Theo's Typesafe Cult•13mo ago

What's the right way to solve this?

<InputRow>
  <Dropdown
    title="Lora Name"
    values={LoraNames.map((t) => ({ name: t, value: t }))}
    value={lora_stack_item.name}
    setValue={(value: z.infer<typeof LoraStackItem>["name"]) => {
      const updatedLoraStack = [...efficiencyLoader.lora_stack.stack];
      if (!updatedLoraStack[index]) return;
      updatedLoraStack[index] = { ...updatedLoraStack[index], name: value };
      setEfficiencyLoader({
        ...efficiencyLoader,
        lora_stack: { ...efficiencyLoader.lora_stack, stack: updatedLoraStack },
      });
    }}
  />
</InputRow>

<InputRow>
  <Dropdown
    title="Lora Name"
    values={LoraNames.map((t) => ({ name: t, value: t }))}
    value={lora_stack_item.name}
    setValue={(value: z.infer<typeof LoraStackItem>["name"]) => {
      const updatedLoraStack = [...efficiencyLoader.lora_stack.stack];
      if (!updatedLoraStack[index]) return;
      updatedLoraStack[index] = { ...updatedLoraStack[index], name: value };
      setEfficiencyLoader({
        ...efficiencyLoader,
        lora_stack: { ...efficiencyLoader.lora_stack, stack: updatedLoraStack },
      });
    }}
  />
</InputRow>

Typescript error: updatedLoraStack[index] might be undefined, but we check it's not in the line before If I set updatedLoraStack[index] as a variable, check if its undefined and then set it, it won't update the original array

Solution:

just use ! to tell ts you're use it's there

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5 Replies

Josh•13mo ago

unfortunately ts doesn't narrow array types like this properly

Solution

Josh•13mo ago

just use ! to tell ts you're use it's there

sonorOP•13mo ago

thx

Rhys•13mo ago

Extract it to a variable and check the variable as another option That lets you skip adding a !

sonorOP•13mo ago

yea, but when I set that variable it wont affect the 1st array

let a = [1, 2, 3];
let b = [...a];
let q = b[2];
q = 5;
console.log(b); // [1, 2, 3]

let a = [1, 2, 3];
let b = [...a];
let q = b[2];
q = 5;
console.log(b); // [1, 2, 3]

for example^ and what i would need is [1, 5, 3] , but q = 5 only affects that original q value, and not the b array

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What's the right way to solve this?