I have an external database that I am trying to access from within a Drupal page, I have successfully queried the database and output data to the page using fetchAssoc(), however this only returns the first row in the database. I would like to return all rows into an array for processing, so I'm attempting to use fetchAllAssoc(), this however results in an exception. The database has the following SQL fields:
id, model, manufacturer, url, date_modified
My test code is as follows:
<?php
db_set_active('product_db');
$query = db_select('product', 'p')->fields('p');
$sqlresults = $query->execute()->fetchAllAssoc('id');
foreach($sqlresults as $sqlresult)
{
printf($sqlresult);
}
db_set_active();
?>
I'm thinking that it is the key field 'id' that I am specifying with fetchAllAssoc() that is the problem, as fetchAssoc() prints values correctly. All documentation I have found seems to say that you pass a database field as the key but I have also passed a numeric value with no success.
Many thanks in advance for any advice, I'm sure I'm just missing something stupid.
I think it should work in this way, but within the foreach you want to print the $sqlresult variable as a string, but it is an object (it causes the error).
printf function needs a string as the first parameter, see:
http://php.net/manual/en/function.printf.php
Use for instance var_dump instead:
var_dump($sqlresult);
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"
);
Having some difficulty pinpointing exactly what is wrong with this block of code. I am expecting it to run through a loop a set number of times and update some rows in the table tbl_games with some values received from the form.
I have tried running the code in phpMyAdmin without variables, which works fine (updates specified row). I assume the problem is something to do with the string in $insert_q.
gamecount will always be an int<30, game_ID will be a unique primary key integer value in tbl_games.
A little background: this code is part of a bigger project - which is centered around football games. An admin adds games to tbl_games (coded and finished), this current file now displays games to the admin which are unplayed (scores for team1 and team2 are NULL) and gives them a space to input scores for each team. This code takes those 2 scores, and the game_ID and updates each row.
It's having no effect on the DB rows though. Please point me in the right direction.
<?php
$lim=$_SESSION['gamecount'];
for ($i=1; $i<$lim; $i++) {
$game_ID = ${"_SESSION['game".$i."_ID']"};
$score_team_1 = ${"_REQUEST['".$i."_team1-score']"};
$score_team_2 = ${"_REQUEST['game".$i."_team2-score']"};
$insert_q = "UPDATE tbl_games SET team1_score = '$score_team_1', team2_score = '$score_team_2' WHERE game_ID = '$game_ID';";
mysql_query($insert_q);
}
session_destroy();
?>
I think the problem is with this line.
$game_ID = ${"POST['game".$i."_ID']"};
It should be something like this.
$game_ID = ${"_POST['game".$i."_ID']"}; or
$game_ID = $_POST['game'.$i.'_ID']; //much cleaner
You need to make use of the mysql reporting. Get it to output any errors, and affected rows. While you may think affected rows will be none, it might not be (always good to check when debugging just so you check everything).
Does your PHP error log have any warnings or other notices that might point to your query being an issue etc?
What is the value you're updating (echo out the var/session) and what is the DB value (look at it in phpmyadmin or mysql command line).
Could be there's nothing to update.
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='".$_SESSION_User."'";`
I have a view that needs updating with a list of id's. So I am storing the values that have been selected to remove from the view in a session variable that then goes into the mySQL query as below. Then when the form is reset the values are also reset out of the array.
But its not working... this is what I've got.
Any help would be appreciated.
if($_POST['flag']=='flag'){
//collect deleted rows
$_SESSION['delete-row'][] = $_POST['idval'];
//Split session array
$idavls = join(',' , $_session['delete-row'];
$sqlDelete = "CREATE OR REPLACE VIEW filtetbl AS SELECT * FROM `".$page['db-name']."`.`leads_tbl` WHERE ".$_SESSION['filter-view']." AND `lead_status` = '1' AND `lead_id` NOT IN (".$idvals.") ORDER BY `lead_added`";
$result = mysql_query($sqlDelete);
if($result){
echo true;
}
else{
echo mysql_error();
}
}
$_session isnt the same as $_SESSION for a start.
Also dont use mysql_query or similar (because it isnt safe) use PDO
This is hard to correct without more information (and there are several errors - probaby cut and paste) so I'll pull apart one by one and you can go from there.
1 - $_SESSION['delete-row'][] = $_POST['idval'];
If 'idval' comes from multiple inputs (i.e. ) then it is already an array, and you should have $_SESSION['delete-row'] = $_POST['idval']; If you are looping in an array of inputs (i.e. trying to append for many posts from then it is correct)
2 - $idavls = join(',' , $_session['delete-row'];
$_SESSION (you said this was a type) and you also need a bracket/bract ar the end
$sqlDelete = "CREATE OR REPLACE VIEW filtetbl AS SELECT * FROM ".$page['db-name'].".leads_tbl WHERE ".$_SESSION['filter-view']." AND lead_status = '1' AND lead_id NOT IN (".$idvals.") ORDER BY lead_added";
Firsly this is very insecure as pointed out by allen213. Even if you don't use PDO to make safe the variable, please cast all the inputs as (int) assuming the IDs are integers, or at least wrap the input in mysql_real_escape_string().
Secondly, the logic in the question doesn't quite make sense. You say you want to remove IDs from the view, but what you are doing is recreating the view with only those IDs in $_SESSION['delete-row'] removed - so this may re-introduce IDs previously removed from the view. You'd actually need to keep $_SESSION['delete-row'] and keep adding to it to ensure the next time the view was created, then all the IDs are removed.
I hope that helps. If not, more code may be required (i.e. the form you are using the send data, anythign else that affects sessions etc.