2011-03-09 146 views
1

......不確定「計算」是否合適......我寫了兩個函數,其中第二個函數需要第一個變量的輸出。我似乎無法讓它通過...我的猜測是,我說錯了,但似乎無法做到正確......可能與我花費在盯着整個事情的時間有關..將計算的變量從一個函數傳遞到另一個函數

我需要通過的變量是subset我想在第二個函數的最後一行使用它。

如果重要,getPos函數從輸入框中獲取它的值。

的JavaScript:

<script> 
var alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]; 

function getPos(value) 
{ 
    var letterPosition = alphabet.indexOf(value); 
    var subset = alphabet.slice(letterPosition+1, 26); 
    document.getElementById('theRest').value = subset; 
} 

function appendTable(id) 
{ 
    var tbody = document.getElementById(id).getElementsByTagName("tbody")[0]; 
    var i = 0; 
    for (var r = 0; r < 4; r++) { 
     var row = tbody.insertRow(r); 
     for (var c = 0; c < 4; c++) { 
      var cell = row.insertCell(c); 
      cell.appendChild(document.createTextNode(subset[i++])); 
     } 
    } 
} 
</script> 

回答

1
<script> 
var alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]; 
var subset; 

function getPos(value) 
{ 
    var letterPosition = alphabet.indexOf(value); 
    subset = alphabet.slice(letterPosition+1, 26); 
    document.getElementById('theRest').value = subset; 
} 

function appendTable(id) 
{ 
    var tbody = document.getElementById(id).getElementsByTagName("tbody")[0]; 
    var i = 0; 
    for (var r = 0; r < 4; r++) { 
     var row = tbody.insertRow(r); 
     for (var c = 0; c < 4; c++) { 
      var cell = row.insertCell(c); 
      cell.appendChild(document.createTextNode(subset[i++])); 
     } 
    } 
} 
</script> 

這應該做的伎倆。
在函數使其成爲全局變量之前聲明子集,如果在函數中將其與var subset一起定義,那麼它將變爲綁定到該函數,刪除變量使其使用全局變量。

+0

現在看起來很明顯..哈哈。你先生是我的英雄。現在我終於可以睡覺了! – tehaaron 2011-03-09 10:49:16

+0

好的,我可以在2分鐘內上牀......在它讓我接受答案之後。 – tehaaron 2011-03-09 10:49:58

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