Posts by ko ko

You've not execution query.

$query = mysql_query($sql);
if(!$query){
   die('Update failed!. Error:' . mysql_error());
}

This one is more clear format.

$sql = "UPDATE `my_db` SET `date_in` = '$today', `customer_name` ="$customer_name", `billing_address` = '$billing_address', `billing_suburb` = '$billing_suburb', `billing_city` = '$billing_city', `same_as` ='$same_as', `shipping_address` = '$shipping_address', `shipping_suburb` = '$shipping_suburb', `shipping_city` = '$shipping_city' WHERE `personID` = '$select'";

Hope this help.

As I mentioned above, it is not quote ('account_type', 'fd_period', 'account_number').
It is groove symbol that is left side of the numeric key 1, mostly known as tilde key.
Copy below and paste in your file:

$query = "UPDATE account SET `account_type` = '$account_type', `fd_period` = '$fd_period' WHERE `account_number` = '$account_number'";

Use single-quote ('') for the value. If not, escape them with slash (\"\").
Hope this help.

Show that line 72 in account.php.

What error ?

Not same like yours.

SET `account_type` = '$account_type'

Use (``) symbol left side of the numeric key 1 for the field name. Use quote ('') for the value.

ႊႊThen, it solved ? If so, mark as Solved with below link.

$query = "UPDATE account SET 'account_type' = '$account_type', 'fd_period' = '$fd_period' WHERE 'account_number' = '$account_number'";

Replace with

$query = "UPDATE account SET `account_type` = '$account_type', `fd_period` = `$fd_period` WHERE `account_number` = '$account_number'";

It is not quote. It is groove left side of the numeric key 1.

What value are you meaning ? This one $sector=$_GET['sec']; ?
Your question is not clear.

Build the require fields in the form. PHP is also ok. You need to put the fields that would process by PHP.

$info = array(
'nic' => $_POST["nic"],
'full_name' => $_POST["full_name"],
'name_with_initials' => $_POST["name_with_initials"],
'phone_number' => $_POST["phone_number"],
'address' => $_POST["address"],
'gender' => $_POST["gender"],
'date_of_birth' => $_POST["date_of_birth"],
'account_type' => $_POST["account_type"],
'fd_period' => $_POST["fd_period"]
);

You must have fields for the above array.

Move this to Javascript forum.

You two are mistaken.
@matricol
Read the thread from the last time, I'm not mentioning about the GLOBAL Selector.

@newbie14
I left the image name with asterisk '*', it's what your image name you want to use. Within your first post, you used 'dark.png' for the tab. Then, I refers 'dark.png' with asterisk '*' or whatever if you have another one. There is no CSS declaration for the background with the value '*'. I showed you as just EXAMPLE.

Hope this help!

You means '$tn' is for the variable from the form ? You form uses POST method, not GET. I'm not sure what you are doing with '$tn'. Maybe, you're wrong with '$_POST' var and '$_GET' vars. Try with below

$tn = $_POST['tn'];

Does your server allow short tag <?= 'something'; ?> Try to replace this statement <?=$tn;?> to <?php echo$tn; ?> , and check does any output returns. If so, your server's PHP setting does not allow for short tag.

I started from there.http://www.html.net/
And here is the good place for CSS design www.csszengarden.com.

Start from basic, and go professional. Below is the simple road map.
HTML+CSS+Js => PHP/ASP.Net/JSP/Framework

Enjoy with your learning.

If you learn PHP without HTML,CSS basic, then, you would have problem in the development cycle.

Post this to the Javascript forum, your javascript functions which does not work.

As @twiss says, position the image on this place and attach the event to the image. Just simple CSS.

Ensure that Rewrite Engine is on.

RewriteEngine On
RewriteCond %{REQUEST_FILENAME} !-d
RewriteCond %{REQUEST_FILENAME}\.php -f
RewriteRule ^/([^/]+)/?$ /$1.php

Better post in HTML,CSS forum. Remove there.

Why not post any code or the link ? We're not sight-seeing men.

Did you already make a search with google ? I'm not sure what you want exactly. Check out there.

$query = "UPDATE account SET `account_type` = '$atype', `fd_period` = 'fdper' WHERE `account_number`= anum";

Are you using this statement ? 'anum' should be '$anum'.

Is '$pd_image' already has absolute URL including your host, domain (e.g., www.example.com/images/image.jpg).If so, let me do it:

<img src="http://www.example.com/resize.php?width=180&height=180&image=<?php echo $pd_image; ?>" alt="<?php echo $pd_name; ?>" />

The above should work if '$pd_image' has absolute path to the image (eg.,http://www.example.com/images/). If not concatenate it by following:

<img src="http://www.example.com/resize.php?width=180&height=180&image=http://www.example.com/<?php echo $pd_image; ?>" alt="<?php echo $pd_name; ?>" />

You're wrong way to use: resize.php?<?php echo $pd_image; ?>? The image file only need to pass after 'image=' parameter.

Mark-up are also messed. Fail DTD http://validator.w3.org.
'<li>' tag is outside of the '<ul>' at line 99. But, I don't think so that probably make the layout flew.
Validate your CSS here.

It doesn't work with relative URL, I tested it. Use by following.

yoursite_url/resize.php?width=180&height=180&image=yoursite_url/image_path/image

No slash '/' after the 'resize.php'. Use '?' and parameters follow by that. The above is correct way.
Hope this help.

@MrDJK, you're right.
@Jaklins, IE has different known for MIME-TYPE. You'll probably know that echo the file type. IE shows 'jpg' as 'pjpeg' and 'png' as 'x-png'. So, the upload will be done with adding a few lines for those MIME types.

$_FILES["fileField"]["type"] == "image/pjpeg" //for jpeg, and jpg
$_FILES["fileField"]["type"] == "image/x-png" //for png

Hope that solve, and credit to MrDJK.

Mark this as 'Solve' if your problem solved.

http://www.example.com/image.php//images/product/ea723c7a2658ac701a45a583986b5da4.jpg?width=180&height=180&image=/images/product/ea723c7a2658ac701a45a583986b5da4.jpg

This one wrong. Should be:

http://www.yoursite.com/image.php?width=180&height=180&image=http://www.yoursite.com/images/product/ea723c7a2658ac701a45a583986b5da4.jpg

You can follow their instruction easily.
Hope this help.

http://www.example.com/image.php?width=180&height=180&image=http://www.example.com/images/product/ea723c7a2658ac701a45a583986b5da4.jpg

How about this one ?

Copy this file to your site and keep under your web folders anywhere. For example, you named this file to 'resize.php' and keep it under the main root. Then call this file with the following:

http://www.yoursite.com/resize.php //absolute path
resize.php                         //relative path

Then, hook with your image as parameter including width,height (etc.), following the references on their site. As following your codes:

<td cless="productWrap">
<td class="productItem" style="opacity: 1;">
<a class="productImage" href="<?php echo $pd_image; ?>" rel="prettyPhoto" title="<?php echo $pd_name;?>" style="opacity: 1;">
<img src="http://www.yoursite.com/resize.php/<?php echo $pd_image; ?>?width=180&height=180&image=<?php echo $pd_image; ?>" alt="<?php echo $pd_name; ?>">
</a>
</td>

You can test manually on the browser with the following URL: http://www.yoursite.com/resize.php/yourimage.jpg?width=110&height=110&image=path_to_image/image.jpg Hope this help!

You may not need to create thumbnails, you can crop the image how much as you need. And you don't also need to include/import any file inside your page. Check out here.

Hook your image with the file and it'll automatically crop your image as thumbnail without cloning or duplicating. Just simply hook with that file like:

<img src="http://yoursite.com/resizer.php?width=110&height=110" />

Easily customize your thumb size as well as you're.
Hope this help!

The thing is, is there a more simple way for creating thumbnails without all of that files?

What do you mean for that question ? Another way to create thumbnail, perhaps with a few line codes ?

And what do you want to combine with those statements ?
Maybe, are you want to create multiple thumbnails and display in the page ?

Query your image from the table and pass this into the function argument where you want to do.

Below is sample:

$sql = "SELECT * FROM image_table WHERE image_id = 123 ORDER BY image_id DESC LIMIT 1";
$row = mysql_query($sql);
$image = mysql_fetch_assoc($row);

//pass the image to the function parameter
createThumbnail($image);

Isn't what you want ?

y.innerHTML="<label class=description for=element_1>Master <font color='red'>*</font></label>y.innerHTML="<label class=description for=element_1>Master <font color='red'>*</font></label><input type=hidden value='' name=hiddenMasterID> ";

I don't know exactly and what's the real problem. But, on the above method, the 'hiddenMasterID' is not already exists in the form. It was manipulated by DOM with the function.
I'm neither Javascript expert nor who are well-known in javascript. And your problem is very relevant with Javascript and not PHP. So you should post this thread into the Javascript forum.
Hope this help.

Show your thumbnail image or what ? What's wrong with the codes ?

Maybe, XMLHttpResponseText wont' work on the form, and the form can't pass the value. You can check with function and attach with 'onsubmit' to the form. I tested with the following:

function check(){
     var el = document.form1.masterID.value;
     alert(el);
}

It does not alert on Firefox and work only on IE. Use hidden field to hold the value of 'masterID' and create function that alter/fill the value to the hidden field when the drop-down list was seleted. You can do that with 'onchange' event.
Below is the clear format of your code.

<!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> 
<link rel="stylesheet" type="text/css" href="my1.css" media="all"> 
<script type="text/javascript">
function getSlaveTrailer(str)
{
	differentiator=Math.floor(Math.random()*50000);
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
      var x=document.getElementById('myTable').insertRow(2); 
      var y=x.insertCell(0); 
      var z=x.insertCell(1);       
      var newdiv = document.createElement("div");
      newdiv.innerHTML = xmlhttp.responseText;     
      y.innerHTML="<label class=\"description\" for=\"element_1\">Master <font color='red'>*</font></label> "; 
      while (newdiv.hasChildNodes()) 
      {    z.appendChild(newdiv.firstChild);
      	
      }      

	
    }
    
  }
xmlhttp.open("GET","getMasterList.php?e=1&d="+differentiator,true);
xmlhttp.send();

}


    </script>
	<script type="text/javascript">
	function check(){
		var el = document.form1.masterID.value;
		alert(el);
	}
	function chk(ck){
		document.form1.msID.value = ck;
	}
	</script>
</head> 
<body> 

<?php
if(isset($_POST['msID'])) {
?>
<h1><?php echo $_POST['msID']; ?></h1>
<?php
}
?>
<table> 
<tr>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" name="form1" id="form1" onsubmit="alert('hi');check();">
<td width="700" valign="top">
	<table id="mainDetails">
  	<tr>
			<td>				
			  <label class="description: for="element_1">A1<font color="red">*</font></label> 
			</td>

			<td>
			  <input class="text" id="asset" name="asset" value="some">  
			</td>
		</tr>
		<tr>
			<td><p class="error" id="assetError"></p></td>
		  <td></td>
		</tr>		
	
		
	</table>
	<table id="myTable">

		
		<tr>
			<td> …

@almostbob
Why :?: ?

<title>June 16, 2011 &raquo; The-Elder-Scrolls-V-Skyrim_2011_06-16-11_002 &mdash; Gematsu Gallery</title>

Check your view source. The title tag is outside of the HTML. You've wrong close tag of '</script>' at line 59 which is not opened.
Validate your markup here. Copy the source codes from the browser 'View Source' and paste into 'Validate by Direct Input'.
Here is CSS validator.

I'm not sure where you want to disable the previous link. Here is easy way to configure the pagination.
To disable previous link, prepare the current page and 1.
To disable next link, prepare the current page and the total numbers of the page.

Hope this help.

PHP is not same like HTML. PHP is server-side and HTML is for client (browser). HTML doesn't need to compile or interpret on the server. It's only interpret by the browser.
Unlike HTML, PHP needs PHP engine which was installed on your web-server that interpret the PHP codes which are delimited by '<?php ?>'.
Find more details here. You can also download the latest PHP version from this site.
Hope this help.

$ud_id = $_POST

Where is 'ORDERING_DEPT' ? That's not found in your form. Ensure that your form has required fields.

$selected2 = ($selected || $day_nam == $day_ini) ? " selected='selected'" : "";

Replace above with the following:

$selected2 = ($selected == $day_nam) ? " selected='selected'" : "";

And first of my post below.

<select name="year_nam" id="year_nam">
<?php
for ($yr_nam = 1950; $yr_nam <= 2020; $yr_nam++)
{
$selected = ($selectedYear || $yr_nam == $year) ? " selected='selected'" : "";
echo "<option value='$yr_nam' $selected>$yr_nam</option>";
}
?>
</select>

I gotta correct the above with the following:

<select name="year_nam" id="year_nam">
<?php
  for ($yr_nam = 1950; $yr_nam <= 2020; $yr_nam++) 
  {
    $selected = ($selectedYear == $yr_nam) ? " selected='selected'" : "";
    echo "<option value='$yr_nam' $selected>$yr_nam</option>";
  }
?>
</select>

The second expression '$day_nam == $day_ini' and '$yr_nam == $year' are not require. Because '$selectedYear' and '$select' assign the specific value with isset($_POST['year_num']) and isset($_POST['day']) methods respectively.
So, '$day_nam' will retains the last value if the form has some problems. If the form has not already submitted, it'll assign default value, i.e., 1. '$yearSelected' has the same case like so.

@madCoder, you're right. You took me the right back. Thanks.

@jacob21
You want to retain the drop-down value, if there is something wrong in submitting the form/validation failed and after the page reloaded, right ?

$selectedYear = isset($_POST['year_nam']) ? $_POST['year_nam'] : $year; // If you use GET method use $_GET['year_nam'] instead of $_POST['year_nam']

This line would keep the selected value if the form has already submitted otherwise it takes default (2011).

No. I represent your image name as '*'. If your image name is 'background.jpg', the star '*' would has 'background'. It is nothing, not syntax nor value, just variable refers to your image name.

<a href="index.html"><input type="button" name="Cancel" /></a>

Better use this one instead above.

<input type="button" name="Cancel" onclick="location.href('index.html');" />

And the codes are not complete. Show the correct one.

<div class="Normal_P">
<span class="Normal_C">
<form action="<?php getLink("http://www.atouchofmoore.net/certificate_stage3.php" array(

<?php

Try this:

.table2 {
    float: left;
    width: 200px;
}
.table3{
    float: right;
    width: 200px;
}
.table1{
    margin-left: 200px; /* requires at least table2 width */
    margin-right: 200px; /* requires at least table3 width */
}

The "url('*.jpg')" means the path of your image, which the star '*' refers to the image name. 'no-repeat' says do not repeat the background-image. '0 50%' is the position of background-image, coordinate point (x, y), which points to 0 at x and 50% at y. This is not familiar with you, you can use 'left center'. Here is short list below:

background-position: [left top] | [left center] | [left bottom] | [right top] | [right center] | [right bottom] | [center top] | [center center] | [center bottom];

equivalent to:

background-position: [0 0] | [0 50%] | [0 100%] | [100% 0] | [100% 50%] | [100% 100%] | [50% 0] | [50% 50%] [50% 100%];

If you want advance, w3school will teach you, and my example below:

<!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" />
<style type="text/css">
<!--
* {
	margin: 0;
	padding: 0;
}
body {
	color: #252525;
	font: normal 0.9em/1.4em Arial, Tahoma, Sans-serif;
}
#wrapper {
	margin: 2em auto;
	width: 770px;
}
#tab {
	
}
#tab-buttons {
	height: 31px;
}
#tab-content {
	background: #f0f0f0;
	border: 1px solid #c0c0c0;
	padding: 1em 1em 0;
}
#tab-content p {
	margin-bottom: 1em;
}
.tab_button {
	background-image:url('dark.png');
	float: left;
	position: relative;
	margin-right: -19px; /* depends on your image size */
	width: 143px;
	height: 31px;
	line-height: 31px; /* give the same values of height, the text will be properly align vertically */
}
.tab1 { …

Try this:

<?php 
  $year = 2011;
  $selectedYear = isset($_POST['year_nam']) ? $_POST['year_nam'] : $year; // If you use GET method use $_GET['year_nam'] instead of $_POST['year_nam'] 
?>
<select name="year_nam" id="year_nam">
<?php
  for ($yr_nam = 1950; $yr_nam <= 2020; $yr_nam++) 
  {
    $selected = ($selectedYear || $yr_nam == $year) ? " selected='selected'" : "";
    echo "<option value='$yr_nam' $selected>$yr_nam</option>";
  }
?>
</select>

background: url('*.jpg')
and no-repeat 0 50%; /* the background-image for the cuurent tab */

That for current opened tab background for showing user which tab is currently opening. If you don't want it, never mind.
Remove 'text-align: center' from my old example. Put the 'padding-left' and take the enough space for the text.

.tab_button {
	padding-left: 15px;
}

Hope this help.

Mark this as 'Solve Thread' then.