在jQuery click（）中使用PHP变量？

Question

I have this,

    $(document).ready(function(){

        $('#link').click(function(){
            var user_login = <?php $base_url; ?>;
            window.location = user_login + '/login';
        });

    });

note that $base_url is global variable. This works in FF but not on Chrome and IE. Thanks.

Answer 1

You need to add quotes around the php tags to designate that the value of $base_url is a string:

$(document).ready(function(){

    $('#link').click(function(){
        var user_login = "<?php echo $base_url; ?>";
        window.location = user_login + '/login';
    });

});

That way, when the browser gets this block, it will look something like:

$(document).ready(function(){

    $('#link').click(function(){
        var user_login = "http://www.example.com"; // after php is executed
        window.location = user_login + '/login';
    });

});

Answer 2

You need to echo the variable, not just simply have it in a code block. And for safety, to prevent introducing JS syntax errors, you should json-encode the value:

        var login = <?php echo json_encode($base_url); ?>;

Answer 3

See below, note the quotes around the PHP echo

$(document).ready(function(){
    $('#link').click(function(){
        var login = '<?php echo $base_url; ?>';
        window.location = user_login + '/login';
    });

});

Answer 4

I suppose its a PHP file, you need to surround your variable contents by quotes to javascript

 $(document).ready(function(){

        $('#link').click(function(){
            var login = <?php '\''.$base_url.'\''; ?>;
            window.location = user_login + '/login';
        });

    });

Answer 5

如其他答案所述，请在PHP文件中而不是.js文件中运行此脚本 ，并且不要忘记在变量中回显并添加引号（在javascript中）！