if 语句不作为对象返回 React JS

Posted 2023-02-22

【中文标题】if 语句不作为对象返回 React JS

【英文标题】：if statement not returning as an object React JS

【发布时间】：2021-04-14 04:36:05

【问题描述】：

我正在尝试在样式标签中编写两个表达式，但它一直给我以下错误The `style` prop expects a mapping from style properties to values, not a string. For example, style=marginRight: spacing + 'em' when using JSX.我该如何解决？

<table id='calendario'>
              <tbody>
                linhas.map((linha) => (
                  <tr key=linha>
                  colunas.map((coluna , e , i) => (
                  <td
                  key=coluna onLoad=() => 
                    if(e < index.length)e = e;
                    if(i >= proximo_mes.length)i = i;
                  
                  style=() => if(linha == 0 && coluna == e)return filter : 'brightness(0.6)';
                  if(linha == 4 && coluna == i)returnfilter : 'brightness(0.6)'
                  elsereturnfilter : 'brightness(1)'
                ></td>
              ))
            </tr>
          ))
        </tbody>
      </table>

【参考方案1】：

像这样使用ternary operator：

style=(linha == 0 && coluna == e) ? filter : 'brightness(0.6)' : (linha == 4 && coluna == i) ? filter : 'brightness(0.6)' :  filter : 'brightness(1)'

或者这个：

style=filter :(linha == 0 && coluna == e) ? 'brightness(0.6)' : (linha == 4 && coluna == i) ?'brightness(0.6)' : 'brightness(1)'

发生错误是因为您返回的是function 而不是object，请查看以下示例：

console.log("Function Style: ", 
  style: () => 
    if (true) 
      return  filter: "brightness(0.6)" ;
    
  
)

console.log("Ternary Style: ", 
  style: true ?  filter: "brightness(0.6)"  :  filter: "brightness(1)" 
)

【讨论】：

非常感谢，但你能告诉我为什么我的没用吗？

没问题，再次检查答案我为这种情况做一个例子。