Loop problem

[gaya]

for($i=0; $i<6;$i++){if($var[$i]==$str){echo "$str".$var[$i];$strfirstlettercaps = strtolower($str);$var1=$strfirstlettercaps.".php";require_once $var1; }else{$strfirstlettercaps="";$var1=fopen("$strfirstlettercaps".".php","x+");echo "<img src='../uc1.jpg'>";} 

The loop is checking for both if and else condition. I dont know why.please anyone tell me what i'm doing wrong here.

[Ingolme]

In some cases, $var[$i] is equal to $str while in others it isn't. What exactly are you doing with that fopen() statement? You're setting $strfirstlettercaps to an empty string and then asking it to open a file called ".php"

[gaya]

The fopen() is for creating an empty php file.

[justsomeguy]

It's creating a file without a name, the file is called ".php". Is that what you want to name the file?

[gaya]

Ya i just want to create a new empty php file like "empty.php". I did it. Now i'm trying to do,by creating that php file, i am trying to show some image.Is that possible.please tell me

Edited February 18, 2013 by gaya

[rootKID]

hmm... im not quite into this, but tell me if im wrong, should the IMG echo not be (outside) the for if/else statement?...unless its the plan... but not sure, still having trouble understanding that kind of stuff when its coming to the part of loops and if/else statements......

Edited February 18, 2013 by rootKID

[justsomeguy]

Ya i just want to create a new empty php file like "empty.php". I did it.

I don't think you're understanding. Look at these two lines: $strfirstlettercaps="";$var1=fopen("$strfirstlettercaps".".php","x+"); The first line sets $strfirstlettercaps to be empty, and then the second line creates a file with that as the filename. The filename of the new file is ".php". It's not "empty.php", it's just ".php" because you just set $strfirstlettercaps to be an empty string.

Now i'm trying to do,by creating that php file, i am trying to show some image.

I don't know what you're asking, how does the file relate to the image?

[gaya]

In the first mentioned code i simply use the if condition alone. Where i'm creating a new type,i included the code as

 $qtname1="$qtname";	   $var1=fopen("$qtname1".".php","x+");$stringData = "<img src='../uc1.jpg'/>";fwrite($var1, $stringData);fclose($var1);

In the above code $qtname is the name what i am entering newly to the database,and open the created file name included image using file write. ie)$qtname=a.php.

[justsomeguy]

I don't think there's a reason for all of the extra variables and quotes, this will work fine:

$var1=fopen($qtname.".php","x+");

Even so, I still don't know what you're asking.

[gaya]

I'm doing a code for creating question bank for all subjects. I've dont the coding for 6different question type,like fillup,choose etc.In future, if any more new question type is added by admin,the page should not be viewed empty. Just it should show an imge like page is under construction.so i'm trying to create a empty php file and the img code sholud be written in that php file using fopen() & fwrite(). This is what i'm trying to do.

[justsomeguy]

It looks like you're doing that, is the HTML not being saved in the file correctly?