Java / Kotlin getting value in for loop for step + one and also a step before if condition in for loop

huangapple
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英文:

Java / Kotlin getting value in for loop for step + one and also a step before if condition in for loop

问题

  2. for (i in 0 until arrTriple!!.size){
  3.    if (arrTriple[i].third >= ordertime){
  4.       
  5.      // 需要获取这个值，同时避免数组越界： arrTriple[i+1].third 
  6.      // 这样我就可以在这里使用我的逻辑。
  8.    }
  9.  }```
  12. <details>
  13. <summary>英文:</summary>
  15. I am currently creating a program to check the order status (pending or confirmed) when a user open the app now the problem is if I need to check after order time and between I am checking time (Current Time) in this timeframe check for how many times I change price and compare if price is match or not.
  17. So how can I check a step before i. like i is 3 and I need to also include 2 in it and compare value of i and i+1 value in every loop without getting index out of bond.
  19. Code : 
  21. ```val arrTriple = PREMIUMHM[itemname.toUpperCase(Locale.getDefault())]
  23.  for (i in 0 until arrTriple!!.size){
  24.    if (arrTriple[i].third &gt;= ordertime){ //arrTriple[i].third is time when when I change price
  25.       
  26.      //Need this value also without getting indexoutofbond  = arrTriple[i+1].third 
  27.      //So I can use my logic here.
  29.    }
  30.  }
  33. </details>
  36. # 答案1
  37. **得分**: 0
  39. 我理解你的问题了。你可以从 i = 1 开始，并且将 i-1 替换为 i+1 吗？类似这样：
  41. ```kotlin
  42. for (i in 1 until arrTriple!!.size) {
  43.    if (arryTriple[i-1].third >= overtime) {
  44.       arrTriple[i].third
  45.    }
  46. }

你会遇到一个特殊情况，即 arrTriple.size == 1，你需要处理它。

编辑：

否则，只需添加一个 if 检查：

  1. for (i in 0 until arrTriple!!.size){
  2.   if (arrTriple[i].third >= ordertime && (i+1) < arrTriple.size) { 
  3.     
  4.   }
  5. }
英文:

Think I understand what you're asking. Can you start with i = to 1 and do i-1 instead of i+1? Something like this:

  1. for (i in 1 until arrTriple!!.size) {
  2.    if (arryTriple[i-1].third &gt;= overtime {
  3.       arrTriple[i].third
  4.    }
  5. }

You'll have an edge case where arrTriple.size == 1 that you'd need to handle still.

EDIT:

Otherwise, just add an if-check.

  1. for (i in 0 until arrTriple!!.size){
  2.   if (arrTriple[i].third &gt;= ordertime &amp;&amp; (i+1) &lt; arrTriple.size){ 
  4.   }
  5. }

答案2

得分: 0

你不一定需要until关键字。你可以像这样编写：

  1. for (element in arrTriple!!) {
  2.     if (element.third >= ordertime) {
  3.         // 在这里添加你的逻辑
  4.     }
  5. }

这是我的方法。

英文:

You don't necessarily need until. You would write it just like this

  1. for(element in arrTriple!!){
  2.     if(element.third &gt;= ordertime){ 
  3.         // Your logic here
  4.     }
  5. }

This would be my approach

huangapple
  • 本文由 发表于 2020年8月24日 02:20:41
  • 转载请务必保留本文链接：https://go.coder-hub.com/63550540.html
  • for-loop
  • java
  • kotlin
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