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keloa

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Posts posted by keloa

  1. Good luck to you "hiring" people who can't work either full OR part time.

     

    That tool you showed seems to be a technical interviewing tool.  It is not a team tool.

     

    I'd suggest you look at github.

    hahahahahaha

     

    you didn't get me.When you are hiring someone online with full time job it's much cheaper than hiring someone who want to get an office ...etc

     

    about the tool I just got a quick look and yes as you said , sadly its an interview tool.still searching for a good tall to work online

  2. Hi

    I have a small company so I cant hire programmers with full time or part time. And I got this idea to start a team online, so is there any software that will make the team work easier.

    I found this software :

    http://codassium.com/

    I didn't get time to take a good look into it but I think it is what I need.

     

    so if there is any software can make this work more easy I hope you tell me about it.espically if it's a free software.

  3. Hi 

    I'm working in a project and I just finished the pagination system (in every page 2 results) but my problem is that I want to add some options beside every result.For example I have to resut which are (john,katy)I want them to be like this :

    John (Edit) (Delete)

    Katy (Edit) (Delete)

     

    This is my code :

    <?php
    include("wasata/inc/config.php");
    mysql_query('SET NAMES "UTF-8"');
    if(!isset($_GET['page'])){
    	$page = 1;
    }else{
    	$page = (int) $_GET['page'];
    	$records_at_page = 2;
    	$q = mysql_query('SELECT * FROM serv');
    	$records_count = @mysql_num_rows($q);
    	@mysql_free_result($q);
    	$page_count = (int) ceil($records_count / $records_at_page);
    	if(($page > $page_count) || ($page <= 0)){
    		mysql_close($con);
    		die('No More Pages');
    	}
    	$start = ($page - 1) * $records_at_page;
    	$end   = $records_at_page;
    
    	if($records_at_page != 0){
    		$q = mysql_query("SELECT * FROM serv ORDER BY id ASC LIMIT $start,$end");
    		while($o = mysql_fetch_object($q)){
    			echo $o->title .'<br />';
    		}
    	}
    	echo '<br />';
    	for($i = 1; $i<= $page_count; $i++){
    		if($page == $i){
    			echo $page;
    		}else{
    			echo '<a href="m.php?page=' .$i . '">' .$i . '</a>';
    			if($i != $page_count){
    				echo ' - ';
    			}
    		}
    
    	}
    }
    mysql_close($con);
    ?> 

    So can any one Help please.

     

  4. New_order.php

    <?php
    session_start();
    include ("inc/config.php");
    include ("inc/header.php");
    if(!isset($_SESSION['user'])){
    	header("Location: login.php");
    }else{
    
    ?>
    <center>
    <form method="POST" action="check_order.php">
    	عنوان الطلب : <input type="text" name="title" /><br />
    	تفاصيل الطلب :<textarea name="describtion"></textarea><br />
    	رابط الطلب ان وجد : <input type="text" size="70" name="url" /><br />
    	<input type="submit" value="تقديم الطلب" />
    </form>
    </center>
    
    <?php
    include("inc/footer.php");	
    }
    	?>
    

    ^ it works fine 

    the second file is :

    check_order.php

    <?php
    session_start();
    include ("inc/config.php");
    include ("inc/header.php");
    if(!isset($_SESSION['user'])){
    	header("Location: login.php");
    }else{
    	$title = $_POST['title'];
    	$describtion = $_POST['describtion'];
    	$url = $_POST['url'];
    	$username = $_SESSION['user'];
    	$track_number = rand(10000, 99999);
    	if(empty($title)or empty($describtion)){
    		echo "الرجاء ملأ جميع الحقول المطلوبة";
    	}else{
    		
    	
    	$sql="SELECT * FROM serv WHERE(track_number='".$track_number."')";
    	$result = mysql_query($sql);
    	$asc = mysql_fetch_assoc($result);
    	$asc['tracknumber'];
    	while ($track_number == $asc['tracknumber']) {
    		$track_number = rand(10000, 99999);
    	}
    $sql = "INSERT INTO serv(username,title,describtion,track_number,url) VALUES('$username','$title','$describtion','$track_number','$url')";
    	$result = mysql_query($sql);
    	if(isset($result)){
    		echo "done"."<br />"."Your tracking number is $track_number";
    	}else{
    		echo "check your code";
    	}
    	}}
    
    ?>
    

    The result:

    Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\Program Files (x86)\EasyPHP-DevServer-13.1VC9\data\localweb\firstproject\check_order.php on line 20
    done
    Your tracking number is 10175

     

    The problems are :

    1-The error on line 20

    2-the data didn't enter the database

  5. Hi

    I'm working in a small project and I need to make a random number that doesn't repeat.The activation codes are stored in the database (mysql database) but the thing is that I want php to do the following:

    1- create a random number (I used the function (rand) and it did exactly what I needed ).

    2-check if the code is already stored in the database and if it's already there it should create another code (which is not stored in the database).

     

    Thanks

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