求救??!無法運(yùn)行啊,沒反應(yīng)
//第一步把之前的數(shù)據(jù)寫成一個(gè)數(shù)組的形式,定義變量為 infos
?var infos=[["小A","女",21,"大一"],["小B","男",23,"大三"],["小C","男",24,"大四"],["小D","女",21,"大一"],["小E","女",22,"大四"],["小F","男",21,"大一"],["小G","女",22,"大二"],["小H","女",20,"大三"],["小I","女",20,"大一"],["小J","男",20,"大三"]];
? //第一次篩選,找出都是大一的信息
?for(var i=0;i<infos.length;i++;){
? ? ?infos[i]=new Array();
? ? ?for(var j=0;j<infos[i].length;j++;){
? ? ? ? ?if(infos[i][3]=="大一" && infos[i][1]=="女"){
? ? ? ? ? ? ?document.write(infos[i]+"<br />");
? ? ? ? ?}
? ? ? ? ?else
? ? ? ? ?{
? ? ? ? ? ? ?continue;
? ? ? ? ?}
? ? ? ?}
?}
? ?//第二次篩選,找出都是女生的信息
</script>
2018-05-29
首先不需要嵌套循環(huán)哦,兩個(gè)篩選條件可以用&&連接那就不需要2次循環(huán)去判斷,其次你內(nèi)部的循環(huán)?
infos[i]=new Array();
這行代碼重新初始化了infos[i],那就不會(huì)有數(shù)據(jù)
2018-05-29
?
<script type="text/javascript">
? var infos=[["小A","女",21,"大一"],["小B","男",23,"大三"],["小C","男",24,"大四"],["小D","女",21,"大一"],["小E","女",22,"大四"],["小F","男",21,"大一"],["小G","女",22,"大二"],["小H","女",20,"大三"],["小I","女",20,"大一"],["小J","男",20,"大三"]];
? //第一次篩選,找出都是大一的信息
?document.write("都是大一的信息:"+"<br/>");
for( var i=0;i<infos.length;i++){
? ? ? if(infos[i][3]=="大一"){
? ? ? ? ? document.write(infos[i]+"<br />");??
? ? ? }
}
document.write("都是女生的信息:"+"<br/>");
for( var i=0;i<infos.length;i++){
? ? ? if(infos[i][1]=="女"){
? ? ? ? ? document.write(infos[i]+"<br />");??
? ? ? }
}
</script>
你在for里面又定義了一個(gè)infos數(shù)組和for外面的數(shù)組有沖突,名字一樣。
2018-05-29
<!DOCTYPE? HTML>
<html >
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>流程控制語句</title>
<script type="text/javascript">
?//第一步把之前的數(shù)據(jù)寫成一個(gè)數(shù)組的形式,定義變量為 infos
?var infos=[['小A','女',21,'大一'],['小B','男',23,'大三'],['小C','男',24,'大四'],['小D','女',21,'大一'],['小E','女',22,'大四'],['小F','男',21,'大一'],['小G','女',22,'大二'],['小H','女',20,'大三'],['小I','女',20,'大一'],['小J','男',20,'大三']];
?
?//第一次篩選,找出都是大一的信息
for(var i=0;i<infos.length;i++){
if(infos[i][3]=="大一" && infos[i][1]=="女"){
document.write(infos[i][0]+"<br>")
}
}
??
?//第二次篩選,找出都是女生的信息
?
?
??
</script>
</head>
<body>
</body>
</html>