Surprisingly the same code with same image is working on my friend's system having same operating system.. How it is possible?? What...
I have already opened the files in binary mode. See the first two if statements
Please tell what is wrong with following code:
if (($rh = fopen('image.jpg', 'rb')) === FALSE)
echo "Cannot open image.jpg";
What is contained in request message sent by browser and what is contained in response message??
One more thing, There could be more than 1...
But why we need to call using System class. Why we not use it using OutputSteam class whose instance is it
'out' is an object of OutputStream class .Then why we use it with system class as we write System.out.print();
I have created a form in html using the code at http://pastebin.com/f11b9724f
The code of process5.php is at http://pastebin.com/f35d8ad2f Now...
But the same warning is still there
$selct="SELECT * FROM stu_info";
while($row = mysql_fetch_array('$result',MYSQL_BOTH))
Separate names with a comma.