:
1) Add a space between the - and $_SESSION_Job101['ecost'].
2) If that doesn't work, change your string to:
mysql_query("UPDATE characters SET currenergy=currenergy-".$_SESSION_Job101['ecost']." WHERE username='".$_SESSION_User."'";`

SELECT Statement with array in WHERE clause

I am trying to track pageviews on my site whether a user is logged in or not and so I'm capturing the sessionId in my log. I only want to show tracking results for session ids that have at some point logged in and so my flow is as follows:
$pagestats = $wpdb->get_results("SELECT * FROM wp_user_stats WHERE uid=".$_GET['viewstats']);
$sessionArray = array();
foreach($pagestats as $checkSession)
{
if( !(in_array($checkSession->sessionId, $sessionArray)))
{
$sessionArray[] = $checkSession->sessionId;
}
}
Next, I am trying to gather all of the stats concerning any session Ids generated by this particular user ...
$sessions = join(',',$sessionArray);
$pagestats = $wpdb->get_results("SELECT * FROM wp_user_stats WHERE `sessionId` IN($sessions)") or die(mysql_error());
This is the part that throws the error. The error is:
Unknown column 'sjhntmqhltknks8pbhr750voe7' in 'where clause'
I don't understand why it's trying to find a column that matches the session Id instead of a result within that column.
The session IDs probably aren't getting quoted so the query looks like WHERE sessionID IN(abc, def, ...).
One way to fix that would be to change a line in the first loop:
$sessionArray[] = "'".$checkSession->sessionId."'";
Or you could create a second array with the quoted values.
The problem is that the session id is not numeric and needs to be wrapped in quotes. Something like the following would add proper quote to your $sessions string:
$sessions = "'" . implode("', '", $sessionArray) . "'";
You should also make sure you are escaping any user supplied input (e.g. $_GET['viewstats']) before using them to help avoid SQL injection attacks. It wouldn't hurt to escape $checkSession->sessionId as you are adding it to the $sessionArray array either.
If you are using a framework (it looks like you may be using WordPress) read the manual for the database component as it may provide functions to handle some of this for you.
I think you need to put quote around your session values
$sessions = "'".join("','",$sessionArray)."'";
Right now it's like WHERE IN (a,b,c) instead of WHERE IN ('a','b','c').
$pagestats = $wpdb->get_results("SELECT * FROM wp_user_stats WHERE `sessionId`=$sessions") or die(mysql_error());
This isn't an answer to the specific issue, but isn't the second SELECT statement unnecessary? If it is all stored in one table (unless there is a typo...) then SELECT * FROM wp_user_stats WHERE uid=$_GET['view_stats'] would retrieve all sessions for that user. Perhaps you need it done for multiple users? Even at that, you could simply do a GROUP BY clause.
Maybe I'm missing something though -- if so, sorry.

SQL Query Not Reading Variables

My query below updates a record using variables to identify the data in the DB. I think my syntax is correct although it might be wrong. Also, I am absolutely sure that the variables have legitimate values in them. Why won't this query work?
UPDATE `databasename`.`".$tablename."` SET `stock` = '".$f."' WHERE `myerspark`.`item_id` ='".$g."' LIMIT 1
Thanks guys. Tom, yes I have tried that and it works fine. But it is frustrating because I echo all three variables at the end of the script and they all display legitimate values.
Hamish, how do I view these errors?
Jon_Darkstar, these variables are assigned in previous lines of code. Here is my entire code block:
//variables $f, $g, and $tablename assigned from POST variables in previous lines
mysql_select_db($database_Yoforia, $Yoforia);
mysql_query("UPDATE `yoforiainventory`.`".$tablename."` SET `stock` = '".$f."' WHERE `".$tablename."`.`item_id` ='".$g."' LIMIT 1 ");
mysql_close($Yoforia);
echo ($f);
echo ($tablename);
echo ($g);
Again, when i echo these variables, they all come out with good values.
I'm kind of confused what belongs to SQL, what belongs to PHP, where that string comes from, etc. What you have might be fine (if there is a double quote in front and end that i dont see.
I'd probably write it like this:
$sql = "UPDATE databasename.$tablename SET stock = '$f' WHERE myerspark.item_id = '$g' LIMIT 1"
$res = mysql_query($sql, $conn).....
you can backtick more stuff (and/or do mysql_real_escape) for 'extra safety;, but that covers the idea.
What is myerspark? i dont see how it relates to the query, that is probably you're real meaningful error, whether there is a syntax error or not. If myerspark is a seperate table from tablename then you've got an issue here, maybe a JOIN you ought to have?

"SELECT * FROM users WHERE id IN ( )" == FAIL

I have a function that I use called sqlf(), it emulates prepared statements. For instance I can do things like:
$sql = sqlf("SELECT * FROM Users WHERE name= :1 AND email= :2",'Big "John"','bj#example.com') ;
For various reasons, I cannot use prepared statements, but I would like to emulate them. The problem that I run into is with queries like
$sql = sqlf("SELECT * FROM Users WHERE id IN (:1)",array(1,2,3) );
My code works, but it fails with empty arrays, e.g. the following throws a mysql error:
SELECT * FROM Users WHERE id IN ();
Does anyone have any suggestions? How should I translate and empty array into sql that can be injected into an IN clause? Substituting NULL will not work.
Null is the only value that you can guarantee is not in the set. How come it is not an option? Anything else can be seen as part of the potential set, they are all values.
I would say that passing an empty array as argument for an IN() clause is an error. You have control over the syntax of the query when calling this function, so you should also be responsible for the inputs. I suggest checking for emptiness of the argument before calling the function.
Is there a possibility that you could detect empty arrays withing sqlf and change the SQL to not have the IN clause?
Alteratively, you could postprocess the SQL before passing it to the "real" SQL executor so that "IN ()" sections are removed although you'd have to do all sorts of trickery to see what other elements had to be removed so that:
SELECT * FROM Users WHERE id IN ();
SELECT * FROM Users WHERE a = 7 AND id IN ();
SELECT * FROM Users WHERE id IN () OR a = 9;
would become:
SELECT * FROM Users;
SELECT * FROM Users WHERE a = 7;
SELECT * FROM Users WHERE a = 9;
That could get tricky depending on the complexity of your SQL - you'd basically need a full SQL language interpreter.
If your prepare-like function simply replaces :1 with the equivalent argument, you might try having your query contain something like (':1'), so that if :1 is empty, it resolves to (''), which will not cause a parse error (however it may cause undesirable behavior, if that field can have blank values -- although if it's an int, this isn't a problem). It's not a very clean solution, however, and you're better off detecting whether the array is empty and simply using an alternate version of the query that lacks the "IN (:1)" component. (If that's the only logic in the WHERE clause, then presumably you don't want to select everything, so you would simply not execute the query.)
I would use zero, assuming your "id" column is a pseudokey that is assigned numbers automatically.
As far as I know, automatic key generators in most brands of database begin at 1. This is a convention, not a requirement (auto-numbered fields are not defined in standard SQL). But this convention is common enough that you can probably rely on it.
Since zero probably never appears in your "id" column, you can use this value in the IN() predicate when your input array is empty, and it'll never match.
The only way I can think to do it would be to make your sqlf() function scan to see if a particular substitution comes soon after an "IN (" and then if the passed variable is an empty array, put in something which you know for certain won't be in that column: "m,znmzcb~~1", for example. It's a hack, for sure but it would work.
If you wanted to take it even further, could you change your function so that there are different types of substitutions? It looks like your function scans for a colon followed by a number. Why not add another type, like an # followed by a number, which will be smart to empty arrays (this saves you from having to scan and guess if the variable is supposed to be an array).

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