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How to fix 'Uncaught TypeError: number is not a function error' and 'Failed to load resource' error

开发者 https://www.devze.com 2023-01-22 05:18 出处:网络
Why does the code have an error? var images = [\'/test/img/Gallery/l开发者_高级运维arge/4cba0c8a-4acc-4f4a-9d71-0a444afdf48d.jpg\',\'/test/img/Gallery/large/4cba0ca8-2158-41af-829a-0a444afdf48d.jpg\'

Why does the code have an error?

var images = ['/test/img/Gallery/l开发者_高级运维arge/4cba0c8a-4acc-4f4a-9d71-0a444afdf48d.jpg','/test/img/Gallery/large/4cba0ca8-2158-41af-829a-0a444afdf48d.jpg','/test/img/Gallery/large/4cbc549a-5228-433f-b0bc-0a444afdf48d.jpg'];
$('.triggerNext').click(function(){
    nextImage();
    return false;
});
function nextImage(){
    currentImage = $('.Pagepage:eq(0)').val();
    nextImage = parseInt(currentImage)+1;
    $('#imageCurrent').attr('src',images[nextImage]);
    $('#imageCurrent') .css('position','absolute').css('left',($(window).width()- $('#imageCurrent').width() )/2);
    $('.Pagepage').val(nextImage);
}

It runs correctly the first time but gets an error after clicking.

Yet, the code below runs fine without any errors:

var images = ['/test/img/Gallery/large/4cba0c8a-4acc-4f4a-9d71-0a444afdf48d.jpg','/test/img/Gallery/large/4cba0ca8-2158-41af-829a-0a444afdf48d.jpg','/test/img/Gallery/large/4cbc549a-5228-433f-b0bc-0a444afdf48d.jpg'];
$('.triggerNext').click(function(){
    currentImage = $('.Pagepage:eq(0)').val();
    nextImage = parseInt(currentImage)+1;
    $('#imageCurrent').attr('src',images[nextImage]);
    $('#imageCurrent') .css('position','absolute').css('left',($(window).width()- $('#imageCurrent').width() )/2);
    $('.Pagepage').val(nextImage);
    return false;
});

How to fix 'Uncaught TypeError: number is not a function error' and 'Failed to load resource' error


It seems like your're overwritting your function here:

nextImage = parseInt(currentImage)+1;

Either change the name from your function or variable. Even better, don't use a global namespace. Anyway, you can't overwrite the name from a function, your're overwritting the function itself.

After the above line, nextImage contains a Number which obviously cannot get executed.

right from the comments

just replace nextImage = parseInt(currentImage)+1; with var nextImage = > parseInt(currentImage)+1; – Alin Purcaru

Using a var statement, also avoids the nextImage from going into the global namespace.


use:

var nextImage = parseInt(currentImage)+1;

so you don't overwrite your function name

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