Javascript变量在php echo中不起作用

Question

My code :我的代码：

<?php echo ' <script>var p=0;for(var i=0;i<=5;i++){p++;}alert("ques".p);? >

The value of p is displayed as 0. p 的值显示为 0。

Answer 1

You need to close your php tag properly as well as the <script> tag like so:您需要正确关闭 php 标签以及<script>标签，如下所示：

<?php echo '<script>var p=0;for(var i=0;i<=5;i++){p++;}alert("ques" +p);</script>'; ?>

Also, change the .此外，更改. to a + as you are concatenating in javascript not PHP到+因为您在 javascript 中而不是 PHP 中连接

Answer 2

正确答案是：

<?php echo '<script>var p=0;for(var i=0;i<=5;i++){p++;}alert("ques" + p)'; ?>

Answer 3

单引号中的字符串将被转义，请改用引号。

<?php echo "<script>var p=0;for(var i=0;i<=5;i++){p++;}alert('ques' +p);</script>"; ?>

Answer 4

The mistakes in your code are,你的代码中的错误是，

Close your php tag properly - ?> instead of ? >正确关闭您的 php 标签 - ?> instead of ? > ?> instead of ? >
Close the script tag in the end - </script>最后关闭脚本标签 - </script>
Close the single quote you started with echo关闭以 echo 开头的单引号

The correct code would be正确的代码是

<?php echo '<script>var p=0;for(var i=0;i<=5;i++){ p++; } alert("ques" + p); </script>'; ?>