Reputation: 35
This is what I'm trying to accomplish:
var result = document.getElementById("o_q1_op1");
document.getElementById("a1_op1").style.height = result.scrollHeight;
But I dont want to rewrite it a dozen times, instead want to use arrays structured like this:
var que = ["o_q1_op1", "o_q1_op2", "o_q2_op3"];
var ans = ["a1_op1","a1_op2","a1_op3","a1_op4"];
This is what I have so Far:
var que = ["o_q1_op1", "o_q1_op2", "o_q2_op3"];
var ans = ["a1_op1","a1_op2","a1_op3","a1_op4"];
var result;
var index;
var text = "";
for (index = 0; index < que.length; index++) {
text += que[index];
var result[index] = document.getElementById(text);
}
var index2;
var text2 = "";
for (index2 = 0; index2 < ans.length; index2++) {
text2 += ans[index2];
document.getElementById(text2).style.height = result[index].scrollHeight;
}
I'v never used 2 arrays to accomplish one task before so I'm not sure how to go about it.
Upvotes: 0
Views: 50
Reputation: 136239
You might like to not use 2 arrays, and instead have one array that ties the two properties together:
var arr = [
{ q:"o_q1_op1",a:"a1_op1"},
{ q:"o_q1_op2",a:"a1_op2"}
];
and then the code becomes much simpler, and you never have to deal with the 2 arrays becoming different lengths
for(var i=0;i<arr.length;i++){
var result = document.getElementById(arr[i].q);
document.getElementById(arr[i].a).style.height = result.scrollHeight;
}
Upvotes: 3
Reputation: 2863
So long as the two arrays are ordered and remain the same length...
for(var i=0;i< que.length;i++){
document.getElementById(ans[i]).style.height =
document.getElementById(que[i]).scrollHeight;
}
Upvotes: 1
Reputation: 17936
well the +=
is wrong because it concatinates string
change and simplify code to
var que = ["o_q1_op1", "o_q1_op2", "o_q2_op3"];
var ans = ["a1_op1","a1_op2","a1_op3","a1_op4"];
var result=[];
for (var index = 0; index < que.length; index++) {
result[index] = document.getElementById(que[index]);
}
for (var index2 = 0; index2 < ans.length; index2++) {
document.getElementById(ans[index2]).style.height = result[index].scrollHeight;
}
Upvotes: 1