我有2个页面,一个PHP和一个Javascript.我想在一个页面上传递这个PHP脚本中的变量:
$strFind="SELECT * FROM cometchat_chatrooms_users WHERE userid=$curmemid";
$result=mysql_query($strFind) or die(mysql_error());
$row=mysql_fetch_array($result);
$room=$row['chatroomid'];
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在另一个页面上的这个Javascript:
var timestamp = 0;
var currentroom = $room;
var heartbeatTimer;
var minHeartbeat = 3000;
var maxHeartbeat = 12000;
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我怎样才能做到这一点?
当PHP生成JavaScript时,您可以执行此操作.给出的例子:
<?php
...
$curmemid = intval($externalValue);
$strFind = 'SELECT * FROM cometchat_chatrooms_users WHERE userid='.$curmemid;
$result = mysql_query($strFind) or die(mysql_error());
if (isset($result)) {
$row = mysql_fetch_array($result);
$room = $row['chatroomid'];
} else
{
echo 'There is something wrong!';
$room = -1;
}
...
?>
<script type="text/javascript">
var timestamp = 0;
var currentroom = <?php echo $room; ?>;
var heartbeatTimer;
var minHeartbeat = 3000;
var maxHeartbeat = 12000;
</script>
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