vikramjeet.singla
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file_exists works same for directories as well. So if you have image name empty in database so that path becomes from this site_url()."theme/default/images/". $pros['image'] to site_url()."theme/default/images/" which is a valid and available directory for file_exists function. So, always use a file_exists function with combination of is_file as below: if(!file_exists(site_url()."theme/default/images/". $pros['image']) && !is_file(site_url()."theme/default/images/". $pros['image'])){ // diplay no image } else { // display actual image }
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use mysql_escape_string to escape slashes or any special characters in your search string
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a bit rusty with my PHP - need help with fopen
vikramjeet.singla replied to simcoweb's topic in PHP Coding Help
just explode the value using pipeline as delimiter explode("|", $value); -
i am agree from "ranjuvs"... moreover please check whether file exists before deleting... try this if(is_file($row["PROPERTY_IMAGE"]) && file_exists($row["PROPERTY_IMAGE"])){ // delete file }
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please try this: <!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> </head> <style type="text/css"> form{ margin:0px; padding:0px; } .lft{ position:relative; float:left; } #changeForm p { color: #999; font-size: .9em; padding: 0 0 8px 12px; } #search { background:transparent url(changesearch.gif) repeat-x scroll 0 0; border:0 none; font-family:Arial,Helvetica,sans-serif; font-weight:bolder; height:30px; width:400px; padding-top:4px; font-size:1.0em; } </style> </head> <body> <form action='#' id='changeForm' method='post'> <input alt='left' border='0' src='left.gif' type='image' id="left" class="lft" /><input id='search' name='name' size='65' type='text' value='' class="lft" /><input alt='Go' border='0' src='right.gif' type='image' id="right" class="lft" /> <p>Type a change in the search bar above or browse others below</p> </form> </body> </html>
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see this: http://www.w3.org/TR/html4/struct/links.html#edef-A if you want a big tooltip then you have to create it by javascript or find any built in code...
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you can have check this by isset and count like this: $arr = array(); if(isset($arr) && count($arr) > 0){ // put your logic here }
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Need some help on this mysql query problem.
vikramjeet.singla replied to mabus's topic in MySQL Help
you have to group by this month and year of your date column.... -
use this: <?php var_dump(pow(2, ); // int(256) echo pow(-1, 20); // 1 echo pow(0, 0); // 1 echo pow(-1, 5.5); // PHP >4.0.6 NAN echo pow(-1, 5.5); // PHP <=4.0.6 1.#IND ?>
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suppose you have these options in select male and female than you can use as mentioned below: <?php $a = array('male'=>'Male', 'female'=>'Female'); ?> <select name="abc"> <?php foreach($a AS $key=>$value){ echo "<option value=\"".$key."\">".$value."</option>"; } ?> </select>
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you can do this by ajax.... you can place onclick event on consultant link and call consultant.php file thorugh ajax call and update information div with the response....
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what you are doing is overriding the value $a_diary each time.... you may try this: // Setup Date Array. Start of week is 03082008 for ($wd = 0; $wd < 7; $wd++) { $col_name = date('dmY', mktime(0, 0, 0, date("m",$StartOfWeek) , date("d",$StartOfWeek)+$wd, date("Y",$StartOfWeek))); $a_diary[$col_name] = array("00" => ".", "01" => ".", "02" => ".", "03" => ".", "04" => ".", "05" => ".", "06" => ".", "07" => ".", "08" => ".", "09" => ".", "10" => ".", "11" => ".", "12" => ".", "13" => ".", "14" => ".", "15" => ".", "16" => ".", "17" => ".", "18" => ".", "19" => ".", "20" => ".", "21" => ".", "22" => ".", "23" => "."); echo '<br>col_name = '.$col_name; echo '<br>$a_diary['.$col_name.']["15"] = '.$a_diary[$col_name]["15"]; } echo '<br>Show diary array for the week'; echo '<br>$a_diary["03082008"]["15"] = '.$a_diary["03082008"]["15"]; echo '<br>$a_diary["04082008"]["15"] = '.$a_diary["04082008"]["15"]; echo '<br>$a_diary["05082008"]["15"] = '.$a_diary["05082008"]["15"]; echo '<br>$a_diary["06082008"]["15"] = '.$a_diary["06082008"]["15"]; echo '<br>$a_diary["07082008"]["15"] = '.$a_diary["07082008"]["15"]; echo '<br>$a_diary["08082008"]["15"] = '.$a_diary["08082008"]["15"];
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please use this <?php $myServer = "localhost"; $myUser = "sa"; $myPass = "mypassword"; $myDB = "mmt"; $dbhandle = mssql_connect($myServer, $myUser, $myPass ) or die("Couldn't connect to SQL Server on $myServer"); $selected = mssql_select_db($myDB, $dbhandle) or die("Couldn't open database $myDB"); mssql_close($dbhandle); ?> you have not mentioned the password while connecting to sql server
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[SOLVED] PHP/MySQL not returning any results
vikramjeet.singla replied to itsjames's topic in PHP Coding Help
as mentioned by Ace always enquote strings with single quotes... -
As mentioned by "ProjectFear" use post as a method in form tag.... double check your form method...