function sumOfColumns(tableID, columnIndex) {
var tot = 0;
$("#" + tableID + " tr").children("td:nth-child(" + columnIndex + ")").each(function () {
tot += parseFloat($(this).html());
});
return tot;
}
var _totalhours = sumOfColumns("tblEmpEarnings", 3);
当我调试_totalhours给我时NaN.有人可以帮忙吗？
我对某些代码有一些问题,其中有一个查询应该返回false并且它返回true.
首先,我有一个不存在的id的SQL查询:
$result = mysqli_query($con,"SELECT * FROM myTable WHERE id='1' "); //There no id with value of 1
然后我添加了一个var_dump来查看是什么:
var_dump($result);
并在一个条件后:
if ($result==0) {
echo 'No results where found';
} else {
echo 'Some results where found';
}
这是var_dump的结果
object(mysqli_result)#2 (5) { ["current_field"]=> int(0) ["field_count"]=> int(12) ["lengths"]=> NULL ["num_rows"]=> int(0) ["type"]=> int(0) }
找到了一些结果
结果应该是:找不到结果,正如您在上面看到的那样,它会返回:找到一些结果
我在这里做错了什么？
我正在使用codeigniter为轮盘赌系统建立一个库.
我希望使代码非常高效和防黑客.
这是代码
<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');
/**
* Name: Prad Roulette
*
* Version: 0.1
*
* Author: Pradyummna Reddy
* prad@hireprad.co.uk
*
*
* Location: https://github.com/pradyummna/prad_roulette
*
* Created: 21/10/2014
*
* Description: The library can be used for roulette system developers
*
* Requirements: PHP5 or above
*
*/
class Prad_roulette
{
/**
*
* Function to find if the number is a valid roulette number
*
* @param …我想显示登录我网站的人的名字.这是我的login.php文件的代码,该文件包含在我网站的一个页面中.
<?php
$connect = mysql_connect("localhost","root","") or die("Error");
mysql_select_db("jpnv_db") or die("Couldn't find db");
function login() {
$username = $_POST['username'];
$password = $_POST['password'];
$query = mysql_query("SELECT * FROM customers WHERE `username`='$username' AND `password`='$password'");
$names = mysql_query("SELECT contactFirstName FROM customers WHERE `username`='$username'");
if (empty($username)) {
$errors[] = 'Please fill in your username. <a href="index.php">Click here to try again.</a>';
}
if (empty($password)) {
$errors[] = 'Please fill in your password. <a href="index.php">Click here to try again.</a>';
}
if ($errors==true) {
foreach ($errors as $error) …我有以下PHP代码:
try{
$stmt = $db->prepare('SELECT nume,prenume FROM candidati');
$stmt->execute();
while($row = $stmt->fetch(PDO::FETCH_OBJ)){
$aux = $row->nume.' '.$row->prenume;
$st = $db->prepare('SELECT COUNT(id) AS total FROM votanti WHERE consiliul_local=?');
$st->execute(array($aux));
while($r = $st->fetch(PDO::FETCH_OBJ)){
$s = $db->prepare("INSERT INTO rezultate SET obtinute=:o WHERE nume=:n AND prenume=:p");
$s->bindParam(':o',$r->total,PDO::PARAM_INT);
$s->bindParam(':n',$row->nume,PDO::PARAM_STR);
$s->bindParam(':p',$row->prenume,PDO::PARAM_STR);
$s->execute();
}
}
} catch(PDOException $e){
echo $e->getMessage();
}
以下课程:
class DB{
public static function connect($engine,$host,$user,$pass,$name){
try{
$dbh = new PDO("$engine:host=$host;dbname=$name;charset=utf8",$user,$pass);
$dbh->setAttribute(PDO::ATTR_ERRMODE,PDO::ERRMODE_EXCEPTION);
$dbh->setAttribute(PDO::ATTR_EMULATE_PREPARES,false);
return $dbh;
} catch(PDOException $e){
echo $e->getMessage();
}
}
}
但是当我执行第一段代码时,它会发送以下错误消息:
SQLSTATE [42000]:语法错误或访问冲突:1064 …
这几乎是我第一次使用MYSQL,我似乎无法解决这个错误.我正在尝试将数据存储到其id(第一列)上有auto_increment的表中.我一直得到的错误是这样的:
"您的SQL语法有错误;请查看与您的MySQL服务器版本对应的手册,以便在'voorletters ='asd'附近使用正确的语法,tussenvoegsel ='',achternaam ='',roepnaam ='',adres ='','在第1行"
我只是填充了一些文本框,没有列也需要数据.这是我使用的代码:
if(isset($_POST['save']))
{
$voorletters = $_POST['voorletters'];
$tussenvoegsel = $_POST['tussenvoegsel'];
$achternaam = $_POST['achternaam'];
$roepnaam = $_POST['roepnaam'];
$adres = $_POST['adres'];
$postcode = $_POST['postcode'];
$plaats = $_POST['plaats'];
$geslacht = $_POST['geslacht'];
$emailadres = $_POST['emailadres'];
$telefoonnummer = $_POST['telefoonnummer'];
$mobielenummer = $_POST['mobielenummer'];
$geboortedatum = $_POST['geboortedatum'];
$bsn = $_POST['bsn'];
mysql_query("INSERT INTO `naw` "
. "voorletters ='$voorletters', "
. "tussenvoegsel ='$tussenvoegsel', "
. "achternaam ='$achternaam', "
. "roepnaam ='$roepnaam', "
. "adres ='$adres', "
. "postcode ='$postcode', "
. "plaats ='$plaats', …我使用以下代码将数组值传递给mysql查询,但数组到字符串转换错误将来
$sql =mysql_query("SELECT userId from groupmembers where groupId='$groupId'");
$bjson = array();
$i = 0;
while($result=mysql_fetch_assoc($sql))
{
$bjson[$i]['userId'] = $result['userId'];
$i++;
}
$query = "SELECT firstName
FROM users
WHERE userId IN('" . implode("','", $bjson) ."')";
我已经在这个问题上苦苦挣扎了几天,因为这是我第一个使用PHP的项目,所以很难自己解决这个问题.我在其他线程的帮助下重写了我的代码,但我仍然找不到解.
所以,我想进行简单的用户注册.我从这个表单中获取用户的输入,该表单包含在registration_page.html中,代码如下:
<form id="registerForm" action="http://localhost//register.php" method="post">
<fieldset align="center">
<legend id="legendText">Register</legend>
<p>Name:
<input type="text" name="fname" value="" required autofocus maxlength="16"></p>
<p>Last name:
<input type="text" name="lname" value="" required maxlength="16"></p>
<p>E-mail:
<input type="email" name="mail" value="" placeholder="@mail.com" maxlength="32" required></p>
<p>Age:
<input type="number" name="age" value="" maxlength="2" max="99" maxlength="2" size="2" min="1" required></p>
<p>Job:
<input type="text" name="job" value="" maxlength="16" required></p>
<table align="center" id="registerPwdAndUsrTable" border="1" width="30%">
<tr>
<td>
<p>Username:
<input type="text" name="username" value="" required maxlength="16"></p>
</td>
<td>
<p>Password:
<input type="password" name="password" value="" required maxlength="16"></p>
</td>
</tr>
</table> …我有3个像这样的输入:
<input type="text" id ="text1"/><br>
<input type="text" id ="text2"/><br>
<input type="text" readonly="readonly" disabled="disabled" id ="text3"/>
的第三输入被禁用,并且我喜欢他的值将等于:(VAL(文本2)/缬氨酸(文本1)*100-100) whene我通过其它两个输入的值
所以我试试这个脚本:
var p1 = document.getElementById('#text1').value;
var p2 = document.getElementById('#text2').value;
document.getElementById('#text3').value = ((p1/p2)*100)-100;
但是不起作用,有人可以帮我吗？
我有以下代码:
$currentPage = isset($_GET['page']) ? $_GET['page'] : 1;
if($currentPage < 1)
{
$currentPage = 1;
}
我想摆脱if声明,并在一行中做所有事情.我怎么能这样做,甚至可以在三元运算符中做到这一点？
php ×8
mysql ×5
html ×3
javascript ×2
jquery ×2
arrays ×1
codeigniter ×1
css ×1
mysqli ×1
sql ×1