Combining strings and numbers in PHP for loop

Question

PHP newbie here

Can anyone please tell me what is wrong with the below syntax. I have a maximum of 4 files - $created_page1, $created_page2 each with a corresponding page title etc and would like to process these in a loop. However PHP throws a wobbly every time I try to concatenate the string and loop number - specifically $created_page.$num_pages doesn't result in sending $created_page1 or $created_page2 to the function, instead it just converts the string and number to an integer. Very basic I am sure but I would be very grateful for any help or a nicer solution that I can easily understand. Thanks in advance!

$addit_pages == 4;

for ($num_pages=1;$num_pages<=$addit_pages ;$num_pages++) { 

replaceFileContent   ($dir,$created_page.$num_pages,"*page_title*",$page_title.$num_pages); 
//replaceFileContent  ($dir,$created_page2,"*page_title*",$page_title2); 
//replaceFileContent  ($dir,$created_page1,"*page_title*",$page_title3); 
//replaceFileContent  ($dir,$created_page3,"*page_title*",$page_title4);    
}

I'm not clear on what you're trying to do. what is "*page_title*"? — Paul Dessert
– Paul Dessert, Commented May 17, 2012 at 23:44
and whats the content of these parameters ? i guess you are mixing up DATE and INTEGER or something like that !? — Sliq
– Sliq, Commented May 17, 2012 at 23:48
and thanks for this too deceze, if there's a more efficient way of doing this than thats the way i want to go, but one step at a time eh? am glad i know that this can be acheived better with arrays. — rookieloops
– rookieloops, Commented May 18, 2012 at 0:56

domvoyt · Accepted Answer · 2012-05-18 00:00:53Z

1

Your code to get the variable name should be:

${'created_page'.$num_pages}

This is because you have to evaluate the string inside the braces before you attempt to access the variable.

Your previous code was trying to access the variables $created_page and $num_pages, and simply concatenate their values into a string.

Of course, the same goes for the page_title variable

${'page_title'.$num_pages}

edited May 18, 2012 at 0:00

answered May 17, 2012 at 23:52

domvoyt

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2 Comments

rookieloops Over a year ago

wow -so many responses so quickly. domvoyt I think you knew what I was getting at best, worked like a dream and i get it. but thanks to everyone- your answers are all helping me to get a grip on this thing.

domvoyt Over a year ago

No problem. @Aknosis had another approach that would work. Basically two different ways of using "Variable variables". But as deceze said, much better done with arrays, so good luck with your learning :)

Wampie Driessen · Accepted Answer · 2012-05-17 23:47:08Z

0

you could try this:

$addit_pages == 4;

for ($num_pages=1;$num_pages<=$addit_pages ;$num_pages++) {

replaceFileContent ($dir,$created_page.strval($num_pages),"*page_title*",$page_title.strval($num_pages)); //replaceFileContent ($dir,$created_page2,"*page_title*",$page_title2); //replaceFileContent ($dir,$created_page1,"*page_title*",$page_title3); //replaceFileContent ($dir,$created_page3,"*page_title*",$page_title4);
}

the PHP strval function makes any integer into a string

answered May 17, 2012 at 23:47

Wampie Driessen

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Comments

aknosis · Accepted Answer · 2012-05-17 23:51:56Z

0

I think what you are asking is you want the variables $created_page1; $created_page2, $created_page3 but php is probably throwing a notice that $created_page doesn't exist.

You need to use variable variables (is this what they're called?)

$addit_pages == 4;

for ($num_pages=1;$num_pages<=$addit_pages ;$num_pages++) { 
    $createdVar = 'created_page'.$num_pages;
    $titleVar = 'page_title'.$num_pages;
    replaceFileContent   ($dir,$$createdVar,"*page_title*",$$titleVar); 
}

When you use $$ this first evaluates the variable $createdVar turns that into created_page1 and then evaluates created_page1 as if you had typed in $created_page1

answered May 17, 2012 at 23:51

aknosis

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1 Comment

aknosis Over a year ago

Link for reference, "Variable variables" is the correct term it seems: us2.php.net/manual/en/language.variables.variable.php

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