What are causes of a connection-timeout with fopen()?
Posted on
16th Feb 2014 07:03 pm by
admin
Hello! Here is the situation: The server I host my website on just upgraded it's PHP build from 4.4.* to 5.2.4. Not just PHP was upgraded, MySQL, postGres, and (I think) Apache were upgraded as well. Before the upgrade, my site worked perfectly.
But now, whenever I access my site, it takes a long time to complete a request and all that is returned is "Warning: fopen(<THE_URL>) [function.fopen]: failed to open stream: Connection timed out in..."
Did you know?Explore Trending and Topic pages for more stories like this.
I know where the error is -- it's in my templating class. I check the remote server's configuration and the directive "allow_url_fopen is enabled and safe_mode is disabled.
The strange thing is that my script works on my local server perfectly which is using PHP build 5.2.5. I looked for any differences between my local-server's PHPINFO() and my remote-server's PHPINFO() and, while I didn't find anything eye-catching yet, I noticed that virtual-directory-support is disabled on my remote-server and it is enabled on my local-server. Would that be what's causing the problem? If not, what are some potential causes for fopen() to behave this way?
No comments posted yet
Your Answer:
Login to answer
228
12
Other forums
help with php
It's not displaying or pulling anything out of the database.
Code: <html><
phpmailer class & pop.gmail.com?
Code: <?php
$mail->IsSMTP();
$mail->Host = "pop.gmail.com";
default SAP userid
hi,, I just like to know if it is ok to use the default SAP user id (SAP*)?
Seperate team from score
I am writing a site that does a NFL Pick 'em type application and I have a feed that gives me the sc
Displaying image from database
Hi,
I've got a site where that's got a database behind it. Currently it has loads of items in
Form a inline view based on the results of the previous query?
Hi Experts,
Can we form a inline view on the results of a previous query (another inline view)?
having a small php error tha deals with the 'foreach'
Last night i was working on a script for a cart and it seemed to working good, until this morning. F
login from external site
Hi my new experience begins, Now what i am trying to do is i make three pages, login.php logout.php
PHP page is blank
Hi Everyone,
I have a site in which I am able to open the first PHP webpage in my browser but
Not Loading Function Into Div
I'm not sure whether to put this under the php forum or ajax forum but because I tink it's more of a