how to put a declaration in a if-function?

BD One

hey community,

I am making a pdf with information from a website. My problem is the image. If there is an image it works with the codes:

const first_picture_url: string = product.pictures[0].url;
const other_picture_urls: string[] = product.pictures.filter(
(_: { url }, index: number) => index > 0).map((picture: { url }) => picture.url);

If there is no image, i get the Error

TypeError: Cannot read property 'url' of undefined.

i found out i could prevent this with

but if i try to put the first codes in this if-function it doesnt work.

My try:

if(product.pictures.length>0){
const first_picture_url: string = product.pictures[0].url;
const other_picture_urls: string[] = product.pictures.filter(
(_: { url }, index: number) => index > 0).map((picture: { url }) => picture.url);
}
else
{
const first_picture_url: string = String(product.airCoolingIntegrated);
}

it doesnt work, i get the message "first_picture_url" is declared but its value is never read"

i just want these lines not be initiated if the length of the attribute is 0. (so no picture existing).

Can anyone help me with that maybe?

Greetings

3 Replies

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Sachin Singh5y ago
Accepted answer
declare the variable outside of the if statement
const first_picture_url: string=null;

roduct.pictures.length !=null){
     first_picture_url: string = product.pictures[0].url;
    other_picture_urls: string[] = product.pictures.filter(

      (_: { url }, index: number) => index > 0).map((picture: { url }) => picture.url);

else

     first_picture_url: string = String(product.airCoolingIntegrated);
BD One5y ago
@Sachin If i use your Code, IT has The Same Error. "First_picture_url ist declared but its value is never read" typescript is telling me If i declare it, i should also use it...
Sachin Singh5y ago
do this instead
if(product.pictures!=null)
{
// your code here
}