Why my parseInt function is not converting my string into integer?

Question

What am I going wrong here? I'm trying to convert a string to an integer to use in addition.

I have an image with a class of portfolio and a src of images/portfolio-images/gbn1.jpg

The first alert says the value of imageInt is 1, but the second alert says the value of newImg is gbn11.jpg when I'm trying to get gbn2.jpg.

I am posting my code below:

var mainImg = $('.active-more').closest(".container").find("img.portfolio:first").attr("src");
var nofilename = mainImg.replace("images/portfolio-images/",""); 
nofilename = nofilename.replace(".jpg",""); 
var imageString = nofilename.replace(/[A-Za-z$-]/g, "");
var imageInt = parseInt(imageString, 10);   
var imageProject = nofilename.replace(/\d+/g, '');  
alert(imageInt);
var newImg= imageProject + imageInt + 1 + ".jpg";
alert(newImg);

+1 for naming variables in sample to show type.

Alexei Levenkov
– Alexei Levenkov

2012-12-20 05:49:33 +00:00
Commented Dec 20, 2012 at 5:49 — Alexei Levenkov
– Alexei Levenkov, Commented Dec 20, 2012 at 5:49

Alexei Levenkov · Accepted Answer · 2012-12-20 05:34:22Z

6

You are doing string concatenation due to order of + - add braces to fix:

 var newImg= imageProject + (imageInt + 1) + ".jpg";

answered Dec 20, 2012 at 5:34

Alexei Levenkov

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4 Comments

Vinoth Babu Over a year ago

@maskedj: try my answer below

Alexei Levenkov Over a year ago

@lwburk, not sure where do you see a problem: string + int or int + string produces string, but int + int produces int (Number to be precise). So when one forces (imageInt + 1) to be executed first it will be int+int, and than go to string+int/ int+string route.

Alexei Levenkov Over a year ago

@lwburk :) no problem, it is actually realatively surprising to see variable name to actually reflects what is stored there.

Wayne Over a year ago

@AlexeiLevenkov - I work on legacy code that extensively uses Hungarian notation, so I actually see it a lot. No excuses here :)

HBP · Accepted Answer · 2012-12-20 09:09:23Z

3

Try this :

var newImg= imageProject + (imageInt + 1) + ".jpg";

The + operator is used to add numbers and to concatenate strings. If the left operand is a string then the right one is converted to a string.

In the above, by enclosing the imageInt + 1 in parentheses we ensure that it is evaluated as an integer before being concatenated as a string.

See the fiddle : http://jsfiddle.net/VQG5j

edited Dec 20, 2012 at 9:09

answered Dec 20, 2012 at 5:35

HBP

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1 Comment

RobG Over a year ago

@lwburk—you can delete comments if you wish.

Vinoth Babu · Accepted Answer · 2012-12-20 05:37:27Z

0

You try the adding stuff separately and concatenate it to the image name, use parseInt on adding too.

Try the code construct given below:

var mainImg = $('.active-more').closest(".container").find("img.portfolio:first").attr("src");
var nofilename = mainImg.replace("images/portfolio-images/",""); 
nofilename = nofilename.replace(".jpg",""); 
var imageString = nofilename.replace(/[A-Za-z$-]/g, "");
var imageInt = parseInt(imageString, 10);   
var imageProject = nofilename.replace(/\d+/g, '');  
alert(imageInt);
imgIntNew = imageInt + parseInt(1);
alert(imageIntNew);
var newImg= imageProject + imgIntNew + ".jpg";
alert(newImg);

edited Dec 20, 2012 at 5:37

answered Dec 20, 2012 at 5:34

Vinoth Babu

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