解决方案:加类型断言 (this.toolsasArray<any>).forEach(item=>{console.log(item.id);})
解决办法就是加一个判断 因为这里find()能够拿到正确的数据,所以else语句不会执行,只要返回非undefined且符合我们定义的泛型类型的数据最后得到的结果就不会包含undefined了
编译器输出this.state.ticketsToShow可能是未定义的 这是我的密码。 浏览0提问于2020-03-05得票数 1 1回答 TypeScript2:导入语句生成TS2532:“对象可能是‘未定义的’”。 、、、 我有一个TypeScript2文件,在该文件中,我试图使用这样的图像:TS2532: Object is possibly 'undefined'. 建议的使用资产的方法是...
The above code will not break like the previous one. It will check if thecityis present and print it. If it is not present, it will printundefined. Using the logical OR (||) operator You can pass a default value to the function if you think the value can be undefined as shown belo...
用watch监听某个一个数组数据,获取其length时,提示Object is possibly ‘null‘ .具体如下: 解决方法如下:就是加null的判断 typescript 提示 Object is possibly ‘null‘ 的N种解决方法 解决方案一 最正确的解决方案,就是加null的判断 const table = document.querySelector('.main-table');if(table) { ...
1.1 忽略 undefined 和 null 类型 functionmyFunc(maybeString:string| undefined |null){// Type 'string | null | undefined' is not assignable to type 'string'.// Type 'undefined' is not assignable to type 'string'.constonlyString:string= maybeString;// ErrorconstignoreUndefinedAndNull:string= ...
Object is possibly null: 对象可能是null 分析:localStorage.getItem(“SET_HISTORY_KEY”) 这个值有可能为空,所以再执行getItem就会报错此刻对象可能为空。 解决: 联合类型 把null的情况写入 类型断言成any类型,any类型上访问任何属性和方法都是被允许的。
this.createdAt??=this.updatedAt=newDate() I think the problem is that arrow functions don't have a 'this' try using an old school function definition? this remains an issue TS2532: Object is possibly 'undefined'. @pre<RoleEntity>('save', (next) => { if (!this.createdAt) { this....
In Typescript, one common error that developers may encounter is the “Object is possibly null” error. This error typically occurs when trying to access a property or method on an object that could potentially be null or undefined. This article will explain the causes of this error, how to...
TypeScript Version: 4.1.2 Search Terms: Object is possibly undefined Code function main(values: Array<string>) { for (let i = 0; i < values.length; i++) { if (typeof values[i] === 'string' && values[i].length > 0) { console.log(i); } } }...