[英]How to show an image after having selected it from a drop down menu?
我有以下代码从下拉菜单中选择图像。
<select name="Image2">
<option value="" selected="selected"></option>
<?php
$dir = "../files/images/product";//your path
$dh = opendir($dir);
while (false !== ($filename = readdir($dh))) {
$files[] = $filename;
echo "<option value='" . $filename . "'>".$filename."</option>"; }
sort($files);
?>
</select>
有什么办法可以预览选定的图像? 我这样尝试过:
<img src="<?php echo $filepath.$filename ?>" />
但是它什么也没显示
尝试这个
<script>
$('#attribute119').change(function () {
var path=$('#filepath').val();
$('#main').attr('src', path+'/'+ $('#attribute119 :selected').text() );
});
</script>
<img src="mainimage.jpg" id="main">
<select name="Image2" id="attribute119">
<option value="" selected="selected"></option>
<?php
$dir = "../files/images/product";//your path
$dh = opendir($dir);
while (false !== ($filename = readdir($dh))) {
$files[] = $filename;
echo "<option value='" . $filename . "'>".$filename."</option>"; }
sort($files);
?>
</select>
<input type=hidden id="filepath" value="<?php echo $filepath ?>"/>
也许你应该试试这个..这可以帮助你..
<select name="Image2" onChange="showImage(this.value)">
<option value="" selected="selected"></option>
<?php
$dir = "../files/images/product";//your path
$dh = opendir($dir);
while (false !== ($filename = readdir($dh))) {
$files[] = $filename;
echo "<option value='".$dir ."/". $filename . "'>" . $filename . "</option>"; }
sort($files);
?>
</select>
<div id="image_div"></div>
<script type="text/javascript">
function showImage(value)
{
var img = "<img src='"+value+"' />";
document.getElementById('image_div').innerHTML = img;
}
</script>
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