小编Dav*_*vid的帖子

我有这个jQuery脚本:

$(document).ready(function() {
    $(':input:enabled:visible:first').focus();
    $('.letters').keyup( function() {
        var $this = $(this);
        if($this.val().length > 1)
            $this.val($this.val().substr(0, 1));
            $(this).next('input').focus();
        });
});

它将重点input='text'放在页面加载的第一个字段上.当用户输入字符时,它会将焦点移动到下一个输入字段.它还将限制每个字段中允许的字符数(当前为1个字符).

我想知道是否可以清除焦点上输入字段的当前值.当用户使用cursror单击以聚焦字段时以及当$(this).next('input').focus();集合关注下一个输入字段时.

还可以验证字符只允许字母字符吗？

我有这样的MySQL查询:

SELECT cp.plan_name, cp.plan_time FROM courses c
  INNER JOIN course_to_plan cpl   ON cpl.course_id = c.course_id
  INNER JOIN courseplans cp       ON cp.plan_id = cpl.plan_id
WHERE cpl.course_id = '$course_id';

这将输出数据,例如:

+----------------------------+-----------+
| plan_name                  | plan_time |
+----------------------------+-----------+
| Plan number one name       |         6 |
| Plan number two name       |         6 |
| Plan number three name     |        10 |
+----------------------------+-----------+

我希望将这些行插入表单提交的新表中.

如何继续编写代码update.php以使其在表中插入值newtable？

if (isset($_POST['submit'])) {

 $course_id = $_POST['course_id'];


 $course_result = mysql_query 
    ("SELECT cp.plan_name, cp.plan_time FROM courses …

<select>我有这个 jQuery 代码，可以使用“添加”按钮在表行中添加元素：

$("#add").click(function() {
    $('#selected > tbody:last').append('<tr><td><select id=\'plan_id\'><option value=\'1\'>One</option><option value=\'2\'>Two</option><option value=\'3\'>Three</option></select></td></tr>');
    });

我需要在下面的代码中修改什么才能将选择元素中的多个值作为数组发布？现在它一次只向我的 MySQL 数据库插入一个值。

$('#course_update').click(function() {

var course_id = $('#course_id').val();
var plan_id = $('#plan_id').val();
var price_id = $('#price_id').val();
var course_name = $('#course_name').val();
var course_isActive = $('#course_isActive').val();
var course_city_id = $('#course_city_id').val();


$('#update_status').html('<img src="../images/ajax-loader.gif" />');
$.post('../update.php', {

    course_id: course_id, 
    plan_id : plan_id,
    price_id: price_id,
    course_name: course_name,
    course_city_id: course_city_id,
    course_isActive: course_isActive

    }, function(data) {
    $('#update_status').html(data);
    return false;
 });
});

我使用以下代码获取工作日的Ymd格式:

$monday = date('Y-m-d', strtotime('Monday'));
$tuesday = date('Y-m-d', strtotime('Tuesday'));
$wednesday = date('Y-m-d', strtotime('Wednesday'));
$thursday = date('Y-m-d', strtotime('Thursday'));
$friday = date('Y-m-d', strtotime('Friday'));

今天是2012-08-01(第31周),我需要的周一价值应该是2012-07-30.

为什么strtotime('Monday')下周一会这样呢？

我有这个 MySQL 查询：

SELECT 
    SUM(scpe.scpe_estemated_days) AS total_days,
    scp.cpl_startdate
FROM
    studentcourseplanelements scpe
        INNER JOIN
    studentcourseplan scp ON scp.cpl_id = scpe.scpe_cpl_id
        INNER JOIN
    (SELECT 
        sd1.student_id, sd1.student_startdate
    FROM
        studentdates sd1
    WHERE
        sd1.student_id = '360'
    LIMIT 1) sd ON sd.student_id = scp.student_id
GROUP BY scp.cpl_id

这输出：

+------------+---------------+
| total_days | cpl_startdate |
+------------+---------------+
|          5 | 2012-11-01    |
|        129 | 2012-11-02    |
+------------+---------------+

我只想选择最高的行total_days，在我的示例 129 中。

我怎样才能做到这一点？

这个查询：

SELECT customer_id, customer_name FROM customers WHERE isActive = '1' ORDER BY customer_name ASC

输出：

+-------------+-----------------------+\n| customer_id | customer_name         |\n+-------------+-----------------------+\n|           1 | \xc3\x84name                 |\n|           2 | Aname                 |\n|           3 | Bname                 |\n+-------------+-----------------------+\n

即使我有排序规则，为什么它不对特殊的瑞典字符进行排序utf8_swedish_ci

SHOW TABLE STATUS FROM myDatabase WHERE name = 'customers';\n\n+-------------+----------------------------+\n| Name        | Engine   | Collation       |\n+-------------+----------------------------+\n| customers   | MyISAM   | utf8_swedish_ci |\n+-------------+----------------------------+\n

我什至尝试将排序规则放入我的查询中：

SELECT * FROM customers WHERE isActive = 1 COLLATE utf8_swedish_ci ORDER BY customer_name ASC\n

但后来我得到：

我有这个MySQL包含单词列表的表:

desc words;
+---------+--------------+------+-----+---------+----------------+
| Field   | Type         | Null | Key | Default | Extra          |
+---------+--------------+------+-----+---------+----------------+
| id      | int(11)      | NO   | PRI | NULL    | auto_increment |
| word    | varchar(255) | NO   |     | NULL    |                |
+---------+--------------+------+-----+---------+----------------+

我还有一个HTML包含三个输入字段的表单,用户应输入三个字母:

<form action='load.php' method='post'>
    <input type='text' name='first_letter'>
    <input type='text' name='second_letter'>
    <input type='text' name='third_letter'>

    <input type='submit' name='submit'>
</form>

是否可以创建一个MySQL查询来按照外观的顺序获取包含三个字母的单词？

例如,如果我们有话

adams
damn
mad

...并且用户提交字母"a","d","m"它应该只给出结果

adams

因为第一个提交的信件是"a",第二个提交的信件是在"a"之后等等(即使其间有其他字母).

或者使用单词排序更容易PHP？如果是这样,怎么样？我是初学程序员.

我有这个jQuery脚本:

$(document).ready(function() {
    //Focus the first field on page load
    $(':input:enabled:visible:first').focus();
    //Clear all fields on page load
    $(':input').each(function() {
        this.value = "";
    });
});
//Clear field on focus
$('input').focus(function() {
    this.value = "";
});
//Allow only alphabetical characters in the fields
$(':input').bind("keypress", function(event) {
    if (event.charCode != 0) {
        var regex = new RegExp("^[a-zA-Z]+$");
        var key = String.fromCharCode(!event.charCode ? event.which : event.charCode);
        if (!regex.test(key)) {
            event.preventDefault();
            return false;
        }
        $(this).next('input').focus();
    }
});
//Enumerate submit click on [ENTER]-keypress
$(':input').keypress(function(e) …

我正在使用此功能(我在此论坛上找到)来计算范围之间的工作天数:

<?php
//The function returns the no. of business days between two dates and it skips the holidays
function getWorkingDays($startDate,$endDate,$holidays){
// do strtotime calculations just once
$endDate = strtotime($endDate);
$startDate = strtotime($startDate);


//The total number of days between the two dates. We compute the no. of seconds and divide it to 60*60*24
//We add one to inlude both dates in the interval.
$days = ($endDate - $startDate) / 86400 + 1;

$no_full_weeks = floor($days / 7);
$no_remaining_days = fmod($days, 7); …

我有这个MySQL问题:

SELECT 
    s.student_id,
    s.student_firstname,
    s.student_lastname,
    sd.student_startdate,
    sd.student_enddate,
    s.isActive,
    c.city_name,
    ctc.category_name
FROM
    students s
        INNER JOIN
    (SELECT sd1.student_id, sd1.student_startdate, sd1.student_enddate FROM studentdates sd1) sd ON sd.student_id = s.student_id
        INNER JOIN
    cityselections c ON c.city_id = s.student_city_id
        INNER JOIN
    coursecategory ctc ON s.student_course_category_id = ctc.category_id
WHERE
    sd.student_enddate BETWEEN CURDATE() AND ADDDATE(CURDATE(), INTERVAL 14 DAY)
        AND s.student_city_id NOT LIKE '1'
        AND s.student_city_id = 18
        AND s.isActive = 1
GROUP BY s.student_id
ORDER BY sd.student_enddate ASC , s.student_lastname , s.student_firstname

在表studentdates …