我有一个每年更新的表单,因此用户输入日期和其他信息,然后提交存储在数据库中的表单。当他们一年后回来时,他们会做同样的事情来覆盖旧记录,问题是,当他们回来时,旧日期正在被回应,这不应该发生。所有其他字段都应该是空白的。我不是要求在代码中回显日期,但它仍然是。
用于定义日期下拉的代码
// parameters: $fname - main name of field
// $date - actual date value from database
// $beginYear - first value in year list
// $endYear - last value in year list
// return: none
function make_date_pulldown($fname, $date, $beginYear, $endYear)
{
// read the date and break it up into $Year, $Month and $Day
// so that we can set the "SELECTED" in the option list
if ($date == ""){
// set some default values to be safe
$Year = 0;
$Month = 0;
$Day = 0;
} else {
$Year = (int) substr($date,0,4);
$Month = (int) substr($date,5,2);
$Day = (int) substr($date,8,2);
}
// need to build a table around these guys so that there won't
// be any word wrap... it's going to be a 1 row by 5 cols.
echo "<table border=0 cellspacing=0 cellpadding=0>\n";
echo "<tr>\n";
// build month list
echo " <td><font class=bluetext face=verdana,arial size=-2>\n";
echo " <center>Month<br>\n";
echo "<select name='month_$fname'>\n";
echo " <option value='00'></option>\n";
for ($i=1;$i<=12;$i++){
printf (" <option value='%02d'",$i);
if ($i == $Month) {
printf (" SELECTED");
}
printf (">%02d</option>\n",$i);
}
echo "</select>\n";
echo " </center>\n";
echo " </td>\n";
echo " <td><font class=bluetext face=verdana,arial size=-2>\n";
echo "/";
echo " </td>\n";
// build day list
echo " <td><font class=bluetext face=verdana,arial size=-2>\n";
echo " <center>Day<br>\n";
echo "<select name='day_$fname'>\n";
echo " <option value='00'></option>\n";
for ($i=1;$i<=31;$i++){
printf (" <option value='%02d'",$i);
if ($i == $Day) {
printf (" SELECTED");
}
printf (">%02d</option>\n",$i);
}
echo "</select>\n";
echo " </center>\n";
echo " </td>\n";
echo " <td><font class=bluetext face=verdana,arial size=-2>\n";
echo "/";
echo " </td>\n";
// build year list
echo " <td><font class=bluetext face=verdana,arial size=-2>\n";
echo " <center>Year<br>\n";
echo "<select name='year_$fname'>\n";
echo " <option value='0000'></option>\n";
for ($i=$beginYear;$i<=$endYear;$i++){
printf (" <option value='%d'",$i);
if ($i == $Year) {
printf (" SELECTED");
}
printf (">%d</option>\n",$i);
}
echo "</select>\n";
echo " </center>\n";
echo " </td>\n";
echo "</tr>\n";
echo "</table>\n";
}
网站上的PHP表格代码
<u>Date Of Submittal</u><br>
<?php
$startYear = date("Y")-1;
$endYear = date("Y")+1;
make_date_pulldown("submitdate", $submitdate, $startYear, $endYear);
?>
答案 0 :(得分:0)
您希望根据年份创建一个下拉菜单....
$date = new DateTime();
//Years
echo makeSelect($date->format("Y")-1, $date->format("Y")+1,$date->format("Y"));
//Month
echo makeSelect(1, 12,$date->format("m"));
//Day
echo makeSelect(1, $date->format("t"),$date->format("d"));
使用的功能
function makeSelect($start, $end, $default = null) {
$content = "<select>";
$range = range($start, $end);
$selected = "";
foreach ( $range as $value ) {
$selected = ($default == $value) ? "selected" : null;
$content .= sprintf('<option value="%s" %s>%1$s</option>', $value, $selected);
}
$content .= "</select>";
return $content;
}
答案 1 :(得分:0)
想出来,在我的get_()函数中调用数据库表我正在读取submitdate的行,它正在拉动并显示基于该命令的日期和make_date_pulldown的函数。
这个函数看起来像是什么(明显减少了):
<?
function get_submit($id, &$submitdate)
{
$query = "select * from submit where id = $id";
$result = mysql_query($query);
if ($result) {
$total_rows = mysql_numrows($result);
} else { $total_rows = 0; }
if ($total_rows > 0) {
$submitdate = trim(mysql_result($result, "", "submitdate"));
} else {
$submitdate = "";
$found = 1;
}
}
?>