[Rewrite Not Working]

mod_rewrite is not a PHP extension! It is an Apache module, which you can enable in the httpd.conf

 

yeah, thanks for correcting

[Rewrite Not Working]

check if mod_rewrite is enabled in php.ini

[Uploading image file from form]

while uploading files use $_FILES instead of $_POST for file data

$fileName = $_POST["uploaded_file"]["name"]; // The file name
$fileTmpLoc = $_POST["uploaded_file"]["tmp_name"]; // File in the PHP tmp folder
$fileType = $_POST["uploaded_file"]["type"]; // The type of file it is
$fileSize = $_POST["uploaded_file"]["size"]; // File size in bytes

the above code should be

$fileName = $_FILES["uploaded_file"]["name"]; // The file name
$fileTmpLoc = $_FILES["uploaded_file"]["tmp_name"]; // File in the PHP tmp folder
$fileType = $_FILES["uploaded_file"]["type"]; // The type of file it is
$fileSize = $_FILES["uploaded_file"]["size"]; // File size in bytes

[Join two tables]

may LEFT JOIN will help you

$sql="SELECT * FROM tweet LEFT JOIN follow ON tweet.user_id=follow.user_id  WHERE follow.user_id = '".$_SESSION['user_id']."' ORDER BY date_time desc";

[Please tell me what's wrong?]

try using this

$num = mysql_num_rows($allusers = mysql_query("SELECT * FROM PMs WHERE `status` = '0' AND ReceiveID ='".$myU->ID."' ORDER BY ID"));
$PMs = mysql_num_rows($allusers = mysql_query("SELECT * FROM PMs WHERE `status` = '0' AND LookMessage = '0' AND ReceiveID = '".$myU->ID."' ORDER BY ID"));

[Sending a URL in my parameter list]

try using urlencode() and urldecode() php functions

[trying to validate my page, send message back to submitter, and send email.]

check this line

mail to($to, $subject, $message, $additional_headers);

 

for this refer

http://php.net/manual/en/function.mail.php

[trying to validate my page, send message back to submitter, and send email.]

check this line

echo " Thank you $_POST[fname] !! Your registration has been submitted!!"; && $sendEmail;

remove "&& $sendEmail;"

 

 

if you want to concat $sendEmail, then you can use like this

echo " Thank you $_POST[fname] !! Your registration has been submitted!!".$sendEmail;

[Help to get inline HTML to show correctly]

"http://" may be missing, please check that

[Anybody knows PHP based dashboard library ?]

check highcharts

 

http://www.highcharts.com/demo/

[codeigniter insert data]

you can check this url for that

 

http://ellislab.com/codeigniter/user-guide/general/models.html#loading

[codeigniter insert data]

http://ellislab.com/codeigniter/user-guide/database/active_record.html#insert

[MySQL error]

To, From are reserved words in MySQL, either you can change the column names or use title (``) character around column names, like `To`,`From`

[#1062 - Duplicate entry Error when trying import database]

you are getting this error because the "id" field is primary key and autoincremented, there may be any old entry with 153 value

[need help php dynamic menu]

may this will help you

 

http://www.php-guru.in/2013/dynamic-menu-in-php-mysql-javascript/

[Notice: Undefined variable: con in F:\xampp\htdocs\l.php on line 223 Warning: mysql_close() expects parameter 1 to be resource]

move your database connection at the top of the page

[JavaScipt Newbie. Why is this code not working?]

are you getting error or validation is not working

 

also check your form action, its "success.asp"

[Do not display Radio Button if its value is in DataBase]

you can disable them, which can be visible but cannot be checked

[Notice: Undefined index: comments in F:\xampp\htdocs\contact.php on line 69]

there is no "comments" element in form, i think its "message"

[Warning: mysql_query() expects parameter 2 to be resource, object given in F:\xampp\htdocs\taranaki2.php on line 69]

there is comma(,) before WHERE, remove it

 

echo $UpdateQuery = "UPDATE taranaki2 SET ID='$_POST[id]', Destinations='$_POST[destinations]', Difficulty='$_POST[difficulty]' WHERE ID='$_POST[hidden]' ";

[Warning: mysql_query() expects parameter 2 to be resource, object given in F:\xampp\htdocs\taranaki2.php on line 69]

how many records are you getting for your SELECT query in phpmyadmin

 

SELECT * FROM taranaki2

[Warning: mysql_query() expects parameter 2 to be resource, object given in F:\xampp\htdocs\taranaki2.php on line 69]

check the query in phpmyadmin for the result

 

also check

http://us1.php.net/mysql_fetch_array

[Warning: mysql_query() expects parameter 2 to be resource, object given in F:\xampp\htdocs\taranaki2.php on line 69]

in this line

 

$con = mysql_connect("localhost","root","","taranaki");

 

what is "taranaki", is this database password?, if yes, then it should be 3rd parameter of mysql_connect

 

also echo your SELECT query and check it in phpmyadmin

[Warning: mysql_query() expects parameter 2 to be resource, object given in F:\xampp\htdocs\taranaki2.php on line 69]

show your final code

[2 Divs side by side]

check this sample CSS and HTML

#maindiv
{
    width:500px;
    background-color:#000;
    height:100px;
    padding:3px;
}
#leftdiv
{
    width:250px;
    background-color:red;
    float:left;
    height:100px;
}
#rightdiv
{
    width:250px;
    background-color:#fff;
    float:right;
    height:100px;
}

<div id="maindiv">
    <div id="leftdiv"></div>
    <div id="rightdiv"></div>
</div>