PHP - Print Forms

can someone please give me some guidance on how to do this please
I am wanting to create a status updating type application on my site
and i have this idea in my head

i want it to retrive and print the last 3 posts (max ids) made by the user
if someone could give me some example code please and i can hopefully work from that.
Thanks
James

Hi,

What is the best way to print the local date and time, my website have customers from 4 or 5 countries.

Saudi - UTC+3
Dubai - UTC+4
India - UTC+5.30


how to print the local time for them

Thanks,

 

hihi,

so I have the following, except it goes all the way up to server 400. Is there a way to make it print after each echo? As it is right now it will not print the entire list until all 400 servers are done

echo "Server 01: " . count($server01->listaccts()) . " / 130" . "<br />" ;
echo "Server 02: " . count($server02->listaccts()) . " / 130" . "<br />" ;
echo "Server 03: " . count($server03->listaccts()) . " / 130" . "<br />" ;
echo "Server 04: " . count($server04->listaccts()) . " / 130" . "<br />" ;
echo "Server 05: " . count($server05->listaccts()) . " / 130" . "<br />" ;
echo "Server 06: " . count($server06->listaccts()) . " / 130" . "<br />" ;
echo "Server 07: " . count($server07->listaccts()) . " / 130" . "<br />" ;
echo "Server 08: " . count($server08->listaccts()) . " / 130" . "<br />" ;


thanks!

Hello.   How would I print the following:     http://site.com/break?return=site2.com   So that I can set a button containing site2.com.   Example: <a href="site2.com"></a>   And so forth:       http://site.com/break?return=site3.com   Would print: <a href="site3.com"></a>

Im trying to print several image url's and names from an sql database into 3 columns,

Aiming to get it to go

1,2,3
4,5,6
7,8,9

etc etc. but for some reason myne is going

1,3,5
2,4,6

etc. Dont bother mentioning that my loops do nothing, i realised that about 10 minutes ago, Any help would be appreciated.



<?php
include 'config.php';
mysql_connect($host, $user, $pass) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());
$result = mysql_query("SELECT * FROM tracks");
echo '<div id="left_wrapper">';
for ($i=0;$i<mysql_num_rows($result);$i+=3) {
    $row = mysql_fetch_array($result);
    $id = $row['id'] + 1;
    echo "<img src='Thumbnails/" . $id .".gif'></img><br>";
    echo $row['name']. "<br>";
}
echo "</div>";
echo '<div id="middle_wrapper">';
for ($i=1;$i<mysql_num_rows($result);$i+=3) {
    $row = mysql_fetch_array($result);
    $id = $row['id'] + 1;
    echo "<img src='Thumbnails/" . $id .".gif'></img><br>";
    echo $row['name']. "<br>";
}
echo "</div>";
echo '<div id="right_wrapper">';
for ($i=2;$i<mysql_num_rows($result);$i+=3) {
    $row = mysql_fetch_array($result);
    $id = $row['id'] + 1;
    echo "<img src='Thumbnails/" . $id .".gif'></img><br>";
    echo $row['name']. "<br>";
}
echo "</div>";
?>
i know each of those loops does nothing, But you can see where they are meant to do, Each div is aligned to different positions, first loop is left, 2nd is center, 3rd is right.

This topic has been moved to Miscellaneous.

http://www.phpfreaks.com/forums/index.php?topic=346728.0

I'm working on a site where I've implemented a simple back end wysiwyg editor for content on a page. Then on the public page I run a php query to pull that content and display it. But doing this cancels out the css I have been using to split content into two columns. Is there a way to do this in php, or is there a way to circumvent the problem?  ( also tried echoing the entire css style along with the query result - that didn't work either)

The <p id='container_sub'> is what is split into two columns. I tried it outside of the query, and inside the query around where I echo results. Neither worked.
Here's the basic php code, and further down the css that makes two columns:
Code: [Select]
<?php

$pageid = '2';
  
// Formulate Query
// This is the best way to perform an SQL query
// For more examples, see mysql_real_escape_string()
$query = sprintf("SELECT content FROM tbl_pages WHERE page_id='%s'",
     mysql_real_escape_string($pageid));

// Perform Query
$result = mysql_query($query);

// Check result
// This shows the actual query sent to MySQL, and the error. Useful for debugging.
if (!$result) {
    $message  = 'Invalid query: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query;
    die($message);
}

// Use result
// Attempting to print $result won't allow access to information in the resource
// One of the mysql result functions must be used
// See also mysql_result(), mysql_fetch_array(), mysql_fetch_row(), etc.
while ($row = mysql_fetch_assoc($result)) {
    echo "<p id='container_sub'>".$row['content']."</p>";

}

// Free the resources associated with the result set
// This is done automatically at the end of the script
mysql_free_result($result);
      
?>

The css....
Code: [Select]
#container_sub {-moz-column-count: 2;
-moz-column-gap: 25px;
-webkit-column-count: 2;
-webkit-column-gap: 20px;
column-count: 2;
column-gap: 20px;}

#container_sub2 {-moz-column-count: 2;
-moz-column-gap: 25px;
-webkit-column-count: 2;
-webkit-column-gap: 20px;
column-count: 2;
column-gap: 20px;}

I currently have a search page on my site that prints the products but it prints the products more than once if its in more than one category I have tried getting distinct item in my SQL. But this doesnt work so im trying an if statement that if there is more than one specific result then to just print this once. I was wondering if anyone had any ideas of how to do this using an if statement I just dont know how to go about just printing the result just once if its greater than 1. The code is below to make it clearer.

$searchterm = $_POST['searchterm'];
trim ($searchterm);

/*check if search term was entered*/
if (!$searchterm){
        echo 'Please enter a search term.';
      echo $searchterm;
}
/*add slashes to search term*/
if (!get_magic_quotes_gpc())
{
$searchterm = addslashes($searchterm);
}


/*query the database*/
$query = "SELECT * from
(products LEFT JOIN categories_products_link ON products.prod_id = categories_products_link.prod_id)
LEFT JOIN categories ON categories_products_link.cat_id = categories.cat_id WHERE prod_title LIKE '%" . $searchterm .  "%' ORDER BY cat_title, prod_title";
$result = mysql_query($query);
/*number of rows found*/
$num_results = mysql_num_rows($result);

echo '<p><h1>Search Results: '.$num_results.'</h1></p><br />';
/*loops through results*/
for ($i=0; $i <$num_results; $i++)
{
 $num_found = $i + 1;   
 $row = mysql_fetch_assoc($result);
 echo "$num_found. "?><a href="store-<?php echo $row['cat_id'];?>-<?php echo $row['prod_id']; ?>/<?php echo seo_makeSafeURI($row['prod_title']); ?>.html"><strong><?php echo $row['prod_title']; ?></strong></a> <br />

Hello guys and gals, I am pretty green to PHP! I have an empty array that that I am trying to put images into. The thing is I have a certain file name in the folder, I want to exclude that file. This is what I have tried, any advice would be appreciated!

Code: [Select]
$thumbImg[] = array();
foreach (glob($DImg) as $PImg)
{
if (!is_file("thumbnail.jpg"))
{
$thumbImg[] = "<img src=\"pathtoimage\">";
}
}

Later on the page I am printing it out with this. It is still including the thumbnail.jpg image. Thank you in advance!!

Code: [Select]
for ($i=0; $i<count($thumbImg); $i++)
print $thumbImg[$i];

Hi,
I am trying to get the form field to echo a php variable. The problem is that the form is 'Printed' via php.

print "<td width=\"300\" valign=\"top\"><input type=\"text\" name=\"usr\"  value=\"i.e. JBloggs \" onfocus=\"if(!this._haschanged){this.value=''};this._haschanged=true;\" tabindex=\"1\"></td>";

Instead of the value being i.e. JBloggs have it echo the $name variable.

Any input would be appreciated.

Probably a simple solution, just one I'm not sure how to do it.

I need the array to print out like this

Code: [Select]
array(
   'name'=>'Store 1',
   'address'=>'LA1'
),

Right now, I got it to look like this

Thanks in advance
Code: [Select]
Array
(
    [name] => 'Albertville Farmers Market',
    [address] => '116 Main Street Albertville, Alabama 35950'
)
,
Array
(
    [name] => 'Alexander City Downtown Market',
    [address] => 'Braod Street Alexander City, Alabama 35010'
)
,

How can I go about accomplishing this?

Here is my code
Code: [Select]
$stores = array('name'=>"'$MktName',",'address'=>"'$address'");   
echo "<pre>";
print_r($stores);   
echo "</pre>";

I am trying to get the code at the bottom of the script to print just once during the loop but it either doesn't print at all or repeats with the loop im am using if (!$i++) to print once and i works the first time i use it.

 
foreach($uploadFilename as $key => $myvar)
{
if (!$i++) print "<!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>Upload Complete....</title>
</head>

<body>

<body onload=\"document.forms.formname.submit);\">

<form id=\"formname\" name=\"form1\" method=\"post\" action=\"reg5.php\">\n";


echo "<input type=\"hidden\" name=\"image$key\" value=\"";
echo end(explode('/',$myvar));
echo "\">\n";

if (!$i++) print "</form>
</body>
</html>\n";


}
 

Should i be using print() or echo() for an xmlhttp.responsetext ?

Code: [Select]
<?php
//================================================================

  // Configure connection parameters
  $connect_string="DRIVER={Sybase SQL Anywhere 5.0}; SERVER=ACMSQL036A; DATABASE=BACKOFF";
  $dbuser="DBA"
  $dbpswd="SQL"

//================================================================
 
  // Connect to DB
  $conn = odbc_connect($connect_string, $dbuser, $dbpswd);
 

  // Query
  $xrefnum = $_GET['ae_xrefnum'];
  $qry = "SELECT xrefnum, plunum FROM DBA.PLU_Cross_Ref where xrefnum = '$xrefnum'";

  // Get Result
  $result = odbc_exec($conn,$qry);
  if (!$result)
  {
   echo odbc_errormsg($conn);
  }

  // Get Data From Result
  $row=odbc_fetch_array($result);
  
  // Show data
  print($row[plunum]);
  }

  // Free Result
  odbc_free_result($result);

  // Close Connection
  odbc_close($conn);

  

//================================================================
?>

Hi

I have this code to echo th econtents of a table, which contains 100 rows.

I want to be able to display all 100 rows.

The problem is that it displays 99 (it excludes the first).

I've tried to backtrack to reconstruct the code to find out where the error is but no joy.

Any ideas?

Thanks in advance!

Code: [Select]

....

// get all entries from table

$sql = "SELECT * FROM 100words ORDER BY word_id";
$result = mysqli_query($dbc, $sql);
$r = mysqli_fetch_row($result);

//print table entries to screen in columns

echo '<div id="container">';

// results presented in html table 5 columns, 20 rows per page
echo '<div id="outerbox">';
                echo '<div class="innerbox">';
echo '<table class="centerresults" border="0">'  . "\n";
echo '<td>'; 
 
$i = 0;   
$max_columns = 5;   

    while ($list = mysqli_fetch_assoc($result)) { 

      extract($list);

// open row if counter is zero       

if($i == 0)
         
echo '<tr>' . "\n";
         
echo '<td width="150px"><a href="word.php?w=' .

$list['word'] . '">' .

$list['word'] . "</a></td> \n "; 
         

// increment counter - if counter = max columns, reset counter and close row       

if(++$i == $max_columns) {
         
echo '</td>' . "\n";
         
$i=0; 
     
}  // END if(++$i == $max_columns) {

} // END while (!empty($myArray)) {

//} END while ($list = mysql_fetch_array($result)) {


 //END if($i < $max_columns) {     

echo '</tr>' . "\n";

echo '</table>' . "\n";

   echo '</div>';
   
      echo '</div>';

I am using function to insert into database.  But the primary key is automatic and I used Quote

$_SESSION['Tes_ID'] = mysql_insert_id();


to retrieve this. But now that I use function method. I am not sure how to retrieve the primary key on to the next page.

Code: [Select]
$value = modulesql($postVar1, $postVar2, $SessionVar1, $SessionVar2);

$_SESSION['Tes_ID'] = mysql_insert_id();



echo $value, $_SESSION['Tes_ID'];


Code: [Select]
<?php
function modulesql($Tes_Name, $Tes_Description, $Use_ID, $Sub_ID){
$con = OpenConnection();

mysql_select_db("examination", $con);
$module = ("INSERT INTO test (`Tes_Name`, `Tes_Description`, `Use_ID`, `Sub_ID`)
VALUES ($Tes_Name, $Tes_Description, $Use_ID, $Sub_ID)")
or die('Cannot Execute:'. mysql_error());


CloseConnection($con);
return $module;
}
?>

Have I lost you with my question??