PravinS
-
Posts
459 -
Joined
-
Last visited
-
Days Won
2
Posts posted by PravinS
-
-
-
mysql_connect() function is not supported in PHP 5.5.0 and your PHP version may have been upgraded to 5.5.0
check this url: http://php.net/manual/en/function.mysql-connect.php
-
echo tha $_POST['who] value at the top and check what value you are getting after submitting the form.
-
Yes, use AJAX to get current quantity($qty) value
-
Try this:
echo date("m",strtotime('2014-08-20 07:32:20')); -
Try TCPDF
-
for PHP SOAP you can try NuSOAP Toolkit
-
you can use AJAX
-
you can check it in jsfiddle for the output
-
search the following line
html += '<td>' + i + '</td>';
and replace it with
if (i == d.getDate()) html += '<td style="color:blue;">' + i + '</td>'; else html += '<td>' + i + '</td>'; -
Use
$result = json_decode($string,true);
you will get $result in array format
-
Check this simple logic
<table border="0" cellspacing="0" cellpadding="5"> <tr> <?php $array = array(1,2,3,4,5,6); $i = 1; foreach($array as $val) { echo '<td>'.$val.'</td>'; if ($i % 2 == 0) echo "</td></tr>"; $i++; } ?> </tr> </table>I hope this will help you
-
you have only 1 while loop and 2 endwhile
-
You can use it like this also
<?php $link = '<a href="post?id='.$id.'&title='.$slug_title.'">'.$title.'</a>'; ?> -
you need to check the MySQL query
$query = urldecode($query);
echo the above line and check the query
-
try FCKeditor or TinyMCE
-
why do you want to split the form, it can be done in single form
-
you have used "Like" as column name, it is reserved word in mysql
-
you can use PHP time zone instead of javascript time zone
-
remove the back ticks (``) from VALUES of INSERT query and use single quotes (' ')
$query = "INSERT INTO user(user,pass,priv,mail,avatar,date) VALUES('$user','$pass','$priv','$mail','$avatar','$date')";may this works for you
-
check the SQL query, whose resource id is passed to mysql_fetch_array() function at mentioned line or else show us the code of C:\xampp\htdocs\index.php
-
use this updated javascript function
function checkForm() { var obj = document.getElementsByName('dest[]'); var destCount = obj.length; var destSel = false; for(var i = 0; i < destCount; i++) { if(obj[i].checked == true) destSel = true; } if(destSel == false) { alert('Select one or more destinations'); } return destSel; } -
use this code
<html> <head> <script language="javascript"> function checkForm(){ var obj = document.getElementsByName('dest[]'); var destCount = obj.length; var destSel = false; for(i = 0; i < destCount; i++){ if(obj.checked == false){ destSel = true; break; } } if(!destSel){ alert('Select one or more destinations'); } return destSel; } </script> </head> <body> Select at least one destination <form action="" method="post"> <input id="cx" type="checkbox" name="dest[]" value="CX" /> <label for="cx">Cox's Bazar</label><br /> <input id="su" type="checkbox" name="dest[]" value="SU" /> <label for="su">Sundarban</label><br /> <input id="sy" type="checkbox" name="dest[]" value="SY" /> <label for="sy">Sylhet</label><br /> <input id="ch" type="checkbox" name="dest[]" value="CH" /> <label for="ch">Chittagong</label><br /> <br /> <input type="submit" name="Go" value=" Go " onclick="return checkForm();"/> </form> </body> </html>also check the jsfiddle
-
Just remove onsubmit="return checkForm(this);" from form tag and add it to button as onclick="return checkForm(this);"
Use this
<form action="" method="post"> <input id="cx" type="checkbox" name="dest[]" value="CX" /> <label for="cx">Cox's Bazar</label><br /> <input id="su" type="checkbox" name="dest[]" value="SU" /> <label for="su">Sundarban</label><br /> <input id="sy" type="checkbox" name="dest[]" value="SY" /> <label for="sy">Sylhet</label><br /> <input id="ch" type="checkbox" name="dest[]" value="CH" /> <label for="ch">Chittagong</label><br /> <br /> <input type="submit" name="Go" value=" Go " onclick="return checkForm(this);"/> </form>
Problem showing top 10 highest score
in PHP Coding Help
Posted
SELECT TOP is not supported in MySQL, you need to use LIMIT
try this