PHP - Php/mysql Usage = Display Full List, Checkbox Each Record Where User Is 'yes'

Hi,

Firstly, I am a complete newcomer to PHP though I am enjoying my first tentative steps into learning the language!

I have been searching Google for hours now to try and find a simple solution to a problem I just can't seem to figure out, though I thought would be easy to find? I have found a lot of very informative solutions to other questions on this forum so figured it was the best one to join?

My question is: I have built and can interact with a simple database and now populate a page with usernames in a list. Each one is a link which, when clicked sends the username to another page. On this page I use the $_GET function and see the username displayed ok. What I can't figure out is how to then display other information about them from the fields in the database identified with that user?

Any help would be greatly appreciated as this is driving me nuts! Also can anyone suggest a good book for learning PHP? There are lots out there but the reviews on Amazon seem to always talk about coding errors in print etc. so I really don't know which one to buy?

FF

I would appreciate your assistance, there are tons of login scripts and they work just fine. However I need my operators to login and then list their activities for the other operators who are logged in to see and if desired send their clients on the desired activity. I have the login working like a charm and the activities are listed just beautifully. How do I combine the two tables in the MySQL with PHP so the operator Logged in can only make changes to his listing but see the others.

FIRST THE ONE script the member logges in here to the one table in MSQL:
<?php
   session_start();
   
   require_once('config.php');
   
   $errmsg_arr = array();
   
   $errflag = false;
   
   $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
   if(!$link) {
      die('Failed to connect to server: ' . mysql_error());
   }
   
   $db = mysql_select_db(DB_DATABASE);
   if(!$db) {
      die("Unable to select database");
   }
   
   function clean($str) {
      $str = @trim($str);
      if(get_magic_quotes_gpc()) {
         $str = stripslashes($str);
      }
      return mysql_real_escape_string($str);
   }
   
   $login = clean($_POST['login']);
   $password = clean($_POST['password']);
   
   if($login == '') {
      $errmsg_arr[] = 'Login ID missing';
      $errflag = true;
   }
   if($password == '') {
      $errmsg_arr[] = 'Password missing';
      $errflag = true;
   }
   
   if($errflag) {
      $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
      session_write_close();
      header("location: login-form.php");
      exit();
   }
   
   $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'";
   $result=mysql_query($qry);
   
   if($result) {
      if(mysql_num_rows($result) == 1) {
         session_regenerate_id();
         $member = mysql_fetch_assoc($result);
         $_SESSION['SESS_MEMBER_ID'] = $member['member_id'];
         $_SESSION['SESS_FIRST_NAME'] = $member['firstname'];
         $_SESSION['SESS_LAST_NAME'] = $member['lastname'];
         session_write_close();
         header("location: member-index.php");
         exit();
      }else {
         header("location: login-failed.php");
         exit();
      }
   }else {
      die("Query failed");
   }
?>

................................................. ................................
Now I need the person who logged in to the table above to be able to make multiple entries to the table below

<?
$ID=$_POST['ID'];
$title=$_POST['title'];
$cost=$_POST['cost'];
$activity=$_POST['activity'];
$ayear=$_POST['aday'];
$aday=$_POST['ayear'];
$seats=$_POST['special'];
$special=$_POST['seats'];
mysql_connect("xxxxxx", "xxx350234427", "========") or die(mysql_error());
mysql_select_db("xxxx") or die(mysql_error());
mysql_query("INSERT INTO `activity` VALUES ('ID','$title', '$cost','$activity', '$aday', '$ayear', '$special', '$seats')");
Print "Your information has been successfully added to the database!"
?>
Click <a href="member-profile.php">HERE</a> to return to the main menu
  <?php      
?>

the difference between memory usage and bandwidth usage?

I did a test with 500 rows in table, got 2.36mb memory usage, and then once without 500 rows in table, i got 2.34 (IN MBS) memory usage, can that translate into bandwidth usage? somone told me that bandwidth usage != memory usage, how so? it does effect it some doesn't it ?

hi,

i have made a website where people resgister their details of them and products.

they have to enter the following details in form

Name of company

name of the product

company address

email id 

password 

mobile number contact

and 

brief details about their company 

after filling all details user is registered on website.

user can then login with email id and pwd.

now after login ..user will get a page where he can upload the photos of products images and their price, so now my question is that when he finishes uploading (|by clicking on upload button) the product images and price text box ..then on final uploaded webspage it should show all other things which he registerd before (company name , mobile number etc) along with images and price...hence the main question that user does not need to enter mobile and address while uploading images and filling proce ..but on the final page it should show mobile and address along with price and images..as user is not going to enter mobile and address again and again as he will have multiple products to upload.

thanks in advance .

1
Total number of likes work, but I'm guessing it can be done differently, any input would be appreciated.
 
/controllers/indexController.php

$posts = Post::find('all', [
      'limit' => 4,
      'include' => ['likes']
    ]);
    
    foreach ($posts as $post) {
      $post->assign_attribute('likeCount', count($post->likes));
    }
}

 
/models/Post.php

class Post extends ActiveRecord\Model {
  
  static $has_many = [
    ['comments', 'class_name' => 'PostComments'],
    ['likes', 'class_name' => 'UserLikePosts']
  ];
  
}

 
/models/UserLikePosts.php

class UserLikePosts extends ActiveRecord\Model {
    
  static $belongs_to = [
    ['user'],
    ['post']
  ];

}

2
The big question is, how can I get a true/false for "if current user has liked post"? I was thinking something like this, but using vars won't work here, and I would have to use a ternary or a function call to do it like this as the session var might not be set (guest).
 

class Post extends ActiveRecord\Model {

  static $has_many = [
    ['comments'],
    ['likes', 'class_name' => 'UserLikePosts'],
    ['user_likes', 'class_name' => 'UserLikePosts', 'conditions' => ['user_id = ?', [$_SESSION['user']['id']]]
  ];

}

Edited by JimL, 21 June 2014 - 03:36 PM.

Hi everyone...

I would like to implement a questionnaire/survey system that has only two Answers (Yes /No).

Basically, this questionnaire system will be widely used on Mobile Phones.

It will be looking something like this below: http://awardwinningfjords.com/2009/06/16/iphone-style-checkboxes.html

Lets say a user selects a response as No (Using the above slider button for a question). That response should be recorded in the database with User Details + Time Stamp.

Also, I would like to generate a report for each question (Responses of Multiple Users) and a report at a User Level for the entire questionnaire/survey (s).

Can some one guide me the fastest & easiest way to achieve this? I'm a new learner of PHP.

Regards
Sandeep

Yesterday, I created a topic about how I could update records and I managed to achieve that successfully.

Now I have another dilemma.

When I have a specific record I want to update, I want to change a category ID of an product (e.g. change it from 1 to 2) but how do I go about doing this?

Here is my code thus far:

Code: [Select]
<?php

require_once ('./includes/config.inc.php');

require_once (MYSQL);

$id=$_GET['prodID'];

$results = mysqli_query($dbc, "SELECT * FROM product WHERE prodID=".$_GET['prodID']."");
$row = mysqli_fetch_assoc($results);

?>
<form action="" method='POST'>
Product ID: <input type="text" value="<?php echo $row['prodID'];?>" name="prodID" /> <br />
Product: <input type="text" value="<?php echo $row['product'] ;?>" name="product" /> <br />
Product Description: <input type="text" value="<?php echo $row['prod_descr'] ;?>" name="prod_descr" /> <br />
Category:
<select name="category">
<option value="<?php echo $row['catID'];?>"></option>
</select>
Price: <input type="text" value="<?php echo $row['price'] ;?>" name="price" /> <br />
In Stock: <input type="text" value="<?php echo $row['stock'] ;?>" name="stock" /> <br />
<br /><input type="submit" value="save" name="save">
</form>

<?php

if(isset($_POST['save']))
 {
    $id = $_POST['prodID'];
    $product = $_POST['product'];
$descr = $_POST['prod_descr'];
$price = $_POST['price'];
$stock = $_POST['stock'];

// Update data
    $update = mysqli_query($dbc, "UPDATE product SET product='$product', prod_descr='$descr', price='$price', stock='$stock' WHERE prodID=".$_GET['prodID']."");
header( 'Location: update_save.php' ) ;
}
?>

Hey guys

I'm not amazing at PHP, but know just enough to be dangerous. 

Basically I'm trying to get these records from my database to display in rows of 4.  For some reason every time, I'm missing a record (the first record returned in my query).  It's probably something simple is my code.  Would be grateful if anyone could lend a suggestion. 

Code: [Select]
<?php 
         $a=0;
         $b=3;
          
          while ($ArrayData = mysql_fetch_array($QueryData)) {
          
          //if a=0, start new row
            if ($a==0) {
            echo '<tr>';
            }
          
          //create cell
          //if thumbnail exists, use it, otherwise throw in the unavailable image
          if ($ArrayData["PhotoThumb"]!="") {
          $ImageToUse = $ArrayData["PhotoThumb"]; 
          } else {
          $ImageToUse = "image_unavailable_thumb.jpg"; }
          echo '
             <td width="33%"  align="center" valign="top"><p><a href="item.php?ItemID='.$ArrayData["ItemID"].'"><img src="images/'.$ImageToUse.'" border="0" /></a></p>
              <p><a href="item.php?ItemID='.$ArrayData["ItemID"].'" class="cart_body_text">'.$ArrayData["ItemName"].'</a><br />
                 <span class="style1"> <strong>$'.sprintf("%01.2f", $ArrayData["UnitPrice"]).'</strong></span><br />
              </p>
              </td>';
        
         //increment by 1
         $a++;
        
        //if a = 3 then close the row using </tr>, ready to start a new one
        if ($a==$b) {
        echo '
        </tr>';
        
        //reset counter, ready to start new row
        $a=0;
        }
        }     
        
        if ($a > 0 ) {
           echo '</tr>';
        } 
         
    ?>
Thanks guys!

Hello all.

I am using an API from jambase.com to display events on a php page. Everything works great, the xml file is read and the data is displayed on the page as it should. My only problem is that when the data is displayed there is always a blank area where the first record of data should go. Any ideas how to get rid of this?

http://www.wddclients.com/mag33/events_search.php (artist search field not enabled yet)

Hopefully this is something pretty minor but if anyone wants to see specific code I can supply that.

Thanks so much for all help!

Greetings!
I have a website www.lanceronlinejobs.com/our_franchises.php I have a franchise images. I want to display each franchise record from database whenever a user click on franchise link. Here is my code.
ourfranchises.php  code:

Code: [Select]
<?php 
include('fconnection.php');

$sql = mysql_query("SELECT * FROM tbl_franchise ORDER BY id DESC") or die(mysql_error());
$row3 = mysql_fetch_array($sql); 
?>
<a href="franchiseDetails.php?city=<?php echo $row3['city'];?>"><img src="images/lbatkhela.jpg" width="150" height="150" alt="Batkhela" /></a>----------------------------------------------------------------------

franchiseDetails.php code:

Code: [Select]
<?php
$id = $_GET['id'];

$query = "SELECT * FROM tbl_franchise WHERE id = '$id' ORDER BY id DESC LIMIT 1";

$result = mysql_query($query);

if (!$result) {
echo "NO RECORD FOUND";
} else {

while($row3 = mysql_fetch_array($result)):
?>

Manager Name: <?php echo $row3['manager_name'];?>
Please help me. Any help would be appreciated.

Thanks.

Hello,

First time posting here, my MYSQL version is 5.1.52 and I have a created a search function to find a user record and if the name is typed correctly the name and details are displayed on the search page, but if the name is typed incorrectly nothing is appearing. Basically I would like to be able to say if no record found and redirect back to the previous page.

Code:
Code: [Select]
<?php
add = $_POST['add'];
        
// Database details
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name="Rewards"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot SELECT DB");

//Add query
$sql = mysql_query("UPDATE Rewards SET Rewards = (Rewards + 10) WHERE Name '%$add%'");



// Database details
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name="UserList"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot SELECT DB");


//Update Total Points query
$sql = mysql_query("UPDATE UserList SET TotalPoints = (TotalPoints + 10) WHERE Name '%$add%'");



// Database details
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name="UserList"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot SELECT DB");



//Search Query
$sql = mysql_query("SELECT * FROM UserList WHERE Name '%$add%'");  

while ($row = mysql_fetch_array($sql)){  

echo '<br/><img src="images/line1.jpg" alt="" width="200" height="2" />';

echo '<br/> Username: '.$row['Username']; 

echo '<br/>';  

echo '<br/><img src="images/line2.jpg" alt="" width="200" height="2" />';

echo '<br/> Name: '.$row['Name'];   

echo '<br/> Year Group: '.$row['YearGroup'];

echo '<br/> Form Class: '.$row['FormClass'];

echo '<br/>';

echo '<br/><img src="images/line3.jpg" alt="" width="200" height="2" />';

}  


// Database details
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name="Rewards"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot SELECT DB");

$sql = mysql_query("SELECT * FROM Rewards WHERE Name '%$add%'");  

while ($row = mysql_fetch_array($sql)){  

echo '<br/> Rewards: '.$row['Rewards'];

}  



// Database details
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name="Sanctions"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot SELECT DB");

$sql = mysql_query("SELECT * FROM Sanctions WHERE Name '%$add%'");  

while ($row = mysql_fetch_array($sql)){  

echo '<br/> Sanctions: '.$row['Sanctions'];

}  


// Database details
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name="UserList"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot SELECT DB");


$sql = mysql_query("SELECT * FROM UserList WHERE Name '%$add%'");  

while ($row = mysql_fetch_array($sql)){  

echo '<br/> Total Points: '.$row['TotalPoints'];

echo '<br/>';

echo '<br/><img src="images/line4.jpg" alt="" width="200" height="2" />';   

}  
Any help would be greatly appreciated.

Many Thanks

P Quinn

MOD EDIT: [code] . . . [/code] tags added.

I am looking to accomplish the following but have been hitting a brick wall:

    * User enters a single keyword into a text field, and conducts a search keyword search.
    * The results are pulled from the campusbooks.com API(API docs attached)
    * result are then output in <div> class and includes all the book details and its corresponding url./img etc

I'm trying to simplify this process but I continue to receive syntax errors.
A step by step practical explanation would do me justice!

Hi guys,

I have a very simple add.php to add data to a mySQL db.

I have a menu/list drop down as one of my fields on my form and this shows an array of results from another table (ranks of the RAF) within my db.

When I click the save button I have it process a INSERT INTO command but all i get inputted into my staff table is the first word... eg if I chose "Pilot Officer" from the list menu and then click save all that would appear in my db is "Pilot".

Any clues? I will paste the php below...

Code: [Select]
<?php
include('config.php');
?>
<form action='' method='POST' enctype='multipart/form-data'>
<p><b>Rank:</b><br />
<select name="rank" id="rank">
<option selected>Please Select</option>
<?php
$query = "SELECT * FROM ranks ORDER BY rank ASC";
$result = mysql_query($query);
while($row = mysql_fetch_array($result))
{
echo "<option value=". $row["rank"] .">". $row["rank"] ."</option>";
}
?>
</select>
  <p><b>Forename:</b><br />
  <input name="forename" type="text" id="forename" value="" size="40">
<p><b>Surname:</b><br /><input name='surname' type='text' id="surname" value='' size="40" />
<p><b>Category:</b><br />
<select name="category" id="category">
<option selected>Please Select</option>
<?php
$query = "SELECT * FROM categories";
$result = mysql_query($query);
while($row = mysql_fetch_array($result))
{
echo "<option value=". $row["category"] .">". $row["category"] ."</option>";
}
?>
</select>
 
  <p><b>Email:</b><br /><input name='email' type='text' id="email" value='' size="50" />
  <p><b>Mobile:</b><br />
    <input name='mobile' type='text' id="mobile" value='' size="40" />
  </p>
  <input type='submit' value='Save' />
  <input type='hidden' value='1' name='submitted' />
</form>
   
<?php

if (isset($_POST['submitted'])) {

$rank = mysql_real_escape_string($_POST['rank']);
$forename = mysql_real_escape_string($_POST['forename']);
$surname = mysql_real_escape_string($_POST['surname']);
$category = mysql_real_escape_string($_POST['category']);
$email = mysql_real_escape_string($_POST['email']);
$mobile = mysql_real_escape_string($_POST['mobile']);

$sql = "INSERT INTO `staff` (`rank` ,  `forename` ,  `surname` ,  `category` ,  `email` ,  `mobile` ) VALUES ( '$rank' ,  '$forename' ,  '$surname' ,  '$category' ,  '$email' ,  '$mobile')"; 
mysql_query($sql) or die(mysql_error()); 

echo (mysql_affected_rows()) ? "Staff Added":"Nothing Added";

}
?>

i am having the database consists of morethan one lac rows. we have a search option
in our website to search the database for required information. the present code is
like this:

form.php

`<input type="radio" name="tag" value="city" /> CITY <br/>`
`<input type="radio" name="tag" value="name" /> NAME OF CUSTOMER <br/>`
`<input type="radio" name="tag" value="amount" /> CHEQUE AMOUNT <br/>`
`<input type="radio" name="tag" value="somethingelse" /> some thing else
`
`Enter the part of any of the above Here :<input type="text" name="value" />`


search.php

`$tag = $_POST['tag'];`
`$value = $_POST['value'];`

`$query = "SELECT * FROM database WHERE $tag LIKE '%$value%' "`

note: we always input the part field only.


with this some times the output comes in thousands of rows. with which we are facing
problems.


we want to search the two or more fields for getting more precise results.

hence i tried this form:

`<h3 align="center">ENTER ALL OR DESIRED ITEMS YOU WANT TO SEARCH</h3>`
`<div width="80%" align="center">`
`<input type="text" name="city" /> CITY <br/>`
`<input type="text" name="name" /> NAME OF THE CUSTOMER <br/>`
`<input type="text" name="amount" /> AMOUNT <br/>`
`<input type="text" name="somethingelse" /> SOME OTHER SEARCH FIELD </div> `

`$query = "SELECT * FROM database WHERE city LIKE %$city%' || name LIKE %$name%' ||
amount LIKE %$amount%' || somethingelse LIKE %$somethingelse%';`

it worked in the mysql console, and even in our website when we give all the
variables. but it displaying the entire database when we dont give even one field in
the search box. i tried to assign NULL to the variable which was not given. it is
also not worked. it works if any variable is replaced with NULL in the query. i
don't know how to do that.


i tried a lot of queries after searching in lot of code provider websites. but none
of them gave the desired results.hence i request you to provide me a sql query code
for search the database using all of the above fields or any two or even with one.
the code must work independent of number of fields we entered.