[Looking for a PHP developer?]

Hi phpfreaks community,

 

My name is ian and I would like to offer my services as a PHP developer.

 

I have 9+ years of experience in PHP/MySQL, basically the WEB in itself + a strong knowledge in Symfony framework and other related framework.

 

I work with HTML, CSS, JavaScript, jQuery, node.js, angular.js, MySQL, PHP, Symfony, CodeIgniter.

 

Feel free to send me an email for more information: irkevin@hotmail.com

 

Have a good day/evening

 

Regards,

Ian

 

[add details into existing column]

"i have a existing table and let the user add another column in the table"

 

You mean you let the user alter you database tables?

 

i  can create the column but do not know how to insert the user input.

 

Maybe and INSERT

 

INSERT INTO tableName(your_fields) VALUES(your_values)

[Help with infinite category]

Hi,

 

I have a function that display an infinite list of categories.. So far this work fine.. But now I want to display only the second level of every main category and not the subs.. Consider this:

 

Main1                Main2                Main3                Main4

  car1                  Bike1                watch1              Game1

    SubSub1            SubSub1            SubSub1            SubSub1

  car2                  Bike2                  watch2            Game1

    SubSub2            SubSub2              SubSub2            SubSub2

 

What I want to achieve is, display on

 

Car1

Car2

Bike1

Bike2

Watch1

Watch2

Game1

Game2

 

Here is the code I have so far:

 

function get_cat()
{
	db_connect();

	$sql = "SELECT * FROM category";

	$result = mysql_query($sql);

	if(mysql_num_rows($result) > 0){
		while($row = mysql_fetch_array($result)){
			$menu_array[$row['id']] = array(
				'name' => $row['name'],
				'parent' => $row['parent_id'],
				'id' => $row['id']
			);
		}	

		return $menu_array;
	}	
}

function generatemenu($parent,$array){
	$has_childs = false;

	//this prevents printing 'ul' if we don't have subcategories for this category

	//use global array variable instead of a local variable to lower stack memory requierment

	foreach($array as $key => $value) {

		if ($value['parent'] == $parent){       
				$has_childs = true;

				if($has_childs == true){

					if($value['parent'] == 0)
						echo '<ul class="main">';
					else
						echo '<ul class="subs">';
				}
				//if this is the first child print '<ul>'              
				if($value['parent'] == 0)         
					echo '<li><a href="index.php?category='.$value['id'].'">' . $value['name'] . '</a>';
				else
					echo '<li><a href="index.php?category='.$value['id'].'">' . $value['name'] . '</a>';
				generatemenu($key ,$array);
				//call function again to generate nested list for subcategories belonging to this category
				echo '</li>';

			if($has_childs == true)
				echo '</ul>';
		}

	}
}

 

Can someone help?

[adding values to cookie on each click]

I need to set a cookie, and later update it. Meaning, the cookie needs to retain its current value, plus add new values to it from GET variable data.

 

 

Nobody to help?

[adding values to cookie on each click]

Hi

 

here is my problem, i have a link like :

 

index.php?id=1

 

When user click on it, a cookie array is set with a value of 1

 

when user clicks on another link with an id of 2, the same cookie is set with another value of 2, at the end , the cookie should read:

 

 

array(    [0] => 1    [1] => 2)

 

Can someone please explain how to do this? im struggling :S

[How do I create an active link in PHP?]

You're welcome miss 

[How do I create an active link in PHP?]

You can try something like this:

 

<?php
   
   $dbHost = 
   $dbUser = 
   $dbPass = 
   $dbDatabase = 
  

$search = $_POST['search'];

if ($search) // perform search only if a string was entered.
{
   $con = mysql_connect($dbHost, $dbUser, $dbPass) or die("Failed to connect to MySQL Server. Error: " . mysql_error());
   mysql_select_db($dbDatabase) or die("Failed to connect to database {$dbDatabase}. Error: " . mysql_error());
   
   
   $query = "SELECT Appr_Value.Appr_ID as id, Appr_Value.Account, Asmnt_Parcel.OwnersName, Asmnt_Situs.Situs
           FROM Appr_Value
           INNER JOIN Asmnt_Parcel
           ON Appr_Value.Account=Asmnt_Parcel.Account
           INNER JOIN Asmnt_Situs
           ON Appr_Value.Account=Asmnt_Situs.Account
           WHERE Appr_Value.Account LIKE '{$search}'
             ORDER BY Appr_Value.Account ASC";
   $result = mysql_query($query, $con) or die(mysql_error().": $query");

   if ($result)
   {
      echo "Results:<br><br>";
      echo "<table width=90% align=center border=1><tr>
      <td align=center bgcolor=#4A6B3F>Account</td>
      <td align=center bgcolor=#4A6B3F>Owners Name</td>
      <td align=center bgcolor=#4A6B3F>Situs</td>
      </tr>";
   
      while ($r = mysql_fetch_array($result))
      { // Begin while
         $act = $r["Account"]; 
         $name = $r["OwnersName"];
         $situs = $r["Situs"];
         echo "<tr>
            <td><a href='formresults.php?id=".$r['id']."'>$act</td>
            <td>$name</td>
            <td>$situs</td>
            </tr>";
      } // end while
      
      echo "</table>";
   }
   else
   {
      echo "Sorry, please try your search again.";
   }
}
else
{
   echo "Start your search";
}
?>

 

Now when user clicks on the link, it takes you to a page similar to this

 

http://www.yourdomain.com/formresults.php?id=1

 

On formresults.php, make a query like this:

 

 $query = "SELECT Appr_Value.Appr_ID as id, Appr_Value.Account, Asmnt_Parcel.OwnersName, Asmnt_Situs.Situs
           FROM Appr_Value
           INNER JOIN Asmnt_Parcel
           ON Appr_Value.Account=Asmnt_Parcel.Account
           INNER JOIN Asmnt_Situs
           ON Appr_Value.Account=Asmnt_Situs.Account
           WHERE Appr_Value.Account LIKE '{$search}'
           AND Appr_value.Appr_ID = '".mysql_real_escape_string($_GET['id'])."'
             ORDER BY Appr_Value.Account ASC";

[PHP Form and MySQL]

Well, as a start, I don't really understand your query.

 

What's this:

 

build_device_test.$'Auto_id'

 

Please clarify. If you could show how your table are set up, that would be a good place to start!

 

[How do I create an active link in PHP?]

In your select statement, you need to select and ID as well, that's what you need.

 

How is your table set up?

[how can i echo an html link?]

Haha, i find myself several time in this situation xD .. You're welcome

[how can i echo an html link?]

Maybe like this:

 

<?php
echo 'welcome   '. $name.'<br />';
echo  "<a href='women/mygirls/index.html'>logout</a>";
?>

[Paging problems]

Try visiting this link

 

I got a similar problem also. Which was fixed by the code in this topic

 

http://www.phpfreaks.com/forums/index.php/topic,154528.msg668929.html#msg668929

[PHP Form and MySQL]

Try this,

 

<?php
$conn = mysql_connect("machine", "", "") or die("Could not connect : " . mysql_error());
   mysql_select_db("my_db", $conn) or die("Could not select database");
   
   $Auto_id = $_GET['Auto_id']; // this shows in url as 1,2,3 is my auto_inc field in database

   $test = "SELECT * FROM build WHERE build_id = '$Auto_id'"; 
   $result = mysql_query($test);

  while($row = mysql_fetch_array($result)){
       $version = $row['version'];
}

echo $version;
?>

 

YOu need to iterate over the data using the while loop. Try it and let me know

[user image]

Basically you will need to look in $_FILES, is_uploaded_file() and move_uploaded_file(), once uploaded, get the URL of the image, or maybe just the image name, and insert it in the Database.

 

 

[How do I create an active link in PHP?]

Countrygirl is trying to get specific result based on the current result. I believe she should query it by the ID of the current result to fetch all the data! Can you post your code?

[How do I create an active link in PHP?]

Basically, you need to create a link along with the ID of the result.. LIke so

 

<a href="formresults.php?id=<?php echo ID_OF_RESULT;?>">Your number</a>

 

On page searchjoin_more.php, your query will be similar to below

 

<?php
$id = isset($_GET['id']) ? $_GET['id'] : '';

$sql = "SELECT * FROM yourtable WHERE id = '".mysql_real_escape_string($id)."'";
?>

 

The page should be in .php , not html !

[joining tables]

What happends if you try a LEFT JOIN?

 

$sql = "Select ProgrammingFoundations.student_id, ProgrammingFoundations.attendance, ProgrammingFoundationsLab.lab_attendance
from ProgrammingFoundations
LEFT JOIN ProgrammingFoundationsLab
on ProgrammingFoundations.student_id = ProgrammingFoundationsLab.student_id";

$result = mysql_query($sql) or die(mysql_error());
if($row = mysql_fetch_array($result))
{
while($row = mysql_fetch_array($result))
{
echo ' lecture attendance : '. $row['attendance'];
echo ' lab attendance : '. $row['lab_attendance'];
}
}
else
{
echo 'not successful';
}

[Please help Fatal error: Function name must be a string]

Oops sorry, was in a hurry when typing lol.. it's 23:35 here.. guess i'm half asleep.. HAHA

[beginner mysql query with php help]

if i understand, you need this line

 

14  I            test-HS-1          TestDialpl

 

You might want to try

 

$Tab = mysql_query("SELECT * from `Tabs` WHERE PID = '14' AND hardware = ' test-HS-1'");

 

But this is not a good approach. You need a unique ID in your table.. Like

 

+---++---++-------++------------+

| ID | PID| |action|  |Hardware|

+---++---++-------++------------+

|  1 | 14  |

|  2 | 14  |

|  3 | 14  |

|  4 | 14  |

 

And so on

 

Then $Tab = mysql_query("SELECT * from `Tabs` WHERE ID = '2'");

 

[Please help Fatal error: Function name must be a string]

Well maybe he didnt make a connection to the database

 

You might need to add those code

 

$link = mysql_connect('localhost','username','password');

mysql_select_db(yourdbname',$link);

[Please help Fatal error: Function name must be a string]

Just append or die(mysql_error()) with it to get a clear explanation of what's happening.. try code below

 

$result = mysql_query($sql, $link) or die('Error : '.mysql_error());

[Please help Fatal error: Function name must be a string]

oops . instead of

 

$result = $mysql_query($sql, $link);

 

Just put the one below

 

$result = mysql_query($sql, $link);

 

There shouldn't be a dollar sign in front of mysql_query(), because it is a function

[Updating based on array of checked boxes]

Did the code I posted work then?

[Please help Fatal error: Function name must be a string]

$result = $msql_query($sql, $link);

 

Should be

 

$result = $mysql_query($sql, $link);

 

What is on line 36?

[[HELP] Table Help]

Why don't you just put a link?

 

Like

 

<?php 
         ## Display Table and database results
    while($results = mysql_fetch_array($build_device_test)){
        
        echo"<tr>";
        echo    "<td class='side' >".$results['version']."</td>";
        echo    "<td class='CBundle' >" .$results['bundle']."</td>";
        echo    "<td class='CDevice' >".$results['devicename']."</td>";
        echo    "<td class='CJava' ></td>";
        echo    "<td class='CMMS' ></td>";
        echo    "<td class='CBranch' ></td>";
        echo    "<td class='CJC' ></td>";
        echo    "<td class='selection' ><a href='delete.php?id='".$results['id']."' onclick='disp_prompt()'>Delete</a></td>"; 
     /* I am creating   the button for each row that is displayed on the table, and the disp_prompt is a basic login javascript thing I did based on alert box for the sake of seeing how to make it work. */
        echo"</tr>";
        }

echo"</tbody>";
echo"</table>";
?>

 

Then on delete.php , run those codes

 

<?php

//connection info first


$id = $_GET['id'];

$sql = "DELETE FROM mytable WHERE id = '$id'";
$result = mysql_query($sql) or die(mysql_error());

if($result){
echo 'Yay, it has been removed';
}
?>

 

Edit : you might want to show your Javascript disp_prompt() function.. I believe it might take the ID as it's argument !