Tanks for your answer. I try it and it does not give me the image in center of the text. Instead, I get the image around left or or right. I use paddings too. Without success. Below, code and result in attachment.
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"
"http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<meta name="description" content="Free Web tutorials on HTML, CSS, XML, and XHTML">
<meta name="keywords" content="HTML, DHTML, CSS, XML, XHTML, JavaScript, VBScript">
<title>image centered around text</title>
</head>
<body>
<b>Without padding</b>
<p><img src="sunset.gif" width="75" height="50" alt="sunset" align="center">Adding your text here will wrap your text around your graphic, with your graphic in the center and your text aligned around the graphicAdding your text here will wrap your text around your graphic, with your graphic in the center and your text aligned around the graphic.Adding your text here will wrap your text around your graphic, with your graphic in the center and your text aligned around the graphic. Adding your text here will wrap your text around your graphic, with your graphic in the center and your text aligned around the graphic.Adding your text here will wrap your text around your graphic, with your graphic in the center and your text aligned around the graphic.Adding your text here will wrap your text around your graphic, with your graphic in the center and your text aligned around the graphic.Adding your text here will wrap your text around your graphic, with your graphic in the center and your text aligned around …
I know how to wrap text around an image by floating it either left or right, etc.. What I try to do is surrounding the image by text as in attachment. This example is made with a page editor (not from a webpage) to show what I need to do.. In fact I'm looking for the corresponding code.
Hi, Need some help to solve this error. Nothing similar in the forum, IMHO.
Environment:
phpMyAdmin – 3.3.9 with this info: # Version MySQL client: 5.0.51a # Version MySQL: 5.0.45 # Server: Localhost via WAMPSERVER 2.1 # PHP 5.2.6 # Apache 2.050
TABLE DROP TABLE IF EXISTS `products_description`; CREATE TABLE IF NOT EXISTS `products_description` ( `products_id` int(11) NOT NULL auto_increment, `language_id` int(11) NOT NULL default '1', `products_name` varchar(64) NOT NULL default '', `products_description` longtext, `products_url` varchar(255) default NULL, `products_viewed` int(5) default NULL, `products_seo_url` varchar(100) NOT NULL, `products_head_title_tag` varchar(80) default NULL, `products_head_desc_tag` longtext, `products_head_keywords_tag` longtext, PRIMARY KEY (`products_id`,`language_id`), KEY `products_name` (`products_name`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;
Error occurs when try to insert the page on the web.
The error: 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '0, products_url, products_head_title_tag, products_head_desc_tag, products_head_' at line 1
insert into products_description (products_name, products_short, products_description, 0, products_url, products_head_title_tag, products_head_desc_tag, products_head_keywords_tag, products_id, language_id) values ('Adotation', '', 'Array', 'L\'Adoration des mages est un thème de l\'iconographie de la peinture chrétienne, etc, ,,.', 'consulter.org', 'Contrairement à beaucoup de tableau', 'seule architecture visible', 'avec des monts au lo', '222', '1')
What looks strange is that: - products_description is populated as …
I'm working on this since 3 days, grabbing the web looking after ideas. Hope someone faced this problem and solved it. All is about a picture site Here how it's doing: There are about 15 items with different options offered to the customers. A combination of size, number of figures, frame and quantity determine the price (see attached file). The customer have to upload an image (500k or less in.jpg/.png) and add comments about his order ( 100 words max). Each uploaded file has to be stored in dbase_MySQL (yymmddhhmmss.jpg/png) with the possibility to insert it in an invoice page. As price will change any time, it will be hard-coded. At last, the number of figures is limited for some small sizes and the framing is not possible for big sizes. codes I can adapt to fit my project. Instead of doing all from scratch, is it possible to indicate me some Even I am very certain its a lot of work, I hope to get help from you. The attached file explain more than 10 000 words what it is.
When sumit button is clicked I get an warning: "mysqi-num_rows(): supplied argument is not a valid mysql resource in line 17" at newdbfunction.php.
Then, when I check the error with the code: "$ ret = mysql-query($query) or dir(mysql_error());", the result is: [Price] Non available / mysql query is empty. See the table price in attachment. My question is: Why it does not "see" the table?
We set up an art website and the table is a kind of price list. So, all data are predetermined and not calculated. It's in fact a 10x10 table.
I exagerrate a bit when I tell I'm completely Null. I can read and understand a little bit a PHP or JS simple program. The fact is that I'm a newby and not really illiterate.
How can the user obtain a specific result by choosing values from 2 dropdown list. As a result, the value obtained can not be modified by the user.The table of ten rows by ten columns have to be stored in a mysql dbase.. It's certanly possible , but tedious and time-consuming by using a bunch of "if ... then statements".
I need some help, as I'm NULL in JavaScript, Ajax and PHP.
An illustration worthing 1000 words ...
From the matrix table, the final result is the coordinate of arow value and a column value.
Sorry, I think I was just trying to confuse the issue because of the asp.net coding base. It was really just a matter of simple javascripting and onclick event, I went back to the basics and I solved the problem.
Hello, I have the same problem embedding upload pictures on table. Can you tell me how you solve the problem.
Hello all, This is my first post. May be my problem is a newby one, but... I try to solve it since 4 full days. So I ask for some help. I picked up an order form from www.dyn-web.com. It works fine and make the job I want except I must to change the shipping policy from this Original shipping options:
// shipping options: radio button text, value //(quantity ordered not taken into account in this form) $ship_options = array ( array ("Pick up locally", "0"), array ("UPS Ground", 5), array ("Fed Ex Overnight", 20) );
to that:
Contry 1st item 2nd items + Canada 6.50$ 6.50$ + 2.50$/additional item USA 8.50$ 8.50$ + 3.50$/additional item Others countries 9.75$ 9.75$ + 4.50/additional item and for more than 9 item, call for prices.
