satishpoul
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About satishpoul
- Birthday 09/22/1991
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PHP,Mysql,c++,java
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mumbai
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in the above code, for (i = 0; i < tr.length; i++) { // for loop is iterating over the each n every row td = tr.getElementsByTagName("td")[0]; // this line is fetching its first column means 0th column to apply the filter in future.. for your need you also need to fetch next column if (td) { if (td.innerHTML.toUpperCase().indexOf(filter) > -1) { // this line is actually important for filtering the result.. you just have to do some code magic with this line according to your need tr.style.display = ""; } else { tr.style.display = "none"; } } }
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use thisprint '<img src="'.$ScanL.'" alt="ScanLine" width="50%" height="50%" border="0" />'; or u can write <img src="<?php echo $ScanL; ?>" alt="ScanLine" width="50%" height="50%" border="0" /> outside php tag
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if you are trying this from localhost server you must have to do some setting like add port for SMTP and all.. but if u use this on live it will work without any settings..
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at the start in php you are checking for $_POST['page_mode'] and if it is not set then ypur are assigning it as a value = ' ' (null/blank) instead of this i wold like suggest if(isset($_POST['$page_mode'])){ $page_mode == $_POST['$page_mode'];if ($page_mode == 'setbio'){ $id = $_SESSION['id'];$find = mysql_query("SELECT bio FROM users WHERE id='$id'");$bio = mysql_fetch_assoc($find);$biotext = $_GET['bio'];$update = db_query("UPDATE users SET bio='$biotext'WHERE id='$id'");}} and check your code by adding echo in each if as well as else loop.. you will get your answer.
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i think according to me you just add one more column to your database as status . when your user process for login.. after successful login just update status as a online otherwise default was offline (you can use true/false or 1/0) or copy logged in users data to your new table where you want to store logged in users and same for offline users.
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when u submit or click on download js function just like in following examplehttp://www.w3schools.com/php/php_ajax_database.aspand put the validation and alert message whatever u want to set.and if all goes well thensend the data for mailing function in php just like in example from above link ===>> xmlhttp.open("GET","mail.php?q="+str,true);(q as a paramete to send mail) and then in mail.php u can do whatever u want to do.or give him link instead of table in above example.
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problem is here action='<?php echo $_SERVER['PHP_SELF'] ?>'> with "" and 'so in action either write "" out side and '' inside or viceversaaction = "<?php echo $_SERVER['PHP_SELF']; ?>"oraction = '<?php echo $_SERVER["PHP_SELF"]; ?>'
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i think you required php_ajax just look at this example:http://www.w3schools.com/php/php_ajax_database.asp
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what exactly u want after filling details and after clicking download buttonand what exactly u want without filling details and clicking on download button.
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try this=><?phpif(isset($_POST['cf_name'],$_POST['cf_email'],$_POST['cf_message'])){$field_name = $_POST['cf_name'];$field_email = $_POST['cf_email'];$field_message = $_POST['cf_message']; $mail_to = 'dedalusonline@yahoo.com';$subject = 'Message from a site visitor '.$field_name;$body_message = 'From: '.$field_name."\n";$body_message .= 'E-mail: '.$field_email."\n";$body_message .= 'Message: '.$field_message;$headers = 'From: '.$field_email."\r\n";$headers .= 'Reply-To: '.$field_email."\r\n";$mail_status = mail($mail_to, $subject, $body_message, $headers);if ($mail_status) { echo "<script>\n"; echo 'var str = \"download\"' . "\n"; echo "document.write(str.link(\"http://www.myshedplans.com/12BY8SHED.pdf\"));\n"; echo "</script>\n";}}else{ echo "<script language=\"javascript\" type=\"text/javascript\">"; echo "alert('Message failed. Please, send an email to gordon@template-help.com');"; echo "window.location = 'contact_page.html';"; echo "</script>";}?>
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becouse your output is Dec not Decmber so specify condition for Dec also
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can't send html link with php value from isset echo to next page
satishpoul replied to thomas_xyz's topic in PHP
you given url and i dont think there is any problem. all the div's and images are very well displayed. -
thankx again everybody who replyed on here.i am gonna refer these links http://coursesweb.ne...ery/jquery-ajax http://jsfiddle.net/YXMPn/ for my concept. and if u also have any other links to refer please post here.
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yes Birbal thankx a lot man i get the entire concept. just nw waiting for technologies or functions to implement on each process.
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