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Hi every one,
I am getting Gateway Timeout Error when running php script.Script was running fine before after some hours it is giving this error.Don't know why.The script was to connect to DB and insert some data into tables.It is getting timeout even before connecting to DB.
Can any one tell me any reason why this happening.
Regards,
Pavan,Independent Programmer.
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a quick google came up with this...
504 Gateway Timeout
ErrorDocument Gateway Timeout | Sample 504 Gateway Timeout
The server was acting as a gateway or proxy and did not receive a timely request from the upstream server.
If the script was working fine and then it is not, I would look into newly installed software and/or changes to the server configuration.
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Hello sir,
i have some knowledge of C
but can u tell me how can i Start programing for software and Internet Application software....and OS making software ? Which programing languages are best for me to start.
and other program....
Please gave me some support.
Thank you for your support
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Buy a book on <insert language="" here="">
I prefer C++, C, C# and VisualBasic.Net
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thank you for your response and help.
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Can anyone show me a short perl test script to print a crystal report?
Thanks,
Steve
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How to do kickstart without using the linux booting cd........?
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Careful you may hurt your toes!
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hi, thnx for ur reply.....
but please do know about the subject before replying......
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hi, thnx for ur reply...
please let me know how could we process from the LAN connections.
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Hi. I a'm writting a simple application on php 5 and zend_framework. I need to create two dropdown lists that depends on each other. Cities and streets for example. I select a new city at my client page and want that my application post me at the other dropdown list streets related to this city in the database. Which would be the better way to do it?
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simple is relative but I would write an ajax call to a php server side script that returned the street data. All of this would started by using a onchange event of the city select from the display page.
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Tnx. Now i know where to look. Big tnx.
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Hi there !
Im using Joomla 1.5 to build my site.
My component(query data from database of website) work well if i don't install joomfish.
After i installed Joom!Fish 2.0.4 Stable into my site for translation my component does not work(with error).
My code of component as bellow:
$query='select * from jos_content where sectionid=11 and state=1
order by catid ';
$database->setQuery( $query );
$result=mysql_query($query);
while($row=mysql_fetch_object($result)){
echo 'data'.$row->title;
}
And i see these bellow warnings if i refer to my component.
Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\so\components\com_products\products.php on line 131
Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\xampp\htdocs\so\components\com_products\products.php on line 131
Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\so\components\com_products\products.php on line 137
Any ideas !
Thanks in regards !
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sharkbc wrote: Access denied for user 'ODBC'@'localhost' (using password: NO)
Check your configuration file as it appears you can't connect to the database using your credentials. Your new install probably overwrote the config file.
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Dear Marc Firth !
The configuration file has no change(I checked).
My website "works well" except my component.
Thanks anyway !
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I found solution. It now work well
sharkbc
Free ware
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Hi Guys,
I am using MySQL Administrator 1.2.17. And my problem is I cannot access the MySQL remotely from the web server. I have the dbHost, dbUsername, dbPassword. But when I tried accessing it through MySQL Administrator tool I got an error that says...
Error:
Could not connect to the specified instance. MySQL Error Number 2005
Unknown MySQL server host '...' (11004)
Is there a way on how can I get through this?
Thanks
hifiger2004
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yeah, fix whatever is in your dbHost variable...
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Hi all.
I am new to PHP and trying to learn it.I have created a web page and defined items as below :
<form method="POST" action="send.php">
<font color="green"> Full Name : </font>
<input type="text" name="name" size="20">
<font color="green"> Gotra(Clan) : </font>
<input type="text" name="gotra" size="20">
<font color="green"> Father's Name : </font>
<input type="text" name="fname" size="20">
<font color="green">Details of Prasad to be Offered : </font>
(eg.: name of temple, quantity of prasad, specific date & any other requirements)
<textarea rows="5" name="prasad" cols="40" ></textarea>
<font color="green"> Any Wish to God : </font>
<textarea rows="5" name="wish" cols="40" ></textarea>
<font color="green"> Postal Address : </font>
<textarea rows="5" name="address" cols="40" ></textarea>
<font color="green"> Phone no. : </font>
<input type="text" name="phone" size="20">
<font color="green"> Preferred Time of Contact : </font>
<input type="text" name="time" size="20">
<font color="green"> Email address : </font>
<input type="text" name="email" size="20">
<br>
<input name="Submit" type="submit" class="input" value="Send">
<input name="Submit2" type="reset" class="input" value="Reset">
</br>
</form>
and the send.php has following block of code :
<?php
if(isset($_POST['submit'])) {
$to = "gagan.u20@gmail.com";
$subject = "E-Prasad Details";
$name_field = $_POST['name'];
$gotra_field=$_POST['gotra'];
$fname_field = $_POST['fname'];
$prasad=$_POST['prasad'];
$wish=$_POST['wish'];
$address=$_POST['address'];
$phone_field=$_POST['phone'];
$time_field=$_POST['time'];
$email_field=$_POST['email'];
$body = "From: $name_field\n Gotra: $gotra_field\n Father's name: $fname_field\n Details of prasad to be offered : $prasad\n Any wish to God: $wish\n Postal address : $address\n Phone no. : $phone_field\n Preferred time of contact : $time_field\n E-mail address : $email_field";
echo "Data has been submitted!";
mail($to, $subject, $body);
} else {
echo "An error occurred! Please try again.";
}
?>
Now the problem is, when I fill the form and click og sublit button, it gives an error that "An error occurred! Please try again."
Tell me where am I wrong in writing the code. Tell me the correct code.
Thanks.
Gagan
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Why do you check
if(isset($_POST['submit'])) ? Obviously that is false and you've trapped the problem and displayed the error message you wrote.
CQ de W5ALT
Walt Fair, Jr., P. E.
Comport Computing
Specializing in Technical Engineering Software
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ok, then tell me what should I write to send e-mail because I am new to PHP.
Help me.
Thanks.
Gagan
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Your email sending part looks fine. I use
mail($to, $subject, $contents, $from_header); with the $from_header, although that might be optional. But that's not the problem you are seeing -- your code isn't even getting there.
Looking at your PHP code, the first thing executed is
if(isset($_POST['submit']))
Since you get the message you showed, obviously that is being evaluated as false, so the else part is executed showing the message.
So my question to you is: Why are you checking to see if the 'submit' POST variable is set? That's where your code is having problems -- before it ever gets to sending the email.
My suggestion is to take the if/else out and run the code to send the email. That should work if email is allowed on your server. If it works, then you can go back and check for whatever you want to check for with the if statement.
See the following links for examples:
http://www.webcheatsheet.com/php/send_email_text_html_attachment.php[^]
http://php.about.com/od/phpapplications/ss/form_mail.htm[^]
CQ de W5ALT
Walt Fair, Jr., P. E.
Comport Computing
Specializing in Technical Engineering Software
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