我无法理解造成这种情况的原因,虽然数学从来都不是我的专业,所以任何帮助都会受到赞赏.我的简单等式分为两部分:
答案1)226 - [值]乘以.9
答案2)226 - [值]乘以.55
如果您使用40作为输入值,我的计算器会说:
答案1)226 - 40 = 186*.9 = 167.4
答案2)226 - 40 = 186*.55 = 102.3
但我的代码说:
答案1)190
答案2)204
我的代码:
HTML:
<input id="HR"size="4" value="" /><span id="number"></span>
Answer 1 <strong><span id="answer1"></span></strong><br>
Answer 2 <strong><span id="answer2"></span></strong>
</p>
<a href="#" class='THR'>click to calculate</a>
Script:
<script type="text/javascript">
$('a.THR').click(function() {
var value = parseInt( $( "#HR" ).val() );
$( "#answer1" ).html( 226 - value * 0.9 );
$( "#answer2" ).html( 226 - value * 0.55 );
return …Run Code Online (Sandbox Code Playgroud) 我有以下ejs模板
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Emasc Editor</title>
<style>
.list{
display: inline-block;
}
</style>
<script type="text/javascript" src="/js/jquery-1.7.1.min.js"></script>
<script type="text/javscript">
$(document).ready(function(){
alert('test');
$("#setMembershipsButton").click(function(e){
// send selected person and groups to server and create memberships
data = {
person: $('#person_list').val(),
groups: $('#group_list').val()
};
$.post('/newConnections', data);
});
$("#person_list").change(function(){
alert('test');
// send selected person to server and retrieve memberships
$.post('/getPersonGroups', { personID: $('#person_list').val() }, function(data, textStatus){
$('#group_list option').removeAttr('selected');
for(var i=0; i<data.ids.length; i++){
$('#group_list option[value=' + data.ids[i] + ']').attr('selected', 'selected');
}
});
});
}); …Run Code Online (Sandbox Code Playgroud) 我在这个练习中遇到了另一个问题:http://lpmj.net/20.php
我的问题是,当我去创建一个新帐户时,我收到错误消息:
'where子句'中的未知列'user'
此时我有很多错别字,我的代码与本书中的代码完全相同.本节中只有一个查询可能是:
$query = "SELECT * FROM rcmembers WHERE user='$user'";
Run Code Online (Sandbox Code Playgroud)
我假设这意味着用户不存在.我该如何创造它?是不是在这些课程的其他方面设置了?
public static double[][] multiplyMatrix(double[][] matrix1, double[][] matrix2) {
// As both arrays are square and the same size, the row size represents the row size and column size for both matrices
int dimension = matrix1.length;
double[][] matrix3 = new double[dimension][dimension];
for (int i = 0; i < dimension-1; i++) {
for (int j = 0; j < dimension-1; j++) {
for (int k = 0; k < dimension-1; j++) {
matrix3[i][j] += matrix1[i][k] * matrix2[k][j];
}
}
}
return matrix3; …Run Code Online (Sandbox Code Playgroud)