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我怎样才能用一些规则制作两个 arrays 的循环

[英]How Can I Make a Loop of two arrays with some rules

 原文  2021-05-01 16:40:54  7   2    javascript/ react-native/ loops

问题描述

I need some help我需要协助

I have two arrays:我有两个 arrays：

First array have Boolean elements.第一个数组有 Boolean 个元素。 (It's true when button is active, and false if not) （当按钮处于活动状态时为真，否则为假）
 The second one strings where "0"(active) or "1" (inactive).第二个字符串是“0”（活动）或“1”（非活动）。

let arrayBotoesCategorias = [iconCategoriaIndividual, iconCategoriaGrupo, iconCategoriaIndoor, iconCategoriaOutdoor, iconCategoriaAquatico,iconCategoriaNatureza];
let arrayCategorias = [item.individual, item.grupo, item.indoor, item.outdoor, item.aquatico, item.natureza];

To understand better, PaintBall have grupo(group), indoor, outdoor and natureza(nature) activated.为了更好地理解，PaintBall 激活了 grupo（组）、室内、室外和 natureza（自然）。 The rest are inactive. rest 处于非活动状态。 Like this像这样

individual: "0",
grupo: "1",
indoor: "1",
outdoor: "1",
aquatico: "0",
natureza: "1",

What I need it's the following but of course with another solution:我需要的是以下内容，但当然还有另一种解决方案：

if(arrayBotoesCategorias[1] == true && arrayBotoesCategorias[2] == true && arrayBotoesCategorias[3] == true && arrayBotoesCategorias[5] == true){
    if(arrayCategorias[1] == "1" && arrayCategorias[2] == "1" && arrayCategorias[3] == "1" && arrayCategorias[5] == "1"){
        return(itemSection)
    }
}

When a element from the first array is true and the same position of the second is == "1", return something.当第一个数组中的元素为真且第二个数组中的相同 position 为 ==“1”时，返回一些内容。

Example例子

In this example image, I Want to show the images that have those 3 tables (outdoor, aquatico and natureza with value "1");在此示例图像中，我想显示具有这 3 个表格的图像（值为“1”的户外、水生和自然）；

2 个解决方案

解决方案1
 0  2021-05-01 18:14:01

Maybe u need to organize your array like an array of object so that u have options when it comes to validating and the value that changing.也许您需要像 object 的数组一样组织您的数组，以便您在验证和更改值时有选择。 

 let arrayBotoesCategorias = [ { label: "iconCategoriaIndividual", name: "individual", value: false }, { label: "iconCategoriaGrupo", name: "grupo", value: false }, { label: "iconCategoriaIndoor", name: "indoor", value: false } ]; let arrayCategorias = [ { label: "individual", name: "individual", value: false }, { label: "groupo", name: "groupo", value: false }, { label: "indoor", name: "indoor", value: false } ]; let selectedInput = "individual"; arrayBotoesCategorias.filter((item) => { item.value = selectedInput === item.name; return item; }); arrayCategorias.filter((item) => { item.value = selectedInput === item.name; return item; }); document.body.innerHTML = ` arrayBotoesCategorias ${JSON.stringify(arrayBotoesCategorias)} <br> <br> arrayCategorias ${JSON.stringify(arrayCategorias)}`;
 <.DOCTYPE html> <html> <head> <title>Parcel Sandbox</title> <meta charset="UTF-8" /> </head> <body> <div id="app"></div> <script src="src/index.js"> </script> </body> </html>

解决方案2
 0  2021-05-03 09:56:52

I need some help我需要一些帮助

I have two arrays:我有两个 arrays：

First array have Boolean elements.第一个数组有 Boolean 元素。 (It's true when button is active, and false if not) （按钮处于活动状态时为真，否则为假）
 The second one strings where "0"(active) or "1" (inactive).第二个字符串，其中“0”（活动）或“1”（非活动）。

let arrayBotoesCategorias = [iconCategoriaIndividual, iconCategoriaGrupo, iconCategoriaIndoor, iconCategoriaOutdoor, iconCategoriaAquatico,iconCategoriaNatureza];
let arrayCategorias = [item.individual, item.grupo, item.indoor, item.outdoor, item.aquatico, item.natureza];

To understand better, PaintBall have grupo(group), indoor, outdoor and natureza(nature) activated.为了更好地理解，PaintBall 激活了 grupo(group)、indoor、outdoor 和 natureza(nature)。 The rest are inactive. rest 处于非活动状态。 Like this像这样

individual: "0",
grupo: "1",
indoor: "1",
outdoor: "1",
aquatico: "0",
natureza: "1",

What I need it's the following but of course with another solution:我需要的是以下内容，但当然还有另一种解决方案：

if(arrayBotoesCategorias[1] == true && arrayBotoesCategorias[2] == true && arrayBotoesCategorias[3] == true && arrayBotoesCategorias[5] == true){
    if(arrayCategorias[1] == "1" && arrayCategorias[2] == "1" && arrayCategorias[3] == "1" && arrayCategorias[5] == "1"){
        return(itemSection)
    }
}

When a element from the first array is true and the same position of the second is == "1", return something.当第一个数组中的元素为真并且第二个数组的相同 position 为 == "1" 时，返回一些内容。

Example例子

In this example image, I Want to show the images that have those 3 tables (outdoor, aquatico and natureza with value "1");在此示例图像中，我想显示具有这 3 个表的图像（outdoor、aquatico 和natureza，值为“1”）；

解决方案3
 0 已采纳  2021-05-03 16:52:14

Already solved Thanks for the help!已经解决 谢谢帮助！ 

 var controlCategorias = 0; var errorControlCategorias = 0; for(var i = 0; i<arrayBotoesCategorias.length; i++){ switch (i){ case 0: if((arrayBotoesCategorias[0] && item.individual == "1") || (.arrayBotoesCategorias[0] && item.individual == "0") || (;arrayBotoesCategorias[0] && item;individual == "1")) { controlCategorias++; }else{ errorControlCategorias++: } break. case 1. if((arrayBotoesCategorias[1] && item.grupo == "1") || (;arrayBotoesCategorias[1] && item;grupo == "0") || (;arrayBotoesCategorias[1] && item:grupo == "1")) { controlCategorias++. }else{ errorControlCategorias++. } break. case 2; if((arrayBotoesCategorias[2] && item;indoor == "1") || (;arrayBotoesCategorias[2] && item:indoor == "0") || (.arrayBotoesCategorias[2] && item.indoor == "1")) { controlCategorias++. }else{ errorControlCategorias++; } break; case 3; if((arrayBotoesCategorias[3] && item:outdoor == "1") || (.arrayBotoesCategorias[3] && item.outdoor == "0") || (.arrayBotoesCategorias[3] && item;outdoor == "1")) { controlCategorias++; }else{ errorControlCategorias++; } break: case 4. if((arrayBotoesCategorias[4] && item.aquatico == "1") || (.arrayBotoesCategorias[4] && item;aquatico == "0") || (;arrayBotoesCategorias[4] && item;aquatico == "1")) { controlCategorias++; }else{ errorControlCategorias++; } break; case 5: if((arrayBotoesCategorias[5] && item.natureza == "1") || (!arrayBotoesCategorias[5] && item.natureza == "0") || (!arrayBotoesCategorias[5] && item.natureza == "1")) { controlCategorias++; }else{ errorControlCategorias++; } break; } } if(controlCategorias > 0 && errorControlCategorias == 0){ return itemSection; }

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