PHP - Select Function Not Showing All

hy

I want to make a select in a class:
Code: [Select]
class test {

var $unu = array();

function __construct(){
$link = mysql_connect("", "", "") or die("Couldn't make connection.");
$db = mysql_select_db("", $link) or die("Couldn't select database");
//$this->unu = $textin;
}


function show($col){
$q = mysql_query("SELECT $col FROM categorie");
$qw = mysql_fetch_array($q);
$this->unu = $qw;
return $this->unu;
}

}


If I want to show how many rows are in that table:
Code: [Select]
$box = new test();

echo count($box->show("titlu"));

it show me that are only 2, although in "categorie" table are 10 rows.

Can you please tell me where is the error ?
Thanks

Hi,
I'm an novice/intermediate PHP prgogrammer, so excuse me if this is a stupid question.

I have one php program that consist of a few functions. Inside the function taht is the starting point I pickup a few entries from a MYSQL database, and display those in my html page.

I wish to dynamically put a link on those entries that makes it possible for the users to click those entries an get directed to other functions inside this very same PHP program WITH a parameter. The parameters that should accompany the url/link into the selected function to be able to select apropriate records from the mysql tables.

Any hints and tips for this would be highly appreciated!

Kind Regsrds,

Terje Hvidsten.

I have a DB

It gets records from a sensor every 5 min

This is then displayed on a Chart

When I request 4 Hrs I get 48 records x 12 Sensors, Not a issue

When I request 12 Hrs I get 144 records x 12 Sensors, A little Slower but Still OK

But as I start to request longer periods Like 7 Days I start getting larger and Larger data sets causing time out issues with the AJAX Calls (4-30MB per sensor downloads of data)




When looking at 4 weeks I really cant see the fine detail so is there a way to request records from MYSQL that drops some records in between

Kinda like the Step function in a For loop


If I request a period that has 2048 records but only want 204 records is there a way to say give me From -> To dates Step 10 and only return every 10th Record in the data set


Thanks

Ive tried to create a function to create a dropdown select box but im getting alot of errors to do with the if statement saying the ($_POST[$name]) value is not set???

function selectBox($name, $firstvalue, $limit, $increment)
{

//    echo "<br>";
//    echo $name;
//    echo "<br>";
//    print_r($_POST[$name]);
//    echo "<br>";
   $select ="selected=\"selected\"";
   $body = "<select name='$name' id='$name' method='POST'>
            <option value=''>$name</option>";
      for ($value = $firstvalue; $value <= $limit; $value += $increment)
      {
         $body .= "<option value= '$value'  "; 
         if ($_POST["$name"] === $value)
         {
            $body .= $select;
         }
         $body .= ">$value</option>";
       }      
          $body .= "</select>";
//    echo $value;
//    echo $_POST[$name];
//    echo $_POST['Width'];
    return $body;
 }

Any help would be greatly appreciated

Hi,
I am making a login page which consists of 2 pages, index.php and config.php

Following is a code of config.php
Code: [Select]
<?php
define("HOST", "localhost");
define("USER", "aaaabbbccc");
define("PASS", "aaabbbccc");
define("DATABASE", "pricetagindia");

$connection = @mysql_connect(HOST, USER, PASS);
$select_db = @mysql_select_db(DATABASE, $connection);

if(!$connection)
{
    echo "<h1>Cannot connect to database</h1>";
    exit(0);
    }
if(!$select_db)
{
    echo "<h1>Cannot select database</h1>";
    exit(1);
    }


?>


Here is a coding of index.php
Code: [Select]
<?php
include('config.php');

$_error = "<p></p>";
if (isset($_POST['username']) || isset($_POST['password']))
{
    $username = $_POST['username'];
    $password = $_POST['password'];
    
    if($username == NULL || $password == NULL)
    {
        $_error = "<p>Username and Password, both fields are required.</p>";
        }
    else
    {
        // Check in DB
        $password = md5($password);
        $checkuser_query = "SELECT * FROM users WHERE username = '$username.' AND password = '$password'";
        $checkuser_result = mysql_query($checkuser_query);
        $checkuser_count = mysql_num_rows($checkuser_result);
        if($checkuser_count == 1)
        {
            echo "match found";
            }
        else
        {
            echo "nothing found error, username was ".$username." and pass is ".$password;
            echo "<p></p>";
            echo $checkuser_result;
            }
    
                
        
        
        
        }
    
    
    
    
    
    }
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Price Tag India</title>
</head>

<body>
<center>
  <h1>Price Tag India</h1>
  <p>&nbsp;</p>
</center>
<center>
<form action="index.php" method="post" target="">
<table width="389" height="168" border="0">
  <tr>
    <td width="141">Username :</td>
    <td colspan="2"><label for="username"></label>
      <input type="text" name="username" id="username" /></td>
  </tr>
  <tr>
    <td>Password :</td>
    <td colspan="2"><label for="password"></label>
      <input type="password" name="password" id="password" /></td>
  </tr>
  <tr>
    <td>&nbsp;</td>
    <td width="56"><input type="submit" name="button" id="button" value="Submit" /></td>
    <td width="223"><input type="reset" name="button2" id="button2" value="Reset" /></td>
  </tr>
  <tr>
    <td>&nbsp;</td>
    <td colspan="2">
    <?php 
    echo $_error;
    ?>
    </td>
  </tr>
</table>
<p>&nbsp;</p>
</form>
</center>
</body>
</html>

I have attached the screenshot of database.

Problem is when I do valid login i.e. with user admin and its password, it is not working, I spent 1 hour for solving this issue but didnt find any solution.
Its frustrating now.....I know its simple but still its not working.

It is giving me error as
Code: [Select]
nothing found error, username was admin and pass is 62cc2d8b4bf2d8728120d052163a77df
Resource id #5

Hi Guys

I don't know if this is possible but can someone point me in the right direction. I have a php function which takes two inputs and returns an output. for simplicity's sake let's say it's an addition function. What I want to do is use a mysql select statement to show all the rows from a database where field1 and field2 equal '3'. Here's the sort of thing I mean.

function addNumbers($one,$two) {
return $one + $two;
}

mysql_query("SELECT * FROM table WHERE 'addNumbers(field1,field2)' = '3'");

What I actually want to do is a lot more complex than this but I am trying to understand how to make the syntax work in simple terms first.

Can anybody help?

Many Thanks
Dan

I am currently using a javascript to redirect the page using a select tag and the onchange property. Each page calls a different function.

I want to be able to pass the <select> value to a variable that I can use within a php condition statement on the same page. This will generate the required output based on the <select> value. This way I dont require a seperate page for each query.

So onchange.... $select_value == this.value?

Then within my page:
Code: [Select]
if ($select_value == 'alpha') {
    get_query_alpha();
}
if ($select_value == 'bravo') {
    get_query_bravo();
}

This is my current set up...
Code: [Select]
<p>Order by:
<select name="query" id="query" onchange="gotourl(this.value)">
<option value="query1.php" >Sort by A</option>
<option value="query2.php" >Sort by B</option>
<option value="query3.php" >Sort by C</option>
<option value="query4.php" >Sort by D</option>
</select></p>

<table> <?php get_query_1();?> </table>

and js...
Code: [Select]
function gotourl(url){
window.location= url;
}
function selectsubject(){
alert("Please select a subject!");
}

Is there a simple way to do this? Thanks.

Here is my query:

Code: [Select]
$db->query("SELECT cache.*,u.username from search_cache cache LEFT Join users as u ON u.id = 'NEED HELP HERE' where ident = '{$username}' ORDER BY DATE DESC") or error('Unable to send the message.', __FILE__, __LINE__, $db->error());
Here is my data inside my search_cache



Here is my while loop:

Code: [Select]
while($recent = $db->fetch_assoc($rec)){
$data = unserialize($recent['search_data']);

}


Now I can use $data['search_type'] as a array and If I use echo $data['search_type']['2'] it echo's out the user_id 32 as seen in the screenshot at the very end.

Problem is, I want to use that  'NEED HELP HERE'  to join with that id 32 dynamically, how the hell is that possible with mysql or is it? 


Or would it be easier to just add a new row/column named user_id and just join of that instead of trying to do all this? would that be faster?

Hi, i got help earlier with the this code-- exploding a field and getting an array, then inserting the array into a multiselect box-- That part is easy and works fine.

Now the explode function gives an array of words with duplicate words in it:
I tried several ways most either give a blank option/value, or the array word..

here is the code:


$query = "SELECT DISTINCT property_functionsexperience FROM #__users_profiles WHERE published = '1' ORDER BY property_functionsexperience ASC";
      $functionsexperiencelistgeneral=doSelectSql($query);      
        foreach($functionsexperiencelistgeneral as $words)
      $property_funcexperilist=$words->property_functionsexperience;      
        $wording = explode(', ', $property_funcexperilist);
        foreach($wording as $funciex)
      $funcexi=array_unique($funciex);
      $funcexpList .= "<option value=\"".$funcexi.','."\">".$funcexi."</option>";
      $funcexpList .= "</optgroup>";
            }
         $funcexpList.="</select>";
         $output['FUNCEXPLIST']=$funcexpList;         


this code is adapted slightly from the working original that gives the double words-- this code gives a blank, but i think its on the correct track!?

PHP date and time function is not showing correct time on my local system

I have the following php code

date_default_timezone_set("Africa/Lagos");

$date = date('d-m-y h:i:s');

echo "Server Time ".$date ."<br>";

echo "The time is " . date("h:i:sa")."<br>";

$current_datetime = date("Y-m-d") . ' ' . date("H:i:s", STRTOTIME(date('h:i:sa')));

echo "Current time1: ".$current_datetime . "<br>";

 

Output

Server Time 21-05-21 09:55:39
The time is 09:55:39am
Current time1: 2021-05-21 09:55:39

 

Expected Output

Server Time 21-05-21 10:55:39
The time is 10:55:39am
Current time1: 2021-05-21 10:55:3

 

Any help would be appreciated.

 

Edited May 21 by Ponel

I am pretty new to PHP and am trying to create a simple (so I assumed) page to takes data from one html page(works fine) and updates a MYSQL Database.  I am getting no error message, but the connect string down to the end of the body section is showing up as plain text in my browser window.  I do not know how to correct this.  I have tried using two different types of connect strings and have verified my names from the HTML page are the same as listed within the php page.  Suggestions on what I need to look for to correct would be great.  I have looked online, but so far all I am getting is how to connect, or how to create a comment, so I thought I would try here. 
Thank you for any assistance I may get!!  - Amy -
 
Code: [Select]
<body><font color="006600">
 <div style="background-color:#f9f9dd;">
  <fieldset>
<h1>Asset Entry Results</h1>
<?php
  // create short variable names
  $tag=$_POST['tag'];
  $serial=$_POST['serial'];
  $category=$_POST['category'];
  $status=$_POST['status'];
  $branch=$_POST['branch'];
  $comments=$_POST['comments'];

  if (!$tag || !$serial || !$category || !$status || !$branch) {
     echo "You have not entered all the required details.<br />"
          ."Please go back and try again.";
     exit;
  }

  if (!get_magic_quotes_gpc()) {
    $tag = addslashes($tag);
    $serial = addslashes($serial);
    $category = addslashes($category);
    $status = addslashes($status);
    $branch = addslashes($branch);
    $comments = addslashes($comments);
  }

  //@ $db = new mysqli('localhost', 'id', 'pw', 'inventory');
  $db = DBI->connect("dbi:mysql:inventory:localhost","id","pw") or die("couldnt connect to database");

  $query = "insert into assets values
            ('".$serial."', '".$tag."', '".$branch."', '".$status."', '".$category."', '".$comments."')";
  $result = $db->query($query);

  if ($result) {
      echo  $db->affected_rows." asset inserted into Inventory.";
  } else {
     echo "An error has occurred.  The item was not added.";
  }

  $db->close();
?>
  </fieldset>
 </div>
</body>

Hi all.
Here is my scripts which allow user to check multiple rows of data and delete it , but it require select data and click for twice to delete the rows , what should be the error?

Code: [Select]
<form name="frmSearch" method="post" action="insert-add.php">

<table width="600" border="1">
  <tr>
    <th width="50"> <div align="center">#</div></th>
    <th width="91"> <div align="center">ID </div></th>
    <th width="198"> <div align="center">First Name </div></th>
    <th width="198"> <div align="center">Last Name </div></th>
    <th width="250"> <div align="center">Mobile Company </div></th>
    <th width="100"> <div align="center">Cell </div></th>
    <th width="100"> <div align="center">Workphone </div></th>
    <th width="100"> <div align="center">Group </div></th>
  </tr>
    </form>
<?

    echo "<form name='form1' method='post' action=''>";

while($objResult = mysql_fetch_array($objQuery))
{


    echo "<tr>";
    echo "<td align='center'><input name=\"checkbox[]\" type=\"checkbox\" id=\"checkbox[]\" value=\"$objResult[addedrec_ID]\"></td>";
    echo "<td>$objResult[addedrec_ID] </td>";
    echo "<td>$objResult[FirstName]</td>";
    echo "<td>$objResult[LastName] </td>";
    echo "<td>$objResult[MobileCompany] </td>";
    echo "<td>$objResult[Cell] </td>";
    echo "<td>$objResult[WorkPhone] </td>";
    echo "<td>$objResult[Custgroup] </td>";

   echo "</tr>";

}


 echo "<td colspan='7' align='center'><input name=\"delete\" type=\"submit\" id=\"delete\" value=\"Delete\">";

 if (isset($_POST['delete']) && isset($_POST['checkbox'])) // from button name="delete"
 {
 $checkbox = ($_POST['checkbox']); //from name="checkbox[]"
     $countCheck = count($_POST['checkbox']);

 for($d=0;$d<$countCheck;$d++)
     {
         $del_id  = $checkbox[$d];

 $sql = "DELETE from UserAddedRecord where addedrec_ID = $del_id";

 $result2=mysql_query($sql)  or trigger_error(mysql_error());;;

     }
         if($result2)
     { 
          $fgmembersite->GetSelfScript();
         }
         else
         {
 echo "Error: ".mysql_error();
         }
 }
  echo "</form>";


Thanks for every reply.

hirealimo.com.au/code1.php

this works as i want it:
Quote

SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1



but apparelty selecting * is not a good thing????

but if I do this:
Quote

SELECT priceID, price FROM price INNER JOIN vehicle....etc


it works but i lose the info from the vehicle table.

but how do i make this work:
Quote

SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc


so that i am specifiying which colums from which tables to query??

thanks

I have 2 queries that I want to join together to make one row  

Dear All,

I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box.

All is working fine except that the locality select box is not being populated.
I know that the problem is in the sql statement

WHERE country_id='$co'

because i am having an error that $co is an undefined variable.

All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement?

Thank you for your help in advance.

MySQL Tables indicated below:

CREATE TABLE countries(
country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT,
country_name VARCHAR(30) NOT NULL,
PRIMARY KEY(country_id),
UNIQUE KEY(country_name),
INDEX(country_id),
INDEX(country_name))
ENGINE=MyISAM;

CREATE TABLE localities(
locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
country_id INT(3) UNSIGNED NOT NULL,
locality_name VARCHAR(50),
PRIMARY KEY (locality_id),
INDEX (country_id),
INDEX (locality_name))
ENGINE=MyISAM;

Extract PHP script included below:

// connect to database
require_once(MYSQL);

if(isset($_POST['submitted']))

{

// trim the incoming data

/* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */
$trimmed=array_map('trim',$_POST);

// clean the data

$co=mysqli_real_escape_string($dbc,$trimmed['country']);

$lc=mysqli_real_escape_string($dbc,$trimmed['locality']);

}

?>

<form action="form.php" method="post">

<p>Country
<select name="country">
<option>Select Country</option>

<?php

$q="SELECT country_id, country_name
FROM countries";

$r=mysqli_query($dbc,$q) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($r))

{

$country_id=$row[0];
$country_name=$row[1];
echo '<option value="' . $country_id . '"';
if(isset($trimmed['country']) && ($trimmed['country']==$country_id))
echo 'selected="selected"';
echo '>' . $country_name . '</option>\n';

}

?>

</select>
</p>

<p>Locality
<select name="locality">
<option>Select Locality</option>

<?php

$ql="SELECT locality_id, country_id, locality_name
FROM localities
WHERE country_id='$co'
ORDER BY locality_name";

$rl=mysqli_query($dbc,$ql) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($rl))

{

$locality_id=$row[0];
$country_id=$row[1];
$locality_name=$row[2];
echo '<option value="' . $locality_id . '"';
if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id))
echo 'selected="selected"';
echo '>' . $locality_name . '</option>\n';

}

// close database connection
mysqli_close($dbc);

?>

</select>
</p>

<p><input type="submit" name="submit" value="Submit" /></p>

<input type="hidden" name="submitted" value="TRUE" />

</form>