将路径为值的php变量传递给javascript [复制]

Question

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• Pass a PHP array to a JavaScript function [duplicate] 4 answers

I have a php variable which has path as value.I am trying to pass this url variable to javascript function.

    private function call_import_function($post_type,$fields,$fields_order,$upld_file)
{
            $uploaded_file  = $upld_file['file'];
            ?>
            <script type="text/javascript">
             jQuery(document).ready(function($) {
             var formdata = {
                            'action': 'get_csv',
                            'post_type' : '<?php echo $post_type;?>',
                            'fields' : '<?php echo $fields;?>',
                            'fields_order' : '<?php echo $fields_order;?>',
                            'uploaded_file' : '<?php echo $uploaded_file;?>',
                        };
             $.post('<?php echo $this->ajax_url;?>',formdata, function( data ) {
                            console.log("Begun!!!");
             }).done(function( data ) {
                            var obj = jQuery.parseJSON(data);

                            if(obj.error)
                            {
                                $("#column2").html(obj.error_msg);
                            }
                            else
                            {
                                console.log(data);

                                //$("#column2").html(obj.output);
                            }
                        });
             });
            </script>
<?php
}

But it gives me an error,

SyntaxError: malformed hexadecimal character escape sequence    
'uploaded_file' : 'E:\xampp\htdocs
ick\projectWed/wp-content/uploads/2'

I have tried php functions like json_encode, urlencode with $url , but non of them provided me good solution for this. I need to solve this error...

</div>

Answer 1

You have forward slashes and backslashes mixed.

$url = ''E:\xampp\htdocs
ick\projectWed/wp-content/uploads/2/3.csv'

should be

$url = "E:\\xampp\\htdocs\
ick\\projectWed\\wp-content\\uploads\\2\\3.csv";

The double backslashes are required in PHP strings. They evaluate to single backslashes. This is to disambiguate them from escape sequences like (new line), \t (tab), \088 (character X), etc.