我正在尝试将一个非常简单的 javascript 变量发布到 php 文件中。
keyinput.php 中的 Jquery 位:
<script type="text/javascript">
var imgArray = [<?php echo implode(',', getImages($cat, $site)) ?>];
$(document).ready(function() {
var img = document.getElementById("showimg");
img.src = imgArray[<?php echo $imgid ?>];
var imgIndex = <?php echo $imgid ?>;
$(document).keydown(function (e) {
var key = e.which;
int rightarrow = 39;
int leftarrow = 37;
int random = 82;
if (key != rightarrow && key != leftarrow && key != random) {
return;
}
else {
//next image: right arrow
if (key == rightarrow)
{
imgIndex++;
if (imgIndex > imgArray.length-1)
{
imgIndex = 0;
}
img.src = imgArray[imgIndex];
}
//last image: left arrow
if (key == leftarrow)
{
if (imgIndex == 0)
{
imgIndex = imgArray.length;
}
img.src = imgArray[--imgIndex];
}
//random: r
if (key == random)
{
imgIndex = Math.floor((Math.random()*(imgArray.length-1))+1);
img.src = imgArray[imgIndex];
}
}
$.post('./templates/viewcomic.php', {variable: imgIndex});
});
});
</script>
<?php
function getImages($catParam, $siteParam) {
include './scripts/dbconnect.php';
if ($siteParam == 'artwork') {
$table = "artwork";
}
else {
$table = "comics";
}
if ($catParam != null) {
$catResult = $mysqli->query("SELECT id, title, path, thumb, catidFK FROM $table WHERE catidFK = $catParam");
}
else {
$catResult = $mysqli->query("SELECT id, title, path, thumb, catidFK FROM $table");
}
$img = array();
while($row = $catResult->fetch_assoc())
{
$img[] = "'" . $row['path'] . "'";
}
return $img;
}
?>
viewcomic.php 中的 PHP 位:
include './scripts/keyinput.php';
$JSIndex = $_POST['variable'];
echo "Index = " . $JSIndex;
//$JSIndex should be equal to the javascript variable imgIndex... but it outputs nothing
任何想法都会非常有帮助!我正在尝试让我的漫画网站上线。
谢谢!