PHP fopen write filename with php variable and file extension

Question

I have a variable $id from URL parameter that contains a name ID-number like 1234.5Z

How can I use fopen to write a file called 1234.5Z.json?

I tried the following syntax:

$fp = fopen($id+'.json', 'w');

But I end up with a file named 1243.5

some one has spent to much time on javascript. in php its a `.` not a `+` — , Mar 28 '17 at 20:37
Possible duplicate of [How to combine two strings together?](http://stackoverflow.com/questions/8336858/how-to-combine-two-strings-together) — user3942918, Mar 28 '17 at 22:08

score 0 · Accepted Answer · answered Mar 28 '17 at 20:38

In PHP, you don't concatenate strings using the + sign, you use the . instead.

So, your code will look more like

$fp = fopen($id . '.json', 'w');

What happens when you were using the +, is that PHP was thinking you were doing an arithmetic operation, and thus, it does it's best to parse the number within the first string, and as a result 1234.5Z -> 1234.5, and .json -> 0, and therein 1234.5 + 0 = 1234.5, which then will be used as the string for the name.

PHP fopen write filename with php variable and file extension

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