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在javascript中使用php_Javascript_Php - Fatal编程技术网

在javascript中使用php

在javascript中使用php,javascript,php,Javascript,Php,这是一个问题。它不起作用。我在重定向前做了一个计时器。 我的代码有错吗 <script> var base_url ="https://www.mydomain.com/"; var redirect = "<?php echo $redirect; ?>"; var targetURL= base_url.concat(redirect); //change the second to start

这是一个问题。它不起作用。我在重定向前做了一个计时器。 我的代码有错吗

     <script>
        var base_url ="https://www.mydomain.com/";
        var redirect = "<?php echo $redirect; ?>";
        var targetURL= base_url.concat(redirect);
        //change the second to start counting down from
        var countdownfrom=9;


        var currentsecond=document.redirect.redirect2.value=countdownfrom+1;
        function countredirect(){
            if (currentsecond!==1){
                currentsecond-=1;
                document.redirect.redirect2.value=currentsecond;
            }
            else{
                window.location=targetURL;
                return;
            }
            setTimeout("countredirect()",1000);
        }

        countredirect();
        //-->
     </script>

var base_url=”https://www.mydomain.com/";
var=”;
var targetURL=base_url.concat(重定向);
//将秒改为从开始倒计时
var countdownfrom=9;
var currentsecond=document.redirect.redirect2.value=countdownfrom+1;
函数countredirect(){
如果(当前秒!==1){
电流秒-=1;
document.redirect.redirect2.value=currentsecond;
}
否则{
window.location=targetURL;
返回;
}
setTimeout(“countredirect()”,1000);
}
countredirect();
//-->
尝试

<script>
var base_url ="https://www.mydomain.com/";
<?php echo 'var redirect = "'.$redirect.'";'; ?>
..
</script>

var base_url=”https://www.mydomain.com/";
..

debug
targetURL
我打赌它看起来像<代码>”https://www.new-i-ching.com/“任何错误异常?经过PHP解析器后呈现的JS是什么?我建议您仔细阅读如何正确地将PHP变量传递给JavaScriptor。这不重要,它是一个字符串,因为我确定这可以吗?