使用 PHP 填充带有年份的选择框
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Populate a Select Box with Years using PHP
提问by Abishek
I would like to fill a select box with year starting from 1950 to the current year. How can I achieve this using PHP? I would not like to use JavaScript for this.
我想用从 1950 年到当年的年份填充选择框。如何使用 PHP 实现这一目标？我不想为此使用 JavaScript。
<select><?php
$currentYear = date('Y');
foreach (range(1950, $currentYear) as $value) {
echo "< option>" . $value . "</option > ";
}
?>
</select>
回答by Chris Baker
Use rangeto create an array containing all the required years, loop that array and print an optionfor each of the values.
使用range创建一个包含所有需要几年的数组，循环数组并打印option每个值。
You can use date('Y')to figure out the current year.
您可以使用date('Y')来计算当前年份。
// use this to set an option as selected (ie you are pulling existing values out of the database)
$already_selected_value = 1984;
$earliest_year = 1950;
print '<select name="some_field">';
foreach (range(date('Y'), $earliest_year) as $x) {
print '<option value="'.$x.'"'.($x === $already_selected_value ? ' selected="selected"' : '').'>'.$x.'</option>';
}
print '</select>';
Try it here: http://codepad.viper-7.com/Pw3U4O
在这里试试：http: //codepad.viper-7.com/Pw3U4O
Documentation
文档
回答by sn0ep
<select name="select">
<?php
for($i = 1950 ; $i < date('Y'); $i++){
echo "<option>$i</option>";
}
?>
</select>
Something like this?
像这样的东西？
回答by Phill Pafford
<?php
$starting_year = 1950;
$ending_year = 2011;
for($starting_year; $starting_year <= $ending_year; $starting_year++) {
$years[] = '<option value="'.$starting_year.'">'.$starting_year.'</option>';
}
?>
<select>
<?php echo implode("\n\r", $years); ?>
</select>
Option #2:
选项#2：
<select>
<?php
foreach(range(1950, (int)date("Y")) as $year) {
echo "\t<option value='".$year."'>".$year."</option>\n\r";
}
?>
</select>
