PHP: putting string and a number with calculation together

In a form I created a loop to show a list of files and next to each file a check box. To give each check biox a unique name, I wanted to add the number of the file to it. So first I did:

<input type="checkbox" name="<?php echo 'upload' . $a-1; ?>">

I made the calculation because I already increased $a before the HTML code came. The result was:

<input type="checkbox" name="-1">

The trick is to put the calculation between brackets, like this:

<input type="checkbox" name="<?php echo 'upload' . ($a-1;) ?>">

And this results in the 'name' being: "upload1", "upload2" etc.

My website: www.dejongfotografie.nl

PHP: comparison of integer with string

PHP does not have explicit declarations of variable types. Whatever you put in a variable or array defines what the variable's type is. String, integer, boolean, float etc.

I have two variables, both containing a value. However somehow one of the variables is seen as a string, so when doing a comparison between the two strings it did not give a result. The variable was not explicitely loaded as a string, a value from the url is put in there using the GET method.

$var1=33 (integer)
$var2="33" (string)

if($var1 == $var2) {echo 'equal';}

This does not give a match. A quick although not very neat resolution is to make sure that var2 is seen as an integer. This can be done by using an arithmatic operator on the variable.

$var2++; //var2=34
$var2--; //var2=33

if($var1 == $var2) {echo 'equal';}

Results in the if-statement being true.

PHP: escaping double quotes

In PHP you sometimes have double quotes (this one: " ) in a string. Because double quotes also indicate the start and end of the string you have to escape them when they are within the string. Today I had the following example. I needed to put a piece of HTML to show an image into a string. So I tried:

$prev = "<img src="leftgr.GIF" width="13" height="13" border="0" >";

ofcourse this did notwork as PHP thinks the string ends with the second double quote "<img src=". It should be:

$prev = "<img src=\"leftgr.GIF\" width=\"13\" height=\"13\" border=\"0\" >";

The backslash (this one: \) escapes characters. PHP knows then then the character does not belong to the PHP code but is just a character in a string.

My website: www.dejongfotografie.nl