bhavin12300

please any one,any kind of help might take me out of this.

sad day, again. anyone please.

any one, any help.plzzz

hi experts, i got stuck during my project, need bit of your advice to find correct path. here is what i am doing in my php code. 1)my script create account on third party website(i have no control over this website) 2)there are 5 forms which i have to submit one by one to create account. 3)ii have successfully created code to submit first two forms and after submitting second form , i am getting third form in my response in script. 4)but when i am submitting data for third form, i am getting response in red text that "Please fill up all fields", i am getting this error message, which means that my code is submitting values to "action" form of third form, but some how its not accepting it or may be possible, my curl doing something wrong while sending data. all 5 forms are accepting data of content-type "multipart/form-data" and thats y i am sending all my data in array and assigning that array to "CURLOPT_POSTFIELDS". following is data send by browser to "action" script of form 3. -----------------------------792694871670 Content-Disposition: form-data; name="1" (Car#21) -----------------------------792694871670 Content-Disposition: form-data; name="2" (Car#22) -----------------------------792694871670 Content-Disposition: form-data; name="3" (Car#23) -----------------------------792694871670 Content-Disposition: form-data; name="4" (Car#24) -----------------------------792694871670 Content-Disposition: form-data; name="5" (Car#25) -----------------------------792694871670 Content-Disposition: form-data; name="6" (Car#26) -----------------------------792694871670 Content-Disposition: form-data; name="7" (Car#27) -----------------------------792694871670 Content-Disposition: form-data; name="8" (Car#28) -----------------------------792694871670 Content-Disposition: form-data; name="9" (Car#29) -----------------------------792694871670 Content-Disposition: form-data; name="10" (Car#30) -----------------------------792694871670 Content-Disposition: form-data; name="19.0.x" 42 -----------------------------792694871670 Content-Disposition: form-data; name="19.0.y" 9 -----------------------------792694871670-- and following is array i am posting using curl. $data_arry3["1"]="(Car#21)"; $data_arry3["2"]="(Car#22)"; $data_arry3["3"]="(Car#23)"; $data_arry3["4"]="(Car#24)"; $data_arry3["5"]="(Car#25)"; $data_arry3["6"]="(Car#26)"; $data_arry3["7"]="(Car#27)"; $data_arry3["8"]="(Car#28)"; $data_arry3["9"]="(Car#29)"; $data_arry3["10"]="(Car#30)"; $data_arry3['19.0.x']="58"; $data_arry3['.0.y']="4"; as far as i know, session,cookies and ref everything seems ok because, i am able to submit first two forms. i think problem is in data being send by curl to that server, something has changed by curl in data before sending to server, the reason i am saying this name of fields are numbers not alphabet. may be i am wrong. so experts any one can let me know whats i am doing wrong?? any advice will sure take me out of this. please if any question do let me know.

its giving be FEB ,19 but when i see real feed in browser, its giving me feb ,20 (this "Saturday, February 20, 2010 3:53 AM" for )

hi, i am parsing one atom feed in my php script. in which there is one tag called <update> which contain following value. 2010-02-19T17:23:53-05:00 in browser its displayed as "Saturday, February 20, 2010 3:53 AM" can any one let me know how to get same value in php script?

hi, issue solved. my hosting team has solved it. the fault was on their side.

hi, i have one simple script which does work fine on local computer but got following problem when i uploaded it server. The problem is, the server genarate 404 error if i pass query string paramter value of this next=http%3A%2F%2Fwww.anas-asarani.com%2Ffb%2F but it work perfectly fine if there is no % in its value. this does not work http://anas-asarani.com/fb/car.php?next=http%3A%2F%2Fwww.anas-asarani.com%2Ffb%2F this works http://anas-asarani.com/fb/car.php?next=asdkj can any one let me know whats wrong on server? waiting for your reply.

first thanks for your reply rajivgonsalves. i added your two lines. i got error. GD was not insalled on server. after installing GD all well. thanks you so much.

please any one , any idea.

hi i have one simple code which add transparent color to image and save it as ".png". it works fine on my local computer using WAMP. but when i ran same code on server it generate no output image. my php version is 5.3 and server php versoin is 5.2 here is my code <?php $info = getimagesize("rose.jpeg"); $width = $info[0]; $height = $info[1]; $im = imagecreatetruecolor($width, $height); $image = @imagecreatefromjpeg("rose.jpeg"); $black = imagecolorallocate($im, 255, 255, 255); imagecolortransparent($im, $black); imagecopy($im, $image, 0, 0,0 ,0,$width, $height); imagepng($im, 'output.png'); imagedestroy($im); ?> can you let me know why whats reason? how can i make it run on server?/

thanks this helped me as well.

hope some will help me out.

yes i did it and restarted it. my curl is enable.but still same problem.

script post data to website with date and time and website rend detail of page of that date and time. script it exact response when i execute query on my server. but when i run same script on local computer, response is not detail page of that date and time i send , but response is same submit of that website.

yes i did many times.

hi, my php version is 5.3.0 and i have already turn on extension for CURL. I will let you know again that my i am getting response from website, but that response is not what is expected. i think something is missing in header or some option for CURL which i didnt set or enable. (MAY BE THAT OPTION IS ALREADY SET OR ENABLE ON MY WEBSERVER). any info which it can be??

any one?

it is enable. script does submit data as well from local pc but response is not what it is expected. does version matter?

hi experts, i need some information from you. i have one curl script which submit data to one of airline website and get result. Problem is: my script works on web server perfectly fine. but when i run script locally it does not work. can any one let me know why is so? its working on my webserver but not on my local pc. I am using WAMP on my local pc. hope someone can take me out of this.

ok. i will check this as well. again thanks. awesome. topic solved

oh wow sir, it worked, but really when i see html source there is nothing ahead and behind "3". may be something invisible on browser???

here is output of above. string(5) "3" string(5) "3" int(0)

sir, still result is 0. with your code as well