art*_*tur 6 php select return option drop-down-menu
我试图使用PHP选择选项,但我的想法用完了!
以下是我迄今为止尝试过的代码:
<select>
<option value="1">Yes</options>
<option value="2">No</options>
<option value="3">Fine</options>
</select>
<input type="text" value="" name="name">
<input type="submit" value="go" name="go">
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那么,我该怎么办?
Chr*_*ian 12
程序员很懒惰......呃....高效....我会这样做:
<select><?php
$the_key = 1; // or whatever you want
foreach(array(
1 => 'Yes',
2 => 'No',
3 => 'Fine',
) as $key => $val){
?><option value="<?php echo $key; ?>"<?php
if($key==$the_key)echo ' selected="selected"';
?>><?php echo $val; ?></option><?php
}
?></select>
<input type="text" value="" name="name">
<input type="submit" value="go" name="go">
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<select>
<option value="1" <?php if ($myVar==1) echo 'selected="selected"';?>>Yes</options>
<option value="2" <?php if ($myVar==2) echo 'selected="selected"';?>>No</options>
<option value="3" <?php if ($myVar==3) echo 'selected="selected"';?>>Fine</options>
</select>
<input type="text" value="" name="name">
<input type="submit" value="go" name="go">
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如果我正确理解你的问题,这是一种非常简单直接的方法.
你可以用这个..
<select name="select_name">
<option value="1"<?php echo(isset($_POST['select_name'])&&($_POST['select_name']=='1')?' selected="selected"':'');?>>Yes</option>
<option value="2"<?php echo(isset($_POST['select_name'])&&($_POST['select_name']=='2')?' selected="selected"':'');?>>No</option>
<option value="3"<?php echo(isset($_POST['select_name'])&&($_POST['select_name']=='3')?' selected="selected"':'');?>>Fine</option>
</select>
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