I have wordpress blog and I have massive comments under some posts. I found recently that some comments are duplicated two or three times. I want to use php code that removes duplicate elements from table wp_comments from mysql database of Wordpress (by opening php file from server). I used this code, but it doesn't work. Any ideas why? When I run this file I have blank page (no errors, but duplicated comments still there).
Code:
<?php
include_once($_SERVER['DOCUMENT_ROOT'].'/wp-load.php' );
global $wpdb;
$comments = $wpdb->get_results("SELECT * FROM ".$wpdb->prefix."_comments"
." ORDER BY comment_post_ID, comment_content");
$prev = NULL;
foreach($comments as $comment) {
if ($prev && $prev->comment_content == $comment->comment_content
&& $prev->comment_post_ID == $comment->comment_post_ID ) {
echo 'It works';
$wpdb->query("DELETE FROM ".$wpdb->prefix."_comments WHERE comment_ID == ".$comment->comment_ID);
}
else
$prev = $comment;
}
$wpdb->print_error();
?>
Maybe you have better idea for this code?
I just spent time formatting your code so I could read it. There is no { after your else. I believe it is required if you include it in the first part of the if statement. In any case, omitting them in one application but using them in another in the same code block no less, is a terrible practice. Makes your code hard to read/maintain.
Additionally, you may not see an error if errors are suppressed.
Maybe I need to brush up on my PHP but will if ($prev) ever be true? It's not a boolean ever is it? It's not like a while when fetching a query. Maybe it is.
Put the following on top of your script, after the opening php tag
error_reporting(E_ALL);
ini_set( 'display_errors','1');
Run it and tell us everything you see.
Related
I'm setting $_SESSION['showroom'] to 'active' when a particular page in Wordpress is displayed:
if(get_the_ID()==6470||get_the_ID()==252){
$_SESSION['showroom']='active';
}
I then set 2 arrays of pages to check against. If the next page displayed is NOT in one of these arrays, $_SESSION['showroom'] gets changed to 'inactive'.
$allowed_templates = array('template-A.php',
'template-B.php',
'template-C.php',
'template-E.php',
'template-G.php');
$allowed_ids = array(6470,252);
$template_name = get_page_template_slug();
$page_id = get_the_ID();
if(in_array($template_name,$allowed_templates)==false && in_array($page_id,$allowed_ids)==false){
$_SESSION['showroom']='inactive';
}
The if statement works most of the time, but sometimes my $_SESSION['showroom'] changes to inactive EVEN though one of the arrays is returning true! After several hours of testing I am unable to locate where the problem is. Echoing out the two parts of the if statement ALWAYS gives me 2 trues or 1 true + 1 false, but never 2 falses:
if(in_array($template_name,$allowed_templates)==false){echo 'TFALSE';}
if(in_array($template_name,$allowed_templates)){echo 'TTRUE';}
if(in_array($page_id,$allowed_ids)==false){echo 'IFALSE';}
if(in_array($page_id,$allowed_ids)){echo 'ITRUE';}
What am I missing here?
Thanks in advance for any help!
EDIT: Have continued testing and found the following anomaly:
if(in_array($template_name,$allowed_templates)==false && in_array($page_id,$allowed_ids)==false){
$_SESSION['showroom']='inactive';
echo 'SET TO INACTIVE';
}
The if statement changes $_SESSION['showroom'] to 'inactive' but DOES NOT echo out 'SET TO INACTIVE'!
There's something strange going on here!
Problem solved. My code was fine. Two missing images files were causing WordPress to crash my sessions. Took 10 hours to find out but happy I found it. Thanks to everyone for their help.
You can try the following;
if(!in_array($template_name,$allowed_templates) && !in_array($page_id,$allowed_ids)){
$_SESSION['showroom']='inactive';
}
Edit: lets try and break it down further... similar to your examples
if(!in_array($template_name,$allowed_templates){
echo "not in templates,";
}
if(!in_array($page_id,$allowed_ids)){
echo "not in ids,";
}
if(!in_array($template_name,$allowed_templates) && !in_array($page_id,$allowed_ids)){
echo "not in both\n";
}
then see if we get a result with not in templates,not in ids, but no trailing not in both
The problem is pure logical. Lets look at this statement:
if (in_array($template_name,$allowed_templates)==false && in_array($page_id,$allowed_ids)==false)
Which translates to "If the template is not valid AND page is not valid"
This means that both statements needs to be fulfilled in order to mark session as inactive. What if the template is fine, but the page is not valid? That definitely should be marked as inactive as well.
By changing the statement to read "If the template is not valid OR page is not valid", we cover up the invalid cases. Because either of them counts as an invalid state, and thus, only one of them needs to be false in order for everything to be false. (the OR-statement)
So code-wise it would be
if (in_array($template_name,$allowed_templates)==false || in_array($page_id,$allowed_ids)==false)
And you are set.
As and addition. I would structure the code as you noted works. Which is more logical. That is, mark it as inactive whenever it's should be treated as inactive, in all other cases mark it as 'active'. Or vice-versa.
im trying to figure out how i can show the last 3-5 or so pages within my site a person has visited. I did some searching, and I couldn't find a WP plugin that does so, if anyone knows of one, please point me in that direction :) if not, I'll have to write it from scratch, and thats where i'll need the help.
I've been trying to understand the DB and how it works. I'm assuming that this is where the magic will happen, with PHP, unless there is a javascript option using cookies to do it.
Im open to all ideas :P & Thank you
If i were to code such a plugin, i'd use the session cookies to populate an array via array_unshift() and array_pop(). it'd be as simple as :
$server_url = "http://mydomain.com";
$current_url = $server_url.$_SERVER['PHP_SELF'];
$history_max_url = 5; // change to the number of urls in the history array
//Assign _SESSION array to variable, create one if empty ::: Thanks to Sold Out Activist for the explanation!
$history = (array) $_SESSION['history'];
//Add current url as the latest visit
array_unshift($history, $current_url);
//If history array is full, remove oldest entry
if (count($history) > $history_max_url) {
array_pop($history);
}
//update session variable
$_SESSION['history']=$history;
Now i've coded this on the fly. There might be syntax errors or typos. If such a mistake appears, just put a notice and i'll modify it. The purpose of this answer is mostly to make a proof of concept. You can adapt this to your liking. Please note that i assume that session_start() is already in your code.
Hope it helps.
===============
Hey! Sorry about the late answer, i was out of town for a couple of days! :)
This addon is to answer your request for a print out solution with LI tags
Here's what i'd do :
print "<ol>";
foreach($_SESSION['history'] as $line) {
print "<li>".$line.</li>";
}
print "</ol>";
Simple as that. you should read the foreach loop here : http://www.php.net/manual/en/control-structures.foreach.php
As for the session_start();, put it before you use any $_SESSION variables.
Hope it helped! :)
I'm going to update and translate the code above for WordPress 5+ because the original question has the wordpress tag. Note that you don't need session_start() anywhere.
Here it goes, add the code below to your singular.php template (or single.php + page.php templates, depending on what you need):
/**
* Store last visited ID (WordPress ID)
*/
function so7035465_store_last_id() {
global $post;
$postId = $post->ID; // or get the post ID from your template
$historyMaxUrl = 3; // number of URLs in the history array
$history = (array) $_SESSION['history'];
array_unshift($history, $postId);
if (count($history) > $historyMaxUrl) {
array_pop($history);
}
$_SESSION['history'] = $history;
}
// Display latest viewed posts (or pages) wherever you want
echo '<ul>';
foreach ($_SESSION['history'] as $lastViewedId) {
echo '<li>' . get_permalink($lastViewedId) . '</li>';
}
echo '</ul>';
You can also store latest viewed custom post types (CPT) by placing the so7035465_store_last_id() function in your single-cpt.php template.
You can also add it to a hook or inject it in your template as an action, but that is beyond the scope of this question.
I apologize if this is a newbie question, but I cannot figure out why this doesn't work - and I can't seem to find anything about it when searching.
Basically, I am trying to scrape some userdetails from our site, that are not available from the sites REST api, so I have to do it manually. I have compiled a textfile with userids, that I use for fetching the wanted details from each user through Simple HTML Dom.
<?php
include('simple_html_dom.php') ;
include('functions.php') ;
$file = fopen("userids2.txt", "r") ;
while(!feof($file)) {
$userid = fgetss($file) ;
$url = 'http://<our URL>/user/'.$userid ;
echo $url ;
webscraper($url) ;
}
fclose($file) ;
?>
and here are the contents of functions.php:
<?php
function webscraper($loopurl) {
$html = new simple_html_dom();
$html->load_file($loopurl);
$test = $html->getElementsById('ctl00_ContentPlaceHolderDefault_UserViewUC_tabContainer_tabProfile_userProfile_ddWork') ;
foreach ($test as $element) {
echo $element ;
}
}
?>
The specific textfile used contains 4 userids that I know contain the information that I want. When I run the script it will only give me the output from the url from the last line in the textfile. It prints out the URLs fine, but refuses to load the remote html for the first three entries. If I delete the last line of the textfile, it then loads the new last line (which it refused to do before).
Any ideas?? Thanks in advance.
Doh.. I found out what the problem was. There was an "invisible" end of line character on all entries in the textfile EXCEPT the last one. So that was why it refused to work. Adding trim when retrieving the line fixed the problem:
$userid = trim(fgetss($file));
I probably should have known this, but at least I won't make this mistake next time :-).
I'm still new to php and working my way around it but i'm stuck at the following piece:
code for deleting a row in my table
i have a link directing towards this piece of my script. i run through the first half just fine but when i press on submit and try to execute my delete query it won't go to my second if statement let alone get to the delete query.
$pgd is the page id
my hunch is there is problem with the action in the form i'm building after my while statement
forgive me for the wierd formatting of my msg but its 2am and very tired, i promise to format my questions in the future better! any help is appreciated
edit: ok other then the obvious mistake of missing method=post #.#;
edit:
hey everyone,
first of all, i'd like to thank everyone for their response.
i just started coding in php last weekend so forgive my messy codes. the code is still running locally and my main goal was to finish the functions and then work on securing my code.
now back to the issue, i'm sorry if i was vague about my problem. i'll try to reiterate it.
my issue isn´t selecting an item i want to delete, the issue is that it won´t get to the 2nd if statement.
Re-edit:
this time with my current code:
if($_GET['delete'] == "y")
{
//content hier verwijderen
$sqlcont1="SELECT * FROM content where id ='".$_GET['id']."'";
echo $sqlcont1;
$resultcont1 = mysql_query($sqlcont1) or die (include 'oops.php');
while($rowcont1= mysql_fetch_array($resultcont1)){
echo '<form class="niceforms" action="?pg='.$pgd.'&delete=y&remove=y&id='.$_GET['id'].'" method="post">';
echo '<h1>'.$rowcont1['Titel'].'</h1>';
echo '<p>'.$rowcont1['Content'].'</p>';
echo '<input type="submit" value="Delete article">';
echo '</form>';
}
if($_GET['remove']=="y"){
echo 'rararara';
$id=$_GET['id'];
$sqlrem="DELETE FROM content WHERE id="$id;
echo $sqlrem;
mysql_query($sqlrem);
}
}
echoing $sqlrem gives me the following now:
DELETE FROM content WHERE id=8
that being my current code, i get in to the second IF statement but now to get it to delete!
#everyone:
ok maybe thinking out loud or following my steps worked but the code works, i know its very messy and it needs fine tuning. i'd like to thank everyone for their help and feedback. i'm liking this and you'll probably see me alot more often with nubby questions and messy codes with no escapes :(
First of all, you have SQL injection vulnerability in your script. Anyone can add some string that will be attached to your query, possibly altering it in a way that can make almost anything with the data from your database.
Escape your values with one of anti-SQL-injection methods. Read more for example on php.net/manual/en/function.mysql-query.php
To the point...
Your deletion code will be executed only if you invoke URL with two params (remove and delete set to y. That means your URL should look similar to something.php?delete=y&remove=y. Maybe you just did not spot it.
Please give details about any errors that occured and tell me whether the above mentioned solution helped.
mysql_fetch_array() returns an array
your while statement acts as an if, and does not iterate thru the array returned as you think it does
you need something like
$all_rows = mysql_fetch_array($result);
foreach ($all_rows as $row) {
$sql = "delete from table where id = " . $row['id'];
}
It looks to me like you're mixing two forms together here: you're wanting to see if you went to the delete row form (the first few lines), and you're trying to present the delete row form (the while loop.) I would break these two things apart. Have a page that simply displays your forms for row deletes, and another page that processes those requests. And another page that brings you to the delete rows page.
For now, just echo all the values you're expecting to receive in $_GET[] and see if they are what you expect them to be.
You have a lot of problems in that script alone, so just to make things easier (considering you uploaded a pic), put an
echo $sqlrem;
in your second if statement, see if the query is displayed. If not, it means it doesn't even get to that part of code, if it gets displayed, copy it and run it in phpmyadmin. That should output a more coherent error message. Tell us what that is and we'll work it through.
I also noticed that your DELETE SQL query might have an issue. If your $pgd' id is a integer, you shouldn't include the ' single quote, that is for string only.
**Correction**
$sqlrem = "DELETE FROM content WHERE id = " . controw1['id'];
EDIT
Anyway, just to help out everyone, I typed out his code for easier viewing.
I think his error is $rowcont1['Tilel'] --> that might caused PHP to have an error because that column doesn't exist. I assumed, it should be `Title' causing an typo error.
if(_$GET['delete'] == "y") {
$sqlcont1 = "SELECT * FROM content where id ='" . $_GET['id'] . "'";
$resultcont1 = mysql_query($sqlcont1) or die (include 'oops.php');
while ($rowcont1 = mysql_fetch_array($resultcont1)) {
echo '<form class = "niceforms" action = "?pg=' .$pgd . '&delete=y&remove=y">';
echo '<h1>' . $rowcont1['Title'] . '<h1>'; // <-- error here
echo '<p>' . $rowcont1['Content'] . '</p>';
echo '<input type = "submit" value = "Delete article">';
echo '</form>';
}
if ($_GET['remove'] == "y"){
$sqlrem = "DELETE FROM content WHERE id = " . $rowcont1['id'];
mysql_query ($sqlrem);
}
}
I'm probably going to make myself look like a fool with my horrible scripting but here we go.
I have a form that I am collecting a bunch of checkbox info from using a binary method. ON/SET=1 !ISSET=0
Anyway, all seems to be going as planned except for the query bit. When I run the script, it runs through and throws no errors, but it's not doing what I think I am telling it tom which is updating the specified fields within the DB.
I've hard coded the desired values into the query and it DOES update the DB. Relying on the variables I believe I've established and am then calling upon in the query does NOT update the DB.
I've also tried echoing all the needed variables after the script runs and exiting right after so I can audit them... and they're all there. Here's an example.
####FEATURES RECORD UPDATE
### HERE I DECIDE TO RUN THE SCRIPT BASED ON WHETHER AN IMAGE BUTTON WAS USED
if (isset($_POST["button_x"])) {
### HERE I AM ASSIGNING 1 OR 0 TO A VAR BASED ON WHTER THE CHECKBOX WAS SET
if (isset($_POST["pool"])) $pool=1;
if (!isset($_POST["pool"])) $pool=0;
if (isset($_POST["darts"])) $darts=1;
if (!isset($_POST["darts"])) $darts=0;
if (isset($_POST["karaoke"])) $karaoke=1;
if (!isset($_POST["karaoke"])) $karaoke=0;
if (isset($_POST["trivia"])) $trivia=1;
if (!isset($_POST["trivia"])) $trivia=0;
if (isset($_POST["wii"])) $wii=1;
if (!isset($_POST["wii"])) $wii=0;
if (isset($_POST["guitarhero"])) $guitarhero=1;
if (!isset($_POST["guitarhero"])) $guitarhero=0;
if (isset($_POST["megatouch"])) $megatouch=1;
if (!isset($_POST["megatouch"])) $megatouch=0;
if (isset($_POST["arcade"])) $arcade=1;
if (!isset($_POST["arcade"])) $arcade=0;
if (isset($_POST["jukebox"])) $jukebox=1;
if (!isset($_POST["jukebox"])) $jukebox=0;
if (isset($_POST["dancefloor"])) $dancefloor=1;
if (!isset($_POST["dancefloor"])) $dancefloor=0;
### I'VE DONE LOADS OF PERMUTATIONS HERE... HARD SET THE 1/0 VARS AND LEFT THE $estab_id TO BE PICKED UP. SET THE $estab_id AND LEFT THE COLUMN DATA TO BE PICKED UP. ALL NO GOOD. IT _DOES_ WORK IF I HARD SET ALL VARS THOUGH
mysql_query("UPDATE thedatabase SET pool_table='$pool', darts='$darts', karoke='$karaoke', trivia='$trivia', wii='$wii', megatouch='$megatouch', guitar_hero='$guitarhero', arcade_games='$arcade', dancefloor='$dancefloor' WHERE establishment_id='22'");
###WEIRD THING HERE IS IF I ECHO THE VARS AT THIS POINT AND THEN EXIT(); they all show up as intended.
header("location:theadminfilething.php");
exit();
THANKS ALL!!!
I recommend you to use something like:
$fields = array('pool', 'darts', 'karaoke', 'trivia', ...);
foreach ( $fields as $field ) {
$$field = isset($_POST[$field]) ? 1 : 0;
}
instead of 20 lines of ifs.
Your columns are ENUM or int type ? If int - drop apostrophes.
Your code could really use some error checking. Make sure you have activated the displaying of errors in your script.
In your testing environment add this at the top of your main script for instance (if you haven't done something equivalent already):
error_reporting( E_ALL | E_STRICT );
ini_set( 'display_errors', 1 );
Then (although not dependant on the above) make sure you probe the result of the query with something like:
if( false === mysql_query( 'UPDATE ...etc' ) )
{
echo 'query failed with error:' . mysql_error();
}
My guess is it will fail with the error that your column name karaoke is mispelled. But there may be more errors.
Also, hsz' suggestions are spot on (though probably not the root of your problem). Makes for easier to maintain code, and significantly reduces code.
Firstly, construct the sql query string in a variable and then pass it to mysql_query(), comment out the header() line and print out the query for debugging. For example:
...
$sql="UPDATE thedatabase SET pool_table='$pool', darts='$darts', karoke='$karaoke', trivia='$trivia', wii='$wii', megatouch='$megatouch', guitar_hero='$guitarhero', arcade_games='$arcade', dancefloor='$dancefloor' WHERE establishment_id='22'";
print("$sql");
mysql_query($sql);
//header("location:theadminfilething.php");
exit();
...
Secondly, even tho you are exiting the script, its good practice to always match your braces. You are missing the end brace for the if statement at the end of your code.
The value of the $sql variable output you can see if it works by executing it 'manually' thru phpmyadmin or the command line. What happens?