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mactron

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Posts posted by mactron

    How to Save Results from Foreach into Txt File

    in PHP Coding Help

    Posted

    I would like to write all files from specific folder into .txt file. The output working just fine, but I'm getting only one file saved into my txt file. Any idea? Thanks! 

    $fileList = glob('C:\users\John\Documents\*pdf');
foreach($fileList as $filename){
if(is_file($filename)){

echo $filename, '<br>'; 

$save_files = fopen("list.txt", "w");
fwrite($save_files,$filename);
fclose($save_files);

}   
}

     

    Multiple returns from a fetch

    in PHP Coding Help

    Posted

    Me again.. I'm fetching data from MySQL but each person in people.php is displayed two times instead one.. The issue must be inside models_category_tbl table, because each person have more then one result. Any advice is appreciated. What should I do?

    I would like to display people something like that: 

    1 John Doe 1 IT developer, 3 mechanic

    models/model = person

    category = occupation/skills

    MySQL: 

    model_index table
+---------+------------+-------------+-------------------+---------------+----------------+
|model_id | model_name | model_title | model_description | model_country | model_slug	  |
+---------+------------+-------------+-------------------+---------------+----------------+
| 1 	  | John Doe   | Johns title | Johns desc        | Hungary       | john-doe 	  |	
+---------+------------+-------------+-------------------+---------------+----------------+

models_category_tbl table
+-------------+---------------------+---------------------+
| category_id | model_category_name | model_category_slug |
+-------------+---------------------+---------------------+
| 1           | it developer        | it-developer        |
| 2           | photographer	    | photographer        |
| 3           | mechanic            | mechanic            |
+-------------+---------------------+---------------------+

models_category_ids table
+----+----------+-------------+
| id | model_id | category_id |
| 1  | 1        | 1           | 
| 1  | 1        | 3           |
+----+----------+-------------+

     

    PHP function to fecth data: 

            public function getAllModels1 () {

        $sql = "SELECT * FROM models_index  JOIN models_category_ids ON models_index.model_id=models_category_ids.model_id JOIN models_category_tbl ON  models_category_tbl.category_id=models_category_ids.category_id" ;
            $stmt = $this->conn->prepare($sql);
            $stmt->execute();
            $result = $stmt->fetchAll((PDO::FETCH_OBJ));
            
            return $result;            
    }

     

    people.php 

    <?php 
declare(strict_types=1);
ob_start();
include "includes/connect.php";
include "includes/functions.php";
 

   	$model = new ModelsData();
	$models = $model->getAllModels1();


	foreach ($models as $model) { ?> 


	<a href="model.php?id=<?= $model->model_id ?>"><img src="<?= $model->model_img_path ?><?=$model->model_img_name ?>"></a>	 
 
	<?php echo $model->model_id . " " .  $model->model_name . " " . $model->category_id .  " " . $model->model_category_name  ; ?>
	

	<h3><?= $model->model_name ?></h3>

 
<?php }  ?>

     

    Output: 

     1 John Doe 1 IT developer 
 1 John Doe 3 mechanic

     

    Insert Array in MySQL

    in PHP Coding Help

    Posted

    From insert query for posts.. 

    $id = $this->conn->lastInsertId();

    I would like something like this.. 

    +----+---------+-------------+
| id | post_id | category_id |
+----+---------+-------------+
| 1  | 2       | 3           |	
| 2  | 2       | 5           |	
| 3  | 3       | 1           |
| 4  | 3       | 2           |
+----+---------+-------------+

    The hole problem is inside  my array. 

     

     

    Insert Array in MySQL

    in PHP Coding Help

    Posted

    I'm trying to insert array of posts categories into MySQL. I'm getting category IDs from drop down list..

    ERR: Notice: Notice: Array to string conversion in C:\laragon\*********\includes\functions.php on line 477

     

    Array 

    Array
(
    [0] => Array
        (
            [0] => 4
            [1] => 5
        )

)

     

    PHP 

    try {
        $sql = "INSERT INTO posts_category_ids (post_id, category_id) VALUES";

        $insertQuery = array();
        $insertData = array();
        foreach ($filter_category as $row) {
            $insertQuery[] = '(?, ?)';
            $insertData[] = $id;
            $insertData[] = $row;
        }

        if (!empty($insertQuery)) {
            $stmt = $this->conn->prepare($sql);
            $stmt->execute($insertData); //477
        }


         }catch (Exception $e){

            echo $e->getMessage();
}
}

     

    Categorizing blog posts

    in MySQL Help

    Posted · Edited by mactron

    I'm working on a simple CMS system. So my question is how to add categories of blogs inside database correctly. If I understand EAV I need to add two tables; one table for names, descriptions of categories and one table for categories IDs and posts IDs.

    I'm correct?

    Cheers

    blog_tbl 

    +------+---------+--------+---------------+-------------------+---------+
| b_id | b_title | b_text | b_seo_title   | b_seo_description | slug    |
+------+---------+--------+---------------+-------------------+---------+
|  1   | Title 1 | txt 1  |  70 charsets  | 170 charsets      | title-1 |
|  2   | Title 2 | txt 2  |  70 charsets  | 170 charsets      | title-2 |
+------+---------+--------+---------------+-------------------+---------+

     

    MySQL architecture

    in PHP Coding Help

    Posted

    Yeah I'm using InnoDB ..

    What about if someday I decided to add likes, dislikes and views for each person. That's mean 3 additional columns inside people_tbl and 13 columns in total. 

    I'm worried about JOINS because I already use three JOINS in one query. I have heard people complain that JOINS may be pretty slow.

    Thanks!

    MySQL architecture

    in PHP Coding Help

    Posted · Edited by mactron

    Done with JOINs and work like a charm. Thanks!

    Just another question ..  I would like to add photo of each person, so I will have two additional columns inside my persons_tbl (person_avatar_path, person_avatar_name). Is it better to create new table for avatars?

    MySQL architecture

    in PHP Coding Help

    Posted · Edited by mactron

    I have three tables, people_tbl, people_category and people_country, but let's get focused on 1st and 2nd table only.

    people_tbl = people and their info (please check MySql architecture above)

    people_category_tbl = occupation

     Witch my code below, I fetch the data between both tables. All the data are correctly fetched and completely relevant. Actually everything works just fine, but I'm curious If I'm on the right path. is there anything that I can fix, improve.. Should I use only one query with JOIN syntax? Thanks! 

    func.php

<?php
class PeopleData extends dbh{

    public function getPeople () {

        $sql = "SELECT * FROM people_tbl";
        $stmt = $this->conn->prepare($sql);
        $stmt->execute();
        $result = $stmt->fetchAll((PDO::FETCH_OBJ));
            
        return $result;
    }

public function getPeopleCategoryName() {

        $sql = "SELECT * FROM people_category_tbl";
        $stmt = $this->conn->prepare($sql);
        $stmt->execute();
        $category = $stmt->fetch(PDO::FETCH_OBJ);
            
        if($category == null) {

            return null;
        }else {
            return $category->people_category_name;
        }
    } 
}



fetch.php
 <?php 


   	$people = new PeopleData();
	$peoples = $people->getPeople();

    $category = new PeopleData();

?>

<?php foreach ($peoples as $people) { ?>

<?php 

	echo $people->people_id . " "; 
    echo $people->people_name . " "; 

    echo $category->getPeopleCategoryName($people->people_category) . " "; 
     

?>

     

    Can't get pop up to work in PHP/JS

    in PHP Coding Help

    Posted · Edited by mactron

    Try 

    $msg = "<script type="text/javascript">
					$(document).ready(function(){

					demo.initChartist();

					$.notify({
					icon: 'pe-7s-gift',
					message: 'TESTING!'				
            },{
                type: 'info',
                timer: 4000
            });

    	});
JS;

print '<script>'.msg.'</script>';

    P.S. Your code is not safe. Use PDO instead MySQLi.

    MySQL architecture

    in PHP Coding Help

    Posted · Edited by mactron

    Hello, I have a table with people and some details insde. So, my question is if my MySQL architecture down below is good or not. Should I devide the table into two tables and kept only p_id, p_name and p_slug inside my table (people_tbl). Thank you!

    p = person
    p_description = bio
    p_category = occupation (people_category_tbl)

    people_tbl
      

    +------+---------+---------+---------------+-------------+-------------------+-----------+------------+---------+
| p_id | p_name  | p_title | p_description | p_seo_title | p_seo_description | p_country | p_category | p_slug  |
+------+---------+---------+---------------+-------------+-------------------+-----------+------------+---------+
|  1   | John Li | title   |  johns bio    | 70 charsets | 170 charsets      | Germany   | 1          | john-li |
|  2   | Bob Stu | title   |  bobs bio     | 70 charsets | 170 charsets      | Italy     | 2          | bob-stu |
+------+---------+---------+---------------+-------------+-------------------+-----------+------------+---------+

     

    Check if person_id already exist in the table

    in PHP Coding Help

    Posted

    4 minutes ago, Barand said:

    But for future reference, as benanamen said, don't waste time checking first. 

    1. Ensure that person_id is either the primary key or is defined as a UNIQUE key. Then it is impossible to insert a duplicate and will throw an error/exception.
    2. Insert the new record and check for a duplicate key error. If there is one, output error message or "found", otherwise it worked and job done.

    Example 

    
CREATE TABLE `user_demo` (
  `userid` int(11) NOT NULL AUTO_INCREMENT,
  `fname` varchar(20) DEFAULT NULL,
  `lname` varchar(20) DEFAULT NULL,
  `username` varchar(20) DEFAULT NULL,
  PRIMARY KEY (`userid`),
  UNIQUE KEY `username` (`username`)
)

DATA:
+--------+-------+----------+----------+
| userid | fname | lname    | username |
+--------+-------+----------+----------+
|      1 | Laura | Norder   | norderl  |
|      2 | Tom   | DiCanari | tomd     |
|      3 | Harry | Potter   | harryp   |
+--------+-------+----------+----------+

     

    Attempting to insert another user with same username gives a 1062 Duplicate key error 

    
mysql> INSERT INTO user_demo (fname, lname, username)
    -> VALUES ('Harry', 'Palmer', 'harryp');
ERROR 1062 (23000): Duplicate entry 'harryp' for key 'username'

    The processing would be 

    
try {
   insert new record
}
catch (exception e) {
   if error number = 1062
       echo "Duplicate!"
   end if
}

     

    I will take a look. Thanks!

    Check if person_id already exist in the table

    in PHP Coding Help

    Posted · Edited by mactron 

    persons_tbl

person_id  |  person_name | person_surname | person_description | person_slug
-----------+--------------+-------------------+--------------------+-------------
    1      |  John		  |	Doe 	 	   | bla bla bla  | john-doe
    1      |  Jane		  |	Doe            | bla bla bla        | jane-doe
    2      |  Greg		  |	Stue           | bla bla bla        | greg-stue
-----------+--------------+----------------+--------------------+-------------

persons_web_tbl

person_web_id  |  person_web_linkedin | person_web_twitter | person_web_facebook | person_web_website | person_id
---------------+----------------------+--------------------+---------------------+--------------------+---------
    1          |  linkedin.com/john   |	twitter.com/john | facebook.com/john    	| www.johndoe.com | 1
    1          |  linkedin.com/jane   |	twitter.com/jane | facebook.com/jane   		| www.janedoe.com | 2
    2          |  linkedin.com/greg   |	twitter.com/greg | facebook.com/greg    	| NULL			  | 3

     

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