PHP - Mysqldump On The Client Side

I want to do simple client side validation but I don't know Javascript.  Can someone point me to an easy example on how to implement or post something here?

We have a website that allows the user to upload his/her picture so we use the ftp_put() command.
I use Code: [Select]
<?php
$path = getcwd();
echo $path;
?>
This is the output:
Code: [Select]
/home/vol10/000a.biz/a000b_7363341/htdocs/gankgame
to determine the absolute path where my website is stored. My website "CAN" connect to the ftp server but it can't upload files. Uploading always fail.

I can't determine the right path.

"My code is attached"

Is it possible adding a button with PHP, which could be used for adding an optional text input field by client side user? My purpose is actually very simple: The client side users could be free creating and adding more and customized variables, which are intended to be used as strings and related or associated variables.
So, am I able to do these kind of tasks simply using PHP alone? Or, should I use some JavaScript like technologies to make those functions available in real world's practices?

Hi All - I'm trying to see if the below is possible using PHP.

is it possible for php to allow a user to check if a port is open/listening on a remote server/PC, from their local machine. I've seen scripts on the web that talk about opening cmd.exe - this won't work as not all of my users will be on windows. Also saw fsock.. but the remote server is not serving http page(s). Also saw telnet, but again, some of my users may not have that windows feature enabled.

Desired workflow:

1. user enters the server hostname/IP and a port

2. my php webapp will emulate checking if that hostname and port is open from the local user's computer.

 

thanks!

<?php
$pathToMysqldump = "C:\xampp\mysql\bin";
$user= "root";
$pass= "";
$namedb= "test";
$output= "C:\xampp\htdocs\test\test.sql

//backup saved
exec("$pathToMysqldump --user=$user--password=$pass $namedb> $output");

//header('Location: http://localhost/test);
exit;

Hi people, i cant do it work, I use xampp and back dont creat the file .sql.

Thanks a lot.

 

 

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=319081.0

Any help on this would be greatly appreciated.  I am trying to run a mysqldump from my site for my client when they want to back up.  They basically will jsut click a button to run the backup. 

With my script below, the backup file is genrated, but there is no table data in the file.  There are the headers in the file with the server IP, linux versin, etc..., but there is no data in the file.  Does this look right?  Thank you for helping me out.

Ryan

Code: [Select]
ini_set ("display_errors", "1");
error_reporting(E_ALL);

$dbhost = 'localhost';
$dbuser = 'databaseuser';
$dbpass = 'password';

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die                      ('Error connecting to mysql');

$dbname = 'database';
mysql_select_db($dbname);

$tableName  = 'lots';
$backupFile1 = '/home/stuff/wwwroot/stuff/appnew/backup/';
$backupFile = $backupFile1.$dbname . date("Y-m-d-H-i-s") . '.gz';
$command = "mysqldump --opt -h$dbhost -u$dbuser -p$dbpass $dbname | gzip > $backupFile";
system($command);
$result = mysql_query($command);

  I’m trying to write a post-panel where the user can see the preview of his post to the right of the
text area he’s writing into.
  I tried the following into a file called writepost.php :
 

<?php

$text=(isset($_GET['message']))?$_GET['message']:'';
$formatted_text=nl2br(stripslashes(htmlspecialchars($text)));

echo '<form method="post" action="proceedtopost.php?>'.
'<table>'.
'<td style width="50%"><tr>'.
'<fieldset> Write your post here : <br> <textarea cols="50" rows="12" '.
'id="message" name="message">'.$text.'</textarea>'.
'</fieldset> <p> '.
'<input type="submit" name="submit" formaction="writepost.php" value="Preview" /> '.
'<input type="submit" name="submit" value="Sent" /> '.
'</p>'.
'</td>'.
'<td>'.
'<fieldset> Your post will appear as follows :<br><p> '.
$formatted_text.
'</fieldset>'.
'</td></tr>'.
'</table>'.
'</form>';

There are several things wrong with this code :

  1) When the user hits the "Preview" button, I expect writepost.php to be reloaded (this is what happens), and the current content of the textarea to be stored in $_GET['message'] (this is not what happens).

  2) Why does my browser output the preview part under the text area (or in other words outputs the HTML table as a single column of two cells), when I insist in my HTML code for the table to be displayed  as a single row ?
 

need a little help guys! I use the script below to display profile images, trouble is it shows 1 on top of the other, and i need it to double up 2 profile images on top of 2 profile images ect any ideas how i can do this. 



require("./include/mysqldb.php");

$con = mysql_connect("$dbhost","$dbuser","$dbpass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$dbame", $con);
$result = mysql_query("SELECT * FROM Search_profiles_up WHERE upgrade_one ='1' ORDER BY RAND() LIMIT 40");


print "<table width=\"293\" height=\"111\" border=\"0\">
  <tr>\n";

while($row = mysql_fetch_array($result))
  {
print "<td width=\"142\"><img src=" . $row['search_small_image'] . " width=\"144\" height=\"169\" /></td>\n";
print "  </tr>\n";
print "  <tr>  \n";
print "<td>" . $row['star'] . "</td>\n";
print "  </tr>\n";
print "  <tr>\n";
print "<td>" . $row['username_search'] . "</td>\n";
print "  </tr>\n";
print "  <tr>  \n";
print "<td>" . $row['phone_search'] . "</td>\n";
print "  </tr>   \n";

  }
print "</table>";

mysql_close($con);
?>


 

I have a few tables in php and want them to be side by side instead of under each other like below. I have spent about 2 days trying to figure it out, Anyone got a clue?


This is the code im using

<?php
$con = mysql_connect("localhost","user","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("People", $con);

$result = mysql_query("SELECT * FROM wang");

echo "<table border='1'>
<tr>
<th>Wang Total Report</th>

</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['got'] . "</td>";
   echo "</tr>";
  }
echo "</table>";

$result2 = mysql_query("SELECT * FROM miah");


echo "<table border='1'>
<tr>
<th>Miah Total Report</th>

</tr>";

while($row = mysql_fetch_array($result2))
  {
  echo "<tr>";
  echo "<td>" . $row['got'] . "</td>";
   echo "</tr>";
  }
echo "</table>";

$result3 = mysql_query("SELECT * FROM vinc");


echo "<table border='1'>
<tr>
<th>Vinc Total Report</th>

</tr>";

while($row = mysql_fetch_array($result3))
  {
  echo "<tr>";
  echo "<td>" . $row['got'] . "</td>";
   echo "</tr>";
  }
echo "</table>";



$result4 = mysql_query("SELECT * FROM ketarie");


echo "<table border='1'>
<tr>
<th>Ketarie Report</th>

</tr>";

while($row = mysql_fetch_array($result4))
  {
  echo "<tr>";
  echo "<td>" . $row['got'] . "</td>";
   echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?> 

Dear all,

I'm working on a website that is hosted on a server with 10Mb of file size limit via HTTP file uploading approach (eg. PHP file uploading form using move_uploaded_file()), although there's no file size limit via FTP file uploading approach (eg. Using FTP client).

Therefore, I was thinking about using PHP FTP functions to bypass the file size limit. However, when I upload file bigger than 10Mb, I still got this error message telling me that my file can't be over 10Mb...

So my question is...what exactly are the differences between PHP FTP and FTP Client, that the web server used to determine whether to block the file bigger than 10Mb or not?

Thank you very much!

Below is my code:

Code: [Select]
<?php
echo '<form action="" method="post" enctype="multipart/form-data">';
echo 'Click the Browse button to find the file you wish to upload';
echo '<input type="file" name="add_file">';
echo '<INPUT TYPE="submit" name="upload" value="upload">';
echo '</form>';

if($_POST['upload']) {

$add_file = $_FILES['add_file'];

//change these values to suit your site
$ftp_user_name = 'username';
$ftp_user_pass = 'password';
$ftp_server = 'ftp.domainname.com';
$ftp_root = '/';

//File upload
$temp_path = $ftp_root . $add_file['tmp_name'];
$target_path = basename($add_file['name']);

$max_size = 26214400;

if($add_file['error'] == 0 && $add_file['size'] < $max_size) {

// set up basic connection
$conn_id = ftp_connect($ftp_server);

// login with username and password
$login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass);

// upload a file
ftp_chdir($conn_id, '/root/second/');
if (ftp_put($conn_id, $target_path, $temp_path, FTP_ASCII)) {
 
echo "successfully uploaded " . $target_path;

} else {

echo "There was a problem while uploading " . $target_path;
}

} else {

echo "There's been an error, please try again later.";

}

// close the connection
ftp_close($conn_id);

}

?>

I'm trying to access webservices using PHP's soap. Here's my code:

<?php

// test connection to agwebservice

     $agLogURL =    "https://www.agemni.com/_snet/AgemniLogin.asmx?WSDL";
     $agDN =        "myDN";
     $agUserName =  "myUserName";
     $agPass =      "myPassword";

     $login = new SoapClient($agLogURL);

      $login->aglogin($agDN, $agUserName, $agPass);

//     $login.service.agLogin($agDN, $agUserName, $agPass);

?>

When I run the script, here's the resulting error:

Fatal error: Uncaught SoapFault exception: [soap:Server] System.Web.Services.Protocols.SoapException: Server was unable to process request. ---> System.NullReferenceException: Object reference not set to an instance of an object. at AgemniLogin.agLogin(String dn, String username, String password) --- End of inner exception stack trace --- in C:\XAMPP\xampp\htdocs\agsync\syncwork.php:12 Stack trace: #0 [internal function]: SoapClient->__call('aglogin', Array) #1 C:\XAMPP\xampp\htdocs\agsync\syncwork.php(12): SoapClient->aglogin('myDN', 'myUserName', 'myPassword') #2 {main} thrown in \htdocs\agsync\syncwork.php on line 12


Not sure if you'll need more info on the SOAP services, but the info is he
http://wiki.agemni.com/Getting_Started/APIs/Agemni_CMS_Sync#Method_details


Really could use any help anyone can offer:

I feel like I was 75% there, and now I'm only 25% there.
I'm trying to put my links onto my page. http://thingsihateaboutyou.net/michelle is my testing page.

So I have register (register.php), login (login.php), logout (logout.php), and the members area. The members area is going to be their username.php, so I assumed this would be $username.php like I have it in the redirect link from the point where you login to the point where you see your client page. This works fine.

The part I'm having trouble with is the visibility of the links on my page to these, AND adding a PHP variable ($username) into the "members area" tag. When you click on members area, it should send you to http://thingsihateaboutyou.net/michelle/*whatever you are logged in as*.php

As for the visibility, if you are not logged in I obviously don't want you to see a link called "log out" or a link called "members area," and vise versa. It's been like 5 hours now, and I keep getting further and further away. I had to bail on my Wordpress so I don't have a Content Management System in place. All of this DID work and it worked excellent, but after I made the switch earlier today it kind of blew up in my face.

I am attaching a ZIP file with everything included... in desperate need of an answer.

Hi, guys. I just developed and published first version of 12framework for iOS which can call also PHP web services. It is web service client which generates dynamic forms from web service results, and enables insert, update and delete of records. You can simply change result field type to text, memo, date, bool. I also made some simple PHP web service to test functionallity of application, so I will really appreciate your feedback. iOS is locked platform, so 12framework provides possibility, to make some useful data forms that works as native application. Link: https://itunes.apple...88083?ls=1&mt=8