how to use the replace function to replace consecutive DIVs with IDs
I'm trying to understand where I went wrong I need to insert by replacing <DIV with <DIV ID="div0", <DIV ID="div1", <DIV ID="div2", <DIV ID="div3",...
PHP Code:
$i=0;
function divCount(){
$i = $i + 1;
$iString = $i + "";
return "div".$iString;
}
$strResult = str_replace("<DIV STYLE=", "<DIV ID='".divCount()."' STYLE=", $result);
echo $strResult;
This is what's happening so far:
Code:
<DIV ID='div1' STYLE="POSITION:ABSOLUTE;TOP:69;LEFT:177" CLASS="APFont00000">...</DIV>
<DIV ID='div1' STYLE="POSITION:ABSOLUTE;TOP:88;LEFT:290" CLASS="APFont00001">...</DIV>
<DIV ID='div1' STYLE="POSITION:ABSOLUTE;TOP:103;LEFT:231" CLASS="APFont00001">...</DIV>
<DIV ID='div1' STYLE="POSITION:ABSOLUTE;TOP:154;LEFT:71" CLASS="APFont00002">...</DIV>
Re: how to use the replace function to replace consecutive DIVs with IDs
try using
$i++; instead of $i = $i + 1
Re: how to use the replace function to replace consecutive DIVs with IDs
Quote:
Originally Posted by dclamp
try using
$i++; instead of $i = $i + 1
Excellent idea. Replace a line of code with an equivalent line of code. I am sure that is going to solve the problem. :rolleyes:
Re: how to use the replace function to replace consecutive DIVs with IDs
Quote:
Originally Posted by gilgalbiblewhee
I'm trying to understand where I went wrong I need to insert by replacing
<DIV with
<DIV ID="div0",
<DIV ID="div1",
<DIV ID="div2",
<DIV ID="div3",...
PHP Code:
$i=0;
function divCount(){
$i = $i + 1;
$iString = $i + "";
return "div".$iString;
}
$strResult = str_replace("<DIV STYLE=", "<DIV ID='".divCount()."' STYLE=", $result);
echo $strResult;
This is what's happening so far:
Code:
<DIV ID='div1' STYLE="POSITION:ABSOLUTE;TOP:69;LEFT:177" CLASS="APFont00000">...</DIV>
<DIV ID='div1' STYLE="POSITION:ABSOLUTE;TOP:88;LEFT:290" CLASS="APFont00001">...</DIV>
<DIV ID='div1' STYLE="POSITION:ABSOLUTE;TOP:103;LEFT:231" CLASS="APFont00001">...</DIV>
<DIV ID='div1' STYLE="POSITION:ABSOLUTE;TOP:154;LEFT:71" CLASS="APFont00002">...</DIV>
Your $i variable is not declared as global in the function. I would advise the you use DOM to make these changes though rather than the more unreliable sting replace.
Re: how to use the replace function to replace consecutive DIVs with IDs
Quote:
Originally Posted by visualAd
Excellent idea. Replace a line of code with an equivalent line of code. I am sure that is going to solve the problem. :rolleyes:
Hey, it is going to help his code look nicer. Maybe PHP didnt like how it looked.
I re-looked your code and i am not seeing anything wrong with it. Is your function being included properly?
edit:
I am not sure, but i think that the $i=0 might me reseting it and it keeps adding 1 to 0.
Re: how to use the replace function to replace consecutive DIVs with IDs
Quote:
Originally Posted by dclamp
I re-looked your code and i am not seeing anything wrong with it. Is your function being included properly?
VisualAd said it all - "Your $i variable is not declared as global in the function"
Re: how to use the replace function to replace consecutive DIVs with IDs
oh. :p ok. didnt know if my little 2 cents helped. Guess not.
thanks for the correction!
Re: how to use the replace function to replace consecutive DIVs with IDs
Quote:
Originally Posted by dclamp
Hey, it is going to help his code look nicer. Maybe PHP didnt like how it looked.
I re-looked your code and i am not seeing anything wrong with it. Is your function being included properly?
edit:
I am not sure, but i think that the $i=0 might me reseting it and it keeps adding 1 to 0.
Don't take this the wrong way; but your suggestion wasn't helpful because it had nothing to do with the problem. I would personally go for ($i+=1) for readability purposes but if you want to use the increment operator on its own use (++$i) which is more efficient.
The difference between ++$i and $i++ is when the variable is increment. The former is incremented before the expression is executed and the latter is executed after the expression is executed.
e.g:
Code:
$i=1;
echo($i++ * 5); // outputs 5
echo(++$i * 5);; // outputs 10
